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Page 1: Christian Parkinson UCLA Basic Exam Solutions: …chparkin/index/BasicExam...Christian Parkinson UCLA Basic Exam Solutions: Analysis 1 Problem F01.1. Let K be a compact set of real

Christian Parkinson UCLA Basic Exam Solutions: Analysis 1

Problem F01.1. Let K be a compact set of real numbers and let f be a continuousfunction on K. Prove that there exists x0 ∈ K such that f(x) ≤ f(x0) for all x ∈ K.

Solution. We first prove that f(K) is compact. Suppose (Uα)α∈I is an open cover of f(K).For any, x ∈ K we have f(x) ∈ f(K). Then f(x) ∈ Uα for some α and so x ∈ f−1(Uα).Hence (f−1(Uα))α∈I is a cover for K. Also each of f−1(Uα) is open since f is continuous so wehave an open cover of K. Then by compactness of K, there is a finite subcover (f−1(Uαi))

Ni=1.

Take y ∈ f(K). Then there is x ∈ K such that f(x) = y. Then x ∈ f−1(Uαi) for somei = 1, . . . , N so y = f(x) ∈ f(f−1(Uαi)) = Uαi . Then (Uαi)

Ni=1 forms a cover of f(K). Hence

any cover of f(K) has a finite subcover, so f(K) is compact.Since f(K) is compact, it is bounded, and thus has a finite supremum M ∈ R. Also f(K)

is closed so it contains its supremum. Hence M ∈ f(K) and so there is x0 ∈ K such thatf(x) = M . Since M is an upper bound for f(K), we have that y ≤ f(x0) for all y ∈ f(K),or put another way, f(x) ≤ f(x0) for all x ∈ K.

Problem F01.4. Let S be the set of all sequences (x1, x2, . . .) such that for all n,

xn ∈ 0, 1.

Prove that there is not a one-to-one mapping from N onto S.

Solution. There is a one-to-one mapping; there is not an onto mapping. We prove this assuch. For any mapping f : N→ S, list the images:

f(1) = (x1,1, x1,2, x1,3, . . .)

f(2) = (x2,1, x2,2, x2,3, . . .)

f(3) = (x3,1, x3,2, x3,3, . . .)

...

Construct the sequence (y1, y2, y3, . . .) so that

y1 6= x1,1, y2 6= x2,2 y3 6= x3,3, . . . .

Then there is no n ∈ N such that f(n) = (y1, y2, y3, . . .) because the nth component of f(n)differs from yn by construction.

Problem W02.6. State some reasonably general conditions on a function f : R2 → Runder which

∂x

(∂f

∂y

)=

∂y

(∂f

∂x

)and prove the formula under the conditions you give.

Solution. We claim that if the mixed partials are continuous, then the formula holds.Indeed, assume the mixed partials are both continuous. Define

F (x, y) =∂2f

∂x∂y(x, y)− ∂2f

∂y∂x(x, y), (x, y) ∈ R2.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 2

Then F is continuous and integrating this function over any rectangle [a, b]× [c, d] in R2, wesee

I ..=

∫ d

c

∫ b

a

F (x, y)dx dy =

∫ d

c

∫ b

a

∂2f

∂x∂y(x, y)dx dy −

∫ d

c

∫ b

a

∂2f

∂y∂x(x, y)dx dy.

Using Fubini’s theorem on the latter integral, we get

I =

∫ d

c

∫ b

a

∂2f

∂x∂y(x, y)dx dy −

∫ b

a

∫ d

c

∂2f

∂y∂x(x, y)dy dx

=

∫ d

c

[∂f

∂y(b, y)− ∂f

∂y(a, y)

]dy −

∫ b

a

[∂f

∂x(x, d)− ∂f

∂x(x, c)

]dx

= f(b, d)− f(b, c)− f(a, d) + f(a, c)− f(b, d) + f(a, d) + f(b, c)− f(a, c) = 0.

Thus F integrates to zero over any rectangle in R2. Assume that F is not identically zero.Then there is (x∗, y∗) ∈ R2 such that (wlog) F (x∗, y∗) = ε > 0. By continuity, we can findδ > 0 so that for

(x, y) ∈ [x∗ − δ, x∗ + δ]× [y∗ − δ, y∗ + δ]

we have F (x, y) > ε/2. Then∫ y∗+δ

y∗−δ

∫ x∗+δ

x∗−δF (x, y)dx dy >

∫ y∗+δ

y∗−δ

∫ x∗+δ

x∗−δ

ε

2dx dy = 2εδ2 > 0.

This contradicts that F integrates to zero over any rectangle, hence F is identically zerowhich implies that

∂2f

∂x∂y(x, y) =

∂2f

∂y∂x(x, y), (x, y) ∈ R2.

Problem W02.7. Suppose F = (F1, F2) : R2 → R2 is everywhere differentiable and thatthe first derivative matrix (

∂F1

∂x∂F1

∂y∂F2

∂x∂F2

∂y

)is continuous and nonsingular everywhere. Suppose also that

||F (x, y)|| ≥ 1 if ||(x, y)|| = 1 and F (0, 0) = (0, 0).

Put U = (x, y) ∈ R2 : x2 + y2 < 1. Prove that U ⊂ F (U).

Solution. Once you have proven that U∩F (U), we know that U∩F (U) = ∅ or U∩F (U) =U since U is connected. However, U ∩ F (U) 6= ∅ because (0, 0) is in the intersection.

To prove that the set is open and closed, appeal to the Inverse Function Theorem. I can’tbe bothered to write down the proof.

Problem F03.2. Let f : Rn → Rn be an infinitely differentiable function and assume thatfor each x ∈ [0, 1], there is a positive integer m such that f (m)(x) 6= 0.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 3

Prove the following stronger statement: there is an integer M such that for each elementx ∈ [0, 1] there is a positive integer m ≤M such that f (m)(x) 6= 0.

Solution. Define

En = x ∈ [0, 1] : there exists k ∈ N, 0 < k ≤ n such that f (k)(x) 6= 0.

By assumption, each x ∈ [0, 1] is in Em where m is as in the statement of the problem.Thus

[0, 1] =∞⋃n=1

En.

Also,

En =n⋃k=1

(f (k))−1

(R− 0) .

Since f is C∞, each of f (k) is continuous and thus(f (k))−1

(R− 0) is open for all k ∈ Nsince it is the pullback of an open set under a continuous function. Thus En is open as aunion of open sets. Thus En, n ∈ N forms and open cover of [0, 1]. By compactness, thereis a finite subcover

[0, 1] =N⋃i=1

Eni .

Putting M = nN gives the result.

Problem S03.3. Find a subset S of R such that both of the following hold for S:

1. S is not a countable union of closed sets

2. S is not a countable intersection of open sets

Solution. Let S be the union of the positive rationals with the negative irrationals. Assumethat S is a countable intersection of open sets Dn, n = 1, 2, 3, . . .. Then Q+ is also a countableintersection of open sets since

Q+ =⋂n

D+n where D+

n = Dn ∩ (0,∞).

Since Q+ is contained in each D+n , we know that each D+

n is dense in R+. Let (qn) denumeratethe positive rationals. Then the sets An = R+ − qn are also all open and dense in R+.Then by the Baire Category Theorem, the intersection of all the An and D+

n sets must beopen and dense in R+. This is impossible because⋂

n

D+n = Q+ whereas

⋂n

An = R+ −Q+.

The contradiction means that (2) is satisfied.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 4

Assume that S is a countable union of closed sets:

S =⋃n

Cn.

ThenR− S =

⋂n

(R− Cn)

which is a countable intersection of open sets. This is impossible by the same reasoning asabove since R− S contains the negative rationals.

Problem S03.4. Consider the following equation for a function F (x, y) on R2:

∂2F

∂x2=∂2F

∂y2.

(a) Show that if a function F has the form F (x, y) = f(x+y)+g(x−y) where f, g : R→ Rare twice differentiable, then F satisfies the equation.

(b) Show that if F (x, y) = ax2 + bxy + cy2 then F (x, y) = f(x + y) + g(x − y) for somepolynomials f, g in one variable.

Solution.

(a) By the chain rule

∂F

∂x= f ′(x+ y)

d

dx(x+ y) + g′(x− y)

d

dx(x− y) = f ′(x+ y) + g′(x− y)

and

∂2F

∂x2= f ′′(x+ y)

d

dx(x+ y) + g′′(x− y)

d

dx(x− y) = f ′′(x+ y) + g′′(x− y).

Likewise∂F

∂y= f ′(x+ y)− g′(x− y)

and∂2F

∂y2= f ′′(x+ y) + g′′(x− y).

Thus∂2F

∂x2=∂2F

∂y2.

(b) The equation immediately yields a = c so F (x, y) = ax2 + bxy + ay2. Then

F (x, y) = a(x2 + y2) + bxy

= a(

12(x+ y)2 + 1

2(x− y)2

)+ b(

12(x+ y)2 − 1

2(x− y)2

)= 1

2(a+ b)(x+ y)2 + 1

2(a− b)(x− y)2.

Thus taking f(x) = 12(a+ b)x2 and g(x) = 1

2(a− b)x2 works.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 5

Problem F04.3. Show that if fn → f uniformly on the closed bounded interval [a, b], then∫ b

a

fn(x)dx→∫ b

a

f(x)dx.

Solution. Let ε > 0. Since fn → f uniformly, N ∈ N such that |fn(x)− f(x)| < ε/2 whenn ≥ N . By definition of the Riemann integral, there are some piecewise constant functionsgn, hn such that gn majorizes fn and hn minorizes fn for all n and∫ b

a

gn(x)dx−∫ b

a

fn(x)dx < ε and

∫ b

a

fn(x)dx−∫ b

a

hn(x)dx < ε.

Since f stays within ε/2 of fn(x) for n suffieciently large, it follows that for such n, gn + ε ispiecewise constant and majorizes f and hn − ε is piecewise constant and minorizes f . Then∫ b

a

f(x)dx ≤∫ b

a

(gn(x) + ε)dx ≤∫ b

a

fn(x)dx+ ε+ ε(b− a) =

∫ b

a

f(x)dx+ (b− a+ 1)ε.

Likewise, we have∫ b

a

f(x)dx ≥∫ b

a

(hn(x)− ε)dx ≥∫ b

a

fn(x)dx− ε− (b− a)ε =

∫ b

a

fn(x)dx− (b− a+ 1)ε.

Since ε is arbitrarily small, this proves that∫ bafn(x)dx→

∫ baf(x)dx.

Problem F04.4. Suppose that (X, d) is a metric space, x ∈ X and (xn) is a sequence inX converging to x. Show that for every y ∈ X, d(xn, y)→ d(x, y).

Solution. From the triangle inequality, we have

d(xn, y) ≤ d(x, y) + d(xn, x) =⇒ d(xn, y)− d(x, y) ≤ d(xn, x)

andd(x, y) ≤ d(xn, y) + d(xn, x) =⇒ d(x, y)− d(xn, y) ≤ d(xn, x)

which imply|d(xn, y)− d(x, y)| ≤ d(xn, x)→ 0 as n→∞.

Problem F04.5. Prove that the space C[0, 1] of continuous functions from [0, 1] to R withthe supremum norm ||f ||∞ = sup

x∈[0,1]

|f(x)| is complete.

Solution. Let (fn) be a Cauchy sequence in C[0, 1]. Let ε > 0. Then there is N ∈ N suchthat m,n > N implies

|fn(x)− fm(x)| ≤ ||fn − fm||∞ < ε

for all x ∈ [0, 1]. This means that for each x ∈ [0, 1], the sequence (fn(x)) is a Cauchysequence in R and thuse converges to some ax ∈ R. Define f : [0, 1]→ R so that f(x) = axfor every x ∈ [0, 1]. It clear that fn → f pointwise, and thus in the supremum norm. Weneed to prove that f ∈ C[0, 1]. Let ε > 0. Then for any x, y ∈ [0, 1], there are M,N ∈ Nand δ > 0 so that

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 6

(1) |f(x)− fn(x)| < ε/3 when n ≥ N ,

(2) |fn(x)− fn(y)| < ε/3 when |x− y| < δ,

(3) |fn(y)− f(y)| < ε/3 when n ≥M .

Setting N∗ = maxM,N and taking n > N∗, we have

|f(x)− f(y)| = |f(x)− fn(x) + fn(x)− fn(y) + fn(y)− f(y)|≤ |f(x)− fn(x)|+ |fn(x)− fn(y)|+ |fn(y)− f(y)|< ε/3 + ε/3 + ε/3 = ε,

which shows that f is continuous and so C[0, 1] is complete.

Problem F04.6. The Bolzano-Weierstrass Theorem in Rn states that if S is a closed,bounded subset in Rn, then every sequence in S has a subsequence converging in S. Assumethe theorem for n = 1 and prove the theorem for n = 2.

Solution. Let (xn) be a sequence in S and let M > 0 be a bound on the norm of the

elements of S. We see for each component x(i)n of xn,∣∣x(i)

n

∣∣ ≤ ||xn|| ≤M.

Thus, the sequence of first components (x(1)n ) is a bounded sequence in R, which, by assump-

tion has a subsequence converging to an element of R, say x(1). Call this subsequence (x(1)nk ).

Then (x(2)nk ) is also a bounded sequence in R and thus has a convergent subsequence (x

(2)nk`

)

which converges to x(2). Since any subsequence of a convergent sequence converges to thesame limit, we know (x

(1)nk`

) still converges to x(1). Then the sequence (xn) converges to the

vector x = (x(1), x(2)). Since S is closed, we know x must be in S.

Problem F04.7. Observe that the point P = (1, 1, 1) belongs to the set S of points in R3

satisfying the equationx4y2 + x2z + yz2 = 3.

Explain how the Implicit Function Theorem allows us to conclude that there is a differen-tiable function f(x, y) such that (x, y, f(x, y)) ∈ S for all (x, y) in a small open neighborhoodof (1, 1).

Solution. The implicit function theorem tells us that is g(x, y, z) is continuously differen-tiable, g(x0, y0, z0) = a ∈ R and dg

dz(x0, y0, z0) 6= 0 then there is a neighborhood U of (x0, y0),

a neighborhood V of z0 and a continuously differentiable function f : U → V such that

(x, y, f(x, y)) : x, y ∈ U = (x, y, z) : g(x, y, z) = a.

Here, if we setg(x, y, z) = x4y2 + x2z + yz2

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 7

and (x0, y0, z0) = (1, 1, 1) then g is smooth, g(x0, y0, z0) = 3 and dgdz

(x0, y0, z0) = 3 6= 0. Thenthere are open neighborhoods U of (1,1), V of 1 and a continuously differentiable functionf : U → V , such that

S = (x, y, z) : g(x, y, z) = 3 = (x, y, f(x, y)) : (x, y) ∈ U.

This is exacly the conclusion we wanted to draw.

Problem F04.8. Let A ∈ Mn(R) be symmetric and let Q(v) = v · Av be the associatedquadratic form defined for v = (v1, . . . , vn) ∈ Rn.

1. Show that ∇Qv = 2Av where ∇Qv is the gradient at v of the function Q.

2. Let M be the minimum value of Q(v) on the unit sphere Sn = v ∈ Rn : ||v|| = 1 andlet u ∈ Sn be a vector such that Q(u) = M . Prove using Lagrange multipliers that uis an eigenvector of A with eigenvalue M .

Solution.

1. Let aj be the columns of A. We see

Q(v) = v · Av = v ·

(n∑j=1

vjaj

)=

n∑j=1

vj(v · aj) =n∑j=1

vj

(n∑i=1

viaij

)

Then∂Q

∂vk(v) =

n∑i=1

viaik +n∑j=1

vjakj.

But since the matrix is symmetric,

∂Q

∂vk(v) = 2

n∑i=1

viaik = 2v · ak

Then∇Q(v) = 2Av.

2. The problem is to minimize Q(u) subject to g(u) = ||u||2 − 1 = 0. Any minimizer umust satisfy

∇Q(u) = λ∇g(u) ⇐⇒ 2Au = λ(2u) ⇐⇒ Au = λu,

for some λ ∈ R. Taking the inner product of the above equation with u, we see

Q(u) = λ(u, u) = λ.

But Q(u) = M so M = λ is an eigenvalue of A.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 8

Problem S04.4. Are there infinite compact subsets of Q? Prove your assertion.

Solution. Yes. The set A = 0 ∪

1n

: n = 1, 2, 3, . . .

is obviously infinite, bounded andcontained in Q. Recall that a subset of R is compact iff it is closed and bounded. A is closedbecause

R− A = (−∞, 0) ∪ (1,∞) ∪

(∞⋃n=1

(1

n+1, 1n

))is a union of open sets and is thus open. This implies that A is infinite and compact.

Problem S04.5. Suppose that G ⊂ Rn is open, f : G→ Rm and that x0 ∈ G.

(a) Carefully define what is meant by f ′(x0) : Rn → Rm.

(b) Suppose that I is a line segment in G such that f ′(x) is defined for x ∈ I. Show thatif f is differentiable at all points of I, then there is a point c ∈ I such that

||f(q)− f(p)|| ≤ ||f ′(c)|| ||q − p||

where p, q are distinct points in I.

Solution.

(a) We say that f : G → Rm is differentiable at a point x0 ∈ G if there is a linear mapT : G→ Rm such that

||f(x0 + h)− f(x0)− T (h)||||h||

→ 0 as ||h|| → 0.

If such a T exists, we define f ′(x0) = T . That is f ′(x0) is a linear map.

(b) If f(q) = f(p), the result is trivial. Otherwise, we see that

||f(q)− f(p)|| = (f(q)− f(p)) · (f(q)− f(p))

||f(q)− f(p)||..= (f(q)− f(p)) · w

where w is a unit vector. Define g : G→ R by

g(p) = f(p) · w.

Since f is differentiable on I, so is g. then for any x ∈ I, g′(x) is a functional on Gand

g′(x)(u) = (f ′(x)(u)) · w.

By the mean value theorem, there is c ∈ I such that

g(q)− g(p) = g′(c)(p− q) = (f ′(c)(p− q)) · w.

But g(q)− g(p) = ||f(q)− f(p)||. Then using Cauchy-Schwarz,

||f(q)− f(p)|| ≤ ||(f ′(c)(p− q))|| ||w|| = ||(f ′(c)(p− q))|| .

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 9

But now we use the fact that ||T (v)|| ≤ ||T || ||v|| for any linear operator T and vectorv. This yields

||f(q)− f(p)|| ≤ ||f ′(c)|| ||q − p|| ,which completes the proof.

Problem S04.6. Let ||·|| be any norm on Rn.

(a) Prove that there is a constant d with ||x|| ≤ d ||x||2 for all x ∈ Rn, and use this toshow that N(x) = ||x|| is continuous in the usual topology.

(b) Prove that there is a constant c with ||x|| ≥ c ||x||2. (Hint: use the fact that N iscontinuous on x : ||x||2 = 1)

(c) Show that if L is an n-dimensional subspace of an arbitrary normed space V , then Lis closed.

Solution.

(a) Take any x ∈ Rn and write x =∑n

i=1 xiei where ei are the standard basis vectors.Then

||x|| =

∣∣∣∣∣∣∣∣∣∣n∑i=1

xiei

∣∣∣∣∣∣∣∣∣∣ ≤

n∑i=1

|xi| ||ei|| ≤

√√√√( n∑i=1

x2i

)(n∑i=1

||ei||2)

= d ||x||2

where d =(∑n

i=1 ||ei||2). Taking x, y ∈ Rn such that ||x− y||2 < δ = ε

d. Then by the

reverse triangle inequality,

|||x|| − ||y||| ≤ ||x− y|| ≤ d ||x− y||2 ≤ d · εd

= ε,

and hence the norm is continuous.

(b) Since the norm is continuous on the unit ball (with respect to the twonorm), and sincethe unit ball is compact, the norm achieves a minumum there. Say c = min||x||2=1 ||x||.Then for any y ∈ Rn,

||y|| = ||y||2

∣∣∣∣∣∣∣∣ y

||y||2

∣∣∣∣∣∣∣∣ ≥ c ||y||2 .

(c) Suppose L is an n-dimensional subspace of a normed space V . We prove that Lcontains its limit points and is thus closed. Let x be a limit point of L. Then there is asequence (xn) in L converging to x. Since L is ndimensional, there is a linear bijectionT : L → Rn. Define the norm ||·|| on Rn by ||y|| = ||T−1y||V , y ∈ Rn. From (a), (b),there are constants c, d such that for all y ∈ Rn

c ||y||2 ≤ ||y|| ≤ d ||y||2 .

Then

c ||Txm − Txn||2 ≤ ||Txm − Txn|| = ||xn − xm||V ≤ d ||Txn − Txm||2

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 10

for all m,n ∈ N. Let ε > 0. Since (xn) is convergent, in particular, it is Cauchy, sothere are m,n ∈ N so that ||xn − xm|| < cε. Choosing these show that (Txn) is Cauchyin Rn with respect to the two norm and so has a limit point in y ∈ Rn since the twonorm is complete. Using the other side of the above inequality, we can show y = Txand so x = T−1y ∈ L. Thus L is closed.

Problem F05.1. A real number α is said to be algebraic if for some finite set of integersa0, . . . , an (not all zero), we have

a0 + a1α + · · ·+ anαn = 0.

Prove that the set of algebraic numbers is countable.

Solution. Let A be the set of algebraic numbers and Z[x] the set of polynomials over Z.We see

A =⋃

p∈Z[x]

x ∈ R : p(x) = 0.

Since Z is countable, so is Z[x]. Thus the union above is a countable union of finite sets andhence countable so A is countable.

Problem F05.3.

(a) Prove that if fj : [0, 1] → R is a sequence of continuous functions which convergeuniformly to F : [0, 1]→ R, then

limj→∞

∫ 1

0

fj(x)2dx =

∫ 1

0

F (x)2dx.

(b) Give an example of a sequence fj : [0, 1]→ R which converge to a continuous functionF : [0, 1]→ R pointwise and for which

limj→∞

∫ 1

0

fj(x)2dx exists, but

limj→∞

∫ 1

0

fj(x)2dx 6=∫ 1

0

F (x)2dx.

Solution.

(a) Let ε > 0. Since fj → F uniformly, we know that F is continuous and hence boundedon [0, 1] by some M ∈ R. Then by uniform convergence, there is N ∈ N such thatj ≥ N implies

|fj(x)− F (x)| <√ε

2and |fj(x)− F (x)| < ε

4M, x ∈ [0, 1].

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 11

Letting j ≥ N , we see∣∣fj(x)2 − F (x)2∣∣ =

∣∣(fj(x)− F (x))2 + 2fj(x)F (x)− 2F (x)2∣∣

≤ |fj(x)− F (x)|2 + 2 |F (x)| |fj(x)− F (x)|

2+ 2M

ε

4M= ε, x ∈ [0, 1].

Thus f 2j → F 2 uniformly and the result follows by the previous problem.

(b) Define

fj(x) =

0, x < 12j,√

4j2(x− 1

2j

), 1

2j≤ x < 1

j,√

4j2(

32j− x), 1

j≤ x < 3

2j,

0, x ≥ 32j

Then each f 2j is a triangular spike centered at 1

jwith base length 1

jand height 2j.

Thus ∫ 1

0

fj(x)2dx =1

2· 1

j· 2j = 1 so lim

j→∞

∫ 1

0

fj(x)2dx = 1.

However, for any x ∈ [0, 1], there is some Jx ∈ N such that 32j< x when j ≥ Jx. For

j ≥ Jx, we have fj(x) = 0. Thus

limj→∞

fj(x)2 = 0, x ∈ [0, 1].

Hence f 2j converge pointwise to the zero function so∫ 1

0

limj→∞

fj(x)2dx = 0.

Problem F05.4. Suppose F : [0, 1]→ [0, 1] is a C2 function with F (0) = 0, F (1) = 0 andF ′′(x) < 0 for all x ∈ [0, 1] Prove that the arc length of the curve (x, F (x)) : x ∈ [0, 1] isless than 3.

Solution. The arc length is given by

L =

∫ 1

0

√1 + F ′(x)2dx.

However,√a2 + b2 ≤ |a|+ |b| for all a, b ∈ R, so

L ≤∫ 1

0

(1 + |F ′(x)|)dx = 1 +

∫ 1

0

|F ′(x)| dx.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 12

Now F ′′(x) < 0 implies that F ′ is a decreasing function of x. Since F (0) = F (1) = 0, weknow F must increase to a maximum and then decrease back to 0. Let the maximum of Foccur at x = xM . Then

L ≤ 1 +

∫ xM

0

F ′(x)dx−∫ 1

xM

F ′(x)dx = 1 + F (xM) + F (xM) = 1 + 2F (xM).

Then since F (xM) ≤ 1, we have L ≤ 3.

Problem F05.8. For a real n× n matrix A set ||A|| = sup||x||=1

||Ax||.

(a) Prove that ||A+B|| ≤ ||A||+ ||B||.

(b) Use part (a) to check that the set M of all n × n matrices is a metric space if thedistance dunction d is defined by

d(A,B) = ||B − A|| .

(c) Prove that (M,d) is complete.

Solution.

(a) For any particular x ∈ Rn with ||x|| = 1, we know

||Ax|| ≤ ||A|| ,

since the norm is the supremum over all such x. Taking ||x|| = 1, we see

||(A+B)x|| = ||Ax+Bx|| ≤ ||Ax||+ ||Bx|| ≤ ||A||+ ||B|| .

Since the inequality holds for all such x, it holds for the supremum. Thus the inequalityis proven.

(b) The non-negativity and symmetry is obvious. If d(A,B) = 0 then ||(B − A)x|| = 0 forall x ∈ Rn, ||x|| = 1. Taking x = ei, for each i = 1, . . . , n shows that each column ofB − A is the zero vector so B − A = 0 and A = B. The triangle inequality followsfrom (a) since for any A,B,C, we have

d(A,B) = ||B − A|| = ||B − C + C − A|| ≤ ||B − C||+ ||C − A|| = d(A,C)+d(B,C).

Thus (M,d) is a metric space.

(c) Let (An) be a Cauchy Sequence in M . We notice that

d(An, Am) = ||Am − An||≥ ||(Am − An)ei||

≥∣∣∣∣∣∣a(m)

i − a(n)i

∣∣∣∣∣∣ (the ith column of each matrix)

≥∣∣∣a(m)ij − a

(n)ij

∣∣∣ .This shows that each entry of each matrix forms a Cauchy sequence in R. So thesequences of each of the entries converge and the matrices converge. Thus (M,d) iscomplete.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 13

Problem S05.7. Let X, Y be topological spaces. We say that a continuous functionf : X → Y is proper if f−1(K) is compact in X whenever K is compact in Y .

(a) Give an example of a function that is proper but not a homeomorphism.

(b) Give an example of a function which is continuous but not proper.

(c) Suppose that f : R→ R is C1 and

|f ′(x)| ≥ 1, for all x ∈ R.

Show that f is proper.

Solution.

(a) Let X be any compact metric space, and let f : X → R be defined f(x) = 0 for allx ∈ X. Then clearly f is continuous. Also, if K ⊂ R is compact, then

f−1(K) = x ∈ X : f(x) ∈ K =

∅, 0 6∈ K,X, 0 ∈ K

is compact. However, the function is not a bijection since it is not a surjection andthus it is not a homeomorphism.

(b) Let f : R→ R be defined so that f(x) = 0 for all x ∈ R. Then clearly f is continuous,but for any compact K ⊂ R such that 0 ∈ K, we have f−1(K) = R which is notcompact. Thus f is not proper.

(c) Suppose K ⊂ R is compact. Let (Uα)α∈I be an open cover of f−1(K). Then (f(Uα))α∈Iis an open cover of f(f−1(K)) = K. Since K is compact, there is finite subcover(f(Ui))

Ni=1 for some N ∈ N.

Let x ∈ f−1(K). Then f(x) ∈ f(Ui) for some i. But this implies there is y ∈ Ui suchthat f(y) = f(x). If we assume that x 6= y (wlog x < y), then by the mean valuetheorem, there is c ∈ (x, y) such that f(x)−f(y) = f ′(c)(x−y). But since f(x) = f(y),this leads to

0 = |f ′(c)| |x− y| ≥ |x− y| ,

which is a contradiction. Thus x = y, so x ∈ Ui for some i. Hence (Ui)Ni=1 is an finite

subcover for f−1(K) so f−1(K) is compact and so f is proper.

Problem S05.8. Suppose f : R→ R is C1. Show that

limn→∞

n∑j=1

∣∣f ( j−1n

)− f

(jn

)∣∣is equal to ∫ 1

0

|f ′(t)| dt.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 14

Solution. Let ε > 0. Since f ′ is continuous on [0, 1] so is |f ′| and since [0, 1] is compact,we know that |f ′| is uniformly continuous there. Let N ∈ N be large enough that for allx, y ∈ [0, 1],

|x− y| < 1

N=⇒

∣∣∣ |f ′(x)| − |f ′(y)|∣∣∣ < ε.

Then for n ≥ N , by the mean value theorem, there is xn ∈ ((j − 1)/n, j/n) such that∣∣f ( j−1n

)− f

(jn

)∣∣ = 1n|f ′(xn)|. Also, there is yn ∈ ((j − 1)/n, j/n) which maximizes |f ′| on

that interval. Then for n > N,∣∣∣∣∣∫ 1

0

|f ′(t)| dt−n∑j=1

∣∣f ( j−1n

)− f

(jn

)∣∣∣∣∣∣∣ =

∣∣∣∣∣n∑j=1

∫ j/n

(j−1)/n

|f ′(t)| dt−∣∣f ( j−1

n

)− f

(jn

)∣∣∣∣∣∣∣≤

∣∣∣∣∣n∑j=1

1

n|f ′(yn)| − 1

n|f ′(xn)|

∣∣∣∣∣≤ 1

n

n∑j=1

∣∣∣ |f ′(yn)| − |f ′(xn)|∣∣∣ < 1

nnε = ε.

Thus limn→∞

n∑j=1

∣∣f ( j−1n

)− f

(jn

)∣∣ =

∫ 1

0

|f ′(t)| dt.

Problem S05.10. Let K be the set of f : [0, 1]→ R that obey

(i) |f(x)− f(y)| ≤ |x− y| ,

(ii)

∫ 1

0

f(x)dx = 1.

Prove that K is a compact subset of C[0, 1].

Solution. Arzela-Ascoli says that a subset of C[0, 1] is compact if and only if it is uniformlybounded and equicontinuous. It is clear that K ⊂ C[0, 1] because the (i) implies continuity.Let f ∈ K. If there was x ∈ [0, 1] such that f(x) ≥ 3, then |f(y)− f(x)| ≤ |y − x| ≤ 1implies that f(y) ≥ 2 for all y ∈ [0, 1] which means that f couldn’t satisfy (ii). Thus K isuniformly bounded. Further, (i) implies equicontinuity directly. Thus K is compact.

Problem S05.11. Make Mn(C) into a metric space in the following fashion

d(A,B) =

√∑i,j

|Aij −Bij|2.

(a) Suppose F : R→Mn(C) is continuous. Show that the set

x : F (x) is invertible

is open in the usual topology on R.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 15

(b) Show that on the set given above, x 7→ [F (x)]−1 is continuous.

Solution.

(a) We seex : F (x) is invertible = x : detF (x) 6= 0.

The latter is open because it is the pullback of an open set under the compositionfunction detF (x) which is continuous since both det and F are continuous.

(b) The maps is continuous because F is continuous and A 7→ A−1 is continuous since theentries of the inverse are polynomials in the entries of A.

Problem S05.12. Let (X, d) be a metric space. Prove that the following are equivalent:

(a) There is a countable dense set in X.

(b) There is a countable basis for the topology on X.

Solution. If there is a countable basis Bnn∈N for the topology, then taking xn ∈ Bn foreach n ∈ N gives a countable dense set x1, x2, x3, . . ..

Conversely, assume that (xn)n∈N is a countable dense set. Consider the set

β = B(xn, 1/m) : n ≥ 1,m ≥ 1

Let U be open in X with x ∈ U . The ball B(x, 2/k) ⊂ U for sufficiently large k. Also thereis some n, such that x ∈ B(xn, 1/k). By the triangle inequality, We have B(xn, 1/k) ⊂ U.Thus for every x ∈ U , there is Bx ∈ β such that x ∈ Bx ⊂ U . Then

U =⋃x∈U

Bx.

Ergo, we can build any open set in X from sets in β so β is a basis for the topology on Xand clearly β is countable since it can be put in bijection with N× N.

Problem W06.1. Show that for each ε > 0, there exists a sequence of intervals (In) withthe properties

Q ⊂∞⋃n=1

In and∞∑n=1

|In| < ε.

Solution. Let (qn)n∈N be a denumeration of the rationals. Define In = (qn − ε/2n+2, qn +ε/2n+2). Then clearly

Q ⊂∞⋃n=1

In

and∞∑n=1

|In| =∞∑n=1

2 · ε

2n+2=∞∑n=1

ε

2n+1=ε

2< ε.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 16

Problem W06.3. Consider a function f : [a, b]→ R which is twice continuously differen-tiable. Let a = x0 < x1 < . . . < xn = b be the uniform partition of [a, b]. Show that thereexists M such that for all n ≥ 1,∣∣∣∣b− an

(1

2f(x0) + f(x1) + . . .+ f(xn−1) +

1

2f(xn)

)−∫ b

a

f(x)dx

∣∣∣∣ ≤ M

n2.

Solution. Call the expression on the left hand side E. Then we rewrite E:

E =

∣∣∣∣∣(n−1∑i=0

b− a2n

(f(xi+1)− f(xi))

)−∫ b

a

f(x)dx

∣∣∣∣∣=

∣∣∣∣∣n−1∑i=0

(b− a2n

(f(xi+1)− f(xi))−∫ xi+1

xi

f(x)dx

)∣∣∣∣∣≤

n−1∑i=0

∣∣∣∣b− a2n(f(xi+1)− f(xi))−

∫ xi+1

xi

f(x)dx

∣∣∣∣ .Integrating by parts, we see∫ xi+1

xi

f(x)dx =

[(x− C)f(x)

∣∣∣xi+1

xi

]−∫ xi+1

xi

(x− C)f ′(x)dx,

where C is a parameter we can choose. Letting C = (xi+1 + xi)/2, we see∫ xi+1

xi

f(x)dx =b− a2n

(f(xi+1)− f(xi))−∫ xi+1

xi

(x− C)f ′(x)dx.

Then if Ei is the ith member of the sum above, we have

Ei =

∣∣∣∣∫ xi+1

xi

(x− xi+1 + xi

2

)f ′(x)dx

∣∣∣∣ .Integrating by parts again, we get

Ei =

∣∣∣∣∣[

1

2

((x− xi+1 + xi

2

)2

−B

)f ′(x)

∣∣∣∣∣xi+1

xi

]−∫ xi+1

xi

1

2

((x− xi+1 + xi

2

)2

−B

)f ′′(x)dx

∣∣∣∣∣ .We choose B =

(b−a2n

)2so that the boundary term goes to zero. Then

Ei =

∣∣∣∣∣∫ xi+1

xi

1

2

((x− xi+1 + xi

2

)2

−(b− a2n

)2)f ′′(x)dx

∣∣∣∣∣≤∫ xi+1

xi

1

2

∣∣∣∣∣(x− xi+1 + xi

2

)2

−(b− a2n

)2∣∣∣∣∣ |f ′′(x)| dx.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 17

Now f ′′ is continuous on [a, b] and thus bounded by some K ∈ R. Hence

Ei ≤K

2

∫ xi+1

xi

(b− a2n

)2

− 1

2

(x− xi+1 + xi

2

)2

dx

=K(b− a)3

12n3.

Then summing over all i, we see

E ≤ K(b− a)3

12n2.

Problem W06.5. Consider a function f(x, y) which is twice continuously differentiable.Suppose that f has a unique minimum at (x, y) = (0, 0). Prove that at (0, 0),

∂2f

∂x2

∂2f

∂y2≥(∂2f

∂x∂y

)2

.

[You may use without proof that the mixed partials are equal for C2 functions.]

Solution. If f has a minimum at a point (x, y) then ∇f(x, y) = ~0. By Taylor’s Theoremin a neighborhood of (0,0),

f(x, y) = f(0, 0) +∇f(0, 0) · (x, y) +1

2

(∂2f

∂x2x2 + 2

∂2f

∂x∂yxy +

∂2f

∂y2y2

)+ E(x, y)

= f(0, 0) +1

2

(∂2f

∂x2x2 + 2

∂2f

∂x∂yxy +

∂2f

∂y2y2

)+ E(x, y)

where E(x, y) → 0 quickly as (x, y) → (0, 0) and the second derivatives are evaluated at(0, 0). We know f(x, y)− f(0, 0) ≥ 0 since f(0, 0) is a minimum. This gives

−E(x, y) ≤ f(x, y)− f(0, 0)− E(x, y) =1

2

(∂2f

∂x2x2 + 2

∂2f

∂x∂yxy +

∂2f

∂y2y2

).

But since E(x, y)→ 0, it follows that, in some neighborhood of (0, 0),(∂2f

∂x2x2 + 2

∂2f

∂x∂yxy +

∂2f

∂y2y2

)≥ 0.

Consider the matrix

H =

(∂2f∂x2

∂2f∂x∂y

∂2f∂x∂y

∂2f∂y2

);

again, with the functions evaluated at (0, 0). The above inequality tells us that this matrix isnon-negative definite. Hence its eigenvalues are non-negative and so its determinant (whichis the product of the eigenvalues) is non-negative. Thus at (0, 0),

∂2f

∂x2

∂2f

∂y2≥(∂2f

∂x∂y

)2

.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 18

Problem S06.3. Prove that

f(x) =∞∑n=1

sin(nx)

n5/2

converges for all x ∈ R and that f(x) is continuous on R with a continuous derivative.

Solution. For any x ∈ R, we have

|f(x)| =

∣∣∣∣∣∞∑n=1

sin(nx)

n5/2

∣∣∣∣∣ ≤∞∑n=1

|sin(nx)|n5/2

≤∞∑n=1

1

n5/2,

which converges. Hence the sum converges for each x ∈ R; in fact, this shows that theconvergence is uniform.

Let (fN) be defined by fN(x) =∑N

n=1sin(nx)

n5/2 . Then each fN is continuous as a finite sumof continuous functions. Also, for any x ∈ R,

|f(x)− fN(x)| =

∣∣∣∣∣∞∑

n=N+1

sin(nx)

n5/2

∣∣∣∣∣ ≤∞∑

n=N+1

1

n5/2

which goes to zero as N →∞. Thus fN → f uniformly. Since the uniform limit of continuousfunctions is continuous, we know that f is continuous.

Further, each fN is actually differentiable, with derivative

f ′N(x) =N∑n=1

cos(nx)

n3/2.

These also converge uniformly by the same reasoning as above, so we know that f is alsodifferentiable with derivative

f ′(x) =∞∑n=1

cos(nx)

n3/2.

Problem S06.5.

(A) Define uniform continuity for a function f on a metric space X with the distancefunction ρ(x, y).

(B) Prove that if 0 < α < 1, then F (x) = xα is uniformly continuous on [0,∞).

Solution.

(A) We say a function f : (X, ρ)→ (Y, σ) is uniformly continuous if for any ε > 0 there isδ = δ(ε) > 0 such that for any x, y ∈ X, ρ(x, y) < δ =⇒ σ(f(x), f(y)) < ε.

(B) We prove that F (x) is uniformly continuous on [0, 2] and on [1,∞). Since F (x) iscontinuous on [0, 2] and [0, 2] is compact, F (x) is uniformly continuous there. Letε > 0 and take δ = ε/α. Then for x, y ∈ [1,∞), by the mean value theorem, there isz ∈ (x, y) such that

|F (x)− F (y)| = |F ′(z)| |x− y| = αz1−α |x− y| ≤ α |x− y| < ε,

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 19

when 0 < |x− y| < δ. Thus F is uniformly continuous on [1,∞).

Now, for ε > 0, there is δ1 > 0 and δ2 > 0 such that for x, y ∈ [0, 2],

|x− y| < δ1 =⇒ |F (x)− F (y)| < ε

and for z, w ∈ [1,∞),

|z − w| < δ2 =⇒ |F (z)− F (w)| < ε.

Take δ = minδ1, δ2,

12

. Then for x, y ∈ [0,∞) with |x− y| < δ we have that

x, y ∈ [0, 2] or x, y ∈ [1,∞). Then since δ is less than both δ1 and δ2, we have that|F (x)− F (y)| < ε. Thus F is uniformly continuous.

Problem S06.6. Let W be the subset of C[0, 1] satisfying

|f(x)− f(y)| < |x− y| and

∫ 1

0

f(x)2dx = 1.

(A) Prove that W is uniformly bounded.

(B) Prove that W is a compact subset of C[0, 1].

Solution.

(A) Suppose that for some f ∈ W , there is an x ∈ [0, 1] such that f(x) > 3. Then

|f(x)− f(y)| < |x− y| ≤ 1

which means that f(y) > 2 for all y ∈ [0, 1]. Then∫ 1

0f(x)2dx >

∫ 1

04dx > 4, a

contradiction. Hence f(x) < 3 for all x ∈ [0, 1]. Likewise, assume there is x ∈ [0, 1]such that f(x) < −3. Then

|f(x)− f(y)| < |x− y| ≤ 1

implies that f(y) < −2 for all y ∈ [0, 1] and thus f(y)2 > 4, y ∈ [0, 1]. Then∫ 1

0f(x)2dx > 4, a contradiction. Thus f(x) ≥ −3, x ∈ [0, 1]. Hence W is uniformly

bounded.

(B) By the Arzela-Ascoli theorem, W is compact if and only if it is equicontinuous anduniformly bounded. We proved the uniform bound in part (A). The first conditiondirectly implies equicontinuity of W . Thus W is compact.

Problem F07.2. Let f : [a, b] → R be continuous and differentiable in (a, b) − c. Iflimx→c f

′(x) = d ∈ R, show that f is differentiable at c and f ′(c) = d.

Solution. It suffices to show that |f(x)− (f(c) + d(x− c))| gets arbitrarily small if x isnear c; this will prove that f ′(c) exists and equals d.

Let ε > 0. Then since limx→c f′(x) = d, there is δ1 > 0, such that |f ′(x)− d| < ε/4 when

|x− c| < δ1.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 20

Also, since f is continuous at c, there is δ2 > 0 such that |f(x)− f(c)| < ε/4 when|x− c| < δ2.

Take y ∈ (a, b), 0 < |y − c| < minδ1, δ2. Then f is differentiable at y and so there isδ3 > 0 such that |x− y| < δ3 implies that |f(x)− (f(y) + f ′(y)(x− y))| < ε/4.

Now taking |x− c| small enough, will ensure that |x− y| < δ4, and so

|f(x)− (f(c) + d(x− c))| ≤ |f(x)− (f(y) + f ′(y)(x− y))|+|(f(y) + f ′(y)(x− y))− (f(c) + d(x− c))|

< ε/4 + |f(y)− f(c)|+ |f ′(y)(x− y)− d(x− c)|< ε/4 + ε/4 + |f ′(y)(x− y)− d(x− y) + d(x− y)− d(x− c)|≤ ε/2 + |f ′(y)− d| |x− y|+ |d| |y − c|< ε/2 + εδ3/4 + |d|minδ1, δ2.

Decreasing each δ (if necessary) will ensure that this is less than ε. Thus f ′(c) = d.

Problem F07.4. Suppose that f : R→ R is twice-differentiable and its second derivativesatisfies |f ′′(x)| ≤ B.

a) Prove that ∣∣∣∣2Af(0)−∫ A

−Af(x)dx

∣∣∣∣ ≤ A3B

3.

b) Use the result from part a) to justify the estimate:∣∣∣∣∣∫ b

a

f(x)dx− b− an

n∑k=1

f

(a+

2k − 1

2n(b− a)

)∣∣∣∣∣ ≤ Cn−2

where C is a constant which does not depend on n.

Solution.

a) Notice that 2Af(0) =∫ A−A f(0)dx.

By Taylor’s theorem, for any x ∈ [−A,A] there is ξx ∈ [−A,A] such that f(x) =

f(0) + f ′(0)x+ f ′′(ξx)2

x2. Then for each x,

|f(x)− f(0)− xf ′(0)| =∣∣∣∣f ′′(ξx)2

x2

∣∣∣∣ ≤ Bx2

2.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 21

From this, we see∣∣∣∣2Af(0)−∫ A

−Af(x)dx

∣∣∣∣ =

∣∣∣∣∫ A

−Af(x)− f(0)dx

∣∣∣∣=

∣∣∣∣∫ A

−Af(x)− f(0)− xf ′(0) + xf ′(0)dx

∣∣∣∣≤∣∣∣∣∫ A

−Af(x)− f(0)− xf ′(0)dx

∣∣∣∣+

∣∣∣∣∫ A

−Af ′(0)xdx

∣∣∣∣≤∫ A

−A|f(x)− f(0)− xf ′(0)| dx

≤∫ A

−A

Bx2

2dx =

A3B

3.

b) Let a = x0 < x1 < · · · < xn = b be the uniform partition of [a, b]. Put Ak =a+ 2k−1

2n(b−a). Then Ak is the midpoint of [xk−1, xk] for each k. Put fk(x) = f(Ak−x)

for each k. Then

f

(a+

2k − 1

2n(b− a)

)= f(Ak) = fk(0)

and ∫ xk

xk−1

f(x)dx =

∫ xk−Ak

xk−1−Akf(x− Ak)dx

=

∫ (b−a)/2n

−(b−a)/2n

f(x− Ak)dx

=

∫ (b−a)/2n

−(b−a)/2n

f(Ak − x)dx

=

∫ (b−a)/2n

−(b−a)/2n

fk(x)dx.

Finally, but h = b−a2n

. Then,∣∣∣∣∣∫ b

a

f(x)dx− b− an

n∑k=1

f

(a+

2k − 1

2n(b− a)

)∣∣∣∣∣ =

∣∣∣∣∣n∑k=1

∫ xk

xk−1

f(x)dx− 2hn∑k=1

fk(0)

∣∣∣∣∣=

∣∣∣∣∣n∑k=1

∫ h

−hfk(x)dx− 2h

n∑k=1

fk(0)

∣∣∣∣∣≤

n∑k=1

∣∣∣∣∫ h

−hfk(x)dx− 2hfk(0)

∣∣∣∣≤

n∑k=1

h3B

3=nh3B

3=B(b− a)3

12n2,

which proves the claim.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 22

Problem F07.5.

a) Show that, given a continuous function f : [0, 1] → R such that f(1) = 0, there is asequence of polynomials which vanish at 1 and converge uniformly to f on [0, 1].

b) If f is continuous on [0, 1] and∫ 1

0

f(x)(x− 1)kdx = 0, for all k = 1, 2, 3, . . . ,

show that f(x) ≡ 0.

Solution.

a) Let ε > 0.

By the Stone-Weierstrass theorem, there is a polynomial p : [0, 1] → R such that||f − p||∞ < ε/3. We show that the sequence p(x)(1 − xn) converges uniformly tof on [0, 1]. Let pn be the nth member of the sequence. It is clear that each pn is apolynomial which vanishes at 1.

Since f(1) = 0, we know |p(1)| < ε/3. Hence by continuity, there is δ > 0 (and wlog,δ < 1), such that |P (x)| < ε/2 whenever x ∈ [1− δ, 1].

Since p is continuous on [0, 1] it is bounded; say |p(x)| ≤ B for all x ∈ [0, 1]. Let N ∈ Nbe such that (1− δ)n < ε/3B, whenever n ≥ N .

Take n ≥ N . Now for any x ∈ [0, 1], we either have x ∈ [0, 1 − δ] or x ∈ (1 − δ, 1].In the former case, |xnp(x)| ≤ |1− δ|n |p(x)| < B · ε/3B = ε/3. In the latter case,|xnp(x)| ≤ |p(x)| < ε/2. Hence for all x ∈ [0, 1], |xnp(x)| < ε/2 and so ||xnp||∞ ≤ ε/2.

For such n, we see

||f − pn||∞ = ||f − p(1− xn)||∞ ≤ ||f − p||∞ + ||pxn||∞ ≤ ε/3 + ε/2 < ε.

Hence, pn → f uniformly.

b) Suppose pn is a sequence of polynomials vanishing at 1 which converge uniformly tof . Then each pn can be written as a finite number of terms of the form (x− 1)k so∫ 1

0

f(x)(x− 1)kdx = 0, for all k =⇒∫ 1

0

f(x)pn(x)dx = 0, for all n.

By uniform convergence, we can switch limits and integrals, so∫ 1

0

f(x)2dx =

∫ 1

0

(limn→∞

pn(x))f(x)dx = lim

n→∞

∫ 1

0

pn(x)f(x)dx = limn→∞

0 = 0.

Hence ||f ||2 = 0 and so f ≡ 0.

Problem F07.9. Suppose F : R2 → R is continuous and bounded and suppose for eachn ∈ N, un : R→ R is differentiable and solves

u′n(x) = F (un(x), x), x ∈ R.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 23

a) Assume that un → u uniformly. Show that u is differentiable and solves

u′(x) = F (u(x), x).

b) Suppose thatu′(x) = F (u(x), x), u(x0) = y0,

has a unique solution u : R→ R and that un(x0)→ y0 as n→∞. Show that un → uuniformly.

Solution.

a) In particular, all un are continuous and thus u is continuous as the uniform limit ofcontinuous functions. For fixed x0 ∈ R, if we can prove that

u(x) = u(x0) +

∫ x

x0

F (u(t), t)dt,

by continuity of u and F , it will actually be the case that u is differentiable and bythe fundamental theorem of calculus, u′(x) = F (u(x), x). It is clearly true that

un(x) = un(x0) +

∫ x

x0

F (un(t), t)dt.

Let ε > 0. Then for x 6= x0, by continuity of F , there is δ > 0 such that

|F (u, t)− F (v, t)| < ε

3 |x− x0|

when ||u− v||∞ < δ. Without loss of generality, we can take δ < ε/3. Further, byuniform convergence of un → u, there is N ∈ N such that ||u− un||∞ < ε/3 for alln ≥ N . For such n,∣∣∣∣u(x)− u(x0)−

∫ x

x0

F (u(t), t)

∣∣∣∣ =

∣∣∣∣u(x)− un(x) + un(x0)− u(x0) +

∫ x

x0

F (un, t)− F (u, t)dt

∣∣∣∣≤ |u(x)− un(x)|+ |un(x0)− u(x0)|+

∫ x

x0

|F (un, t)− F (u, t)| dt

= ε/3 + ε/3 +

∫ x

x0

ε

3 |x− x0|dt = ε.

Since ε > 0 was arbitrary, this implies that

u(x) = u(x0) +

∫ x

x0

F (u(t), t)dt,

which completes the proof.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 24

b) We simply need to show that un converge uniformly to some function u∗ because thenby the above work, u∗ will satisfy

u′∗(x) = y0 +

∫ x

x0

F (u∗(t), t)dt

and by uniqueness, we will have u ≡ u∗.

Consider, since F is bounded (by, say L > 0), we see

|un(x)− un(y)| =∣∣∣∣∫ x

y

F (un(t), t)dt

∣∣∣∣ ≤ ∣∣∣∣∫ x

y

Ldt

∣∣∣∣ = L |x− y| .

Hence the sequence (un) is equicontinuous.

Further, since un(x0) → y0, there is N ∈ N such that |un(x0)− y0| ≤ 1, n ≥ N . Thenfor any n ≥ N ,

|un(x)| =∣∣∣∣un(x0) +

∫ x

x0

F (un(t), t)dt

∣∣∣∣≤ 1 + |y0|+ L |x− x0|

which shows that (un) is pointwise bounded. The set is also closed by the work frompart a).

Thus by Arzela-Ascoli, there is a subsequence of (un) which converges uniformly onany compact subset of R. The same reasoning would work for any subsequence of (un).Thus (un) converges uniformly on any compact subset of R.

Problem F07.11. Let f be a bounded real function on [0, 1]. Show that f is Riemannintegrable if and only if f 3 is Riemann integrable.

Solution. Let M ∈ R be the bound for f on [0, 1]. Suppose that f is Riemann integrable.Let ε > 0. Then there is a piecewise constant function g which majorizes f (and without

loss of generality, we can take g to be bounded by M as well) for which∫ 1

0(g(x)− f(x))dx <

ε/3M2. Since x3 is non decreasing on R, it follows that g3 majorizes f 3. Further∫ 1

0

(g(x)3 − f(x)3)dx =

∫ 1

0

(g(x)− f(x))(g(x)2 + g(x)f(x) + f(x)2)dx

≤∫ 1

0

(g(x)− f(x))(M2 +M ·M +M2)dx

= 3M2

∫ 1

0

(g(x)− f(x))dx < ε.

Similarly, if we take a piecewise constant h which minorizes f and such that∫ 1

0(f(x) −

h(x))dx < ε/3M2, we find that h3 minorizes f 3 and∫ 1

0

(f(x)3 − h(x)3)dx < ε.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 25

Since ε was arbitrary, this implies that f 3 is Riemann integrable.I can’t figure out an elementary proof in the other direction.[Note: the easiest soluton is as follows. The functions f and f 3 have the same points

of continuity (since x 7→ x3 is a homeomorphism from R to R). A function is Riemannintegrable if and only if it is continuous allmost everywhere. Hence f is Riemann integrableiff f is continuous almost everywhere iff f 3 is continuous almost everywhere iff f 3 is Riemannintegrable.]

Problem S07.6. Consider the integral equation

y(t) = y0 +

∫ t

0

f(s, y(s))ds (∗)

where f(t, y) is continuous on [0, T ] × R and is Lipschitz in y with Lipschitz constant K.Assume you have shown that the iterates defined by

y0(t) ≡ y0, yn(t) = y0 +

∫ t

0

f(s, yn−1(s))ds, t ∈ [0, T ], n ∈ N

converge uniformly to a solution y of (∗). Show that if Y (t) is a solution of (∗) and satisfies|Y (t)− y0| ≤ C for some constant C and all t ∈ [0, T ], then Y (t) ≡ y(t) on [0, T ].

Solution. We prove by induction that |Y (t)− yn(t)| ≤ C(tK)n

n!, t ∈ [0, T ]. The base case:

|Y (t)− y0| ≤C(tK)0

0!= C, t ∈ [0, T ]

is given by assumption. Assume that

|Y (t)− yn(t)| ≤ C(tK)n

n!, t ∈ [0, T ].

Then

|Y (t)− yn+1(t)| =∣∣∣∣y0 +

∫ t

0

f(s, Y (s))ds− y0 −∫ t

0

f(s, yn(s))ds

∣∣∣∣=

∣∣∣∣∫ t

0

f(s, Y (s))− f(s, yn(s))ds

∣∣∣∣≤∫ t

0

|f(s, Y (s))− f(s, yn(s))| ds

≤ K

∫ t

0

|Y (s)− yn(s)| ds

≤ K

∫ t

0

C(sK)n

n!ds =

C(tK)n+1

(n+ 1)!, t ∈ [0, T ].

However, C(tK)n

n!≤ C(TK)n

n!→ 0 as n → ∞. Thus yn converge uniformly to Y on [0, T ].

Hence, since limits are unique, Y (t) = y(t) on [0, T ].

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 26

Problem S07.8. Suppose the functions fn are twice continuously differentiable on [0, 1]and satisfy

limn→∞

fn(x) = f(x), for all x ∈ [0, 1] and

|f ′n(x)| ≤ 1, |f ′′n(x)| ≤ 1, for all x ∈ [0, 1], n ≥ 1.

Prove that f(x) is continuously differentiable on [0, 1].

Solution. Since each fn(x) is differentiable with derivative bounded by 1, we know that forany x, y ∈ [0, 1] by the mean value theorem, there is c ∈ (x, y) such that

|fn(x)− fn(y)| = |f ′(c)| |x− y| ≤ |x− y|

Note, this holds for all n, so this shows that (fn) is equicontinuous. Also, since (fn(0))converges as a sequence in R, in particular, the sequence is bounded. Say |fn(0)| ≤ M ,n ∈ N. Then for all x ∈ [0, 1], by the triangle inequality,

|fn(x)| ≤ |fn(x)− fn(0)|+ |fn(0)| =∣∣∣∣∫ x

0

f ′n(t)dt

∣∣∣∣+M

≤∫ x

0

|f ′n(x)| dx+M ≤∫ x

0

1 dx+M = x+M ≤ 1 +M.

Since this holds for all x ∈ [0, 1], n ∈ N, we see that (fn) is uniformly bounded. Henceby Arzela-Ascoli, there is a subsequence of (fn) which converges uniformly (uniformly since[0, 1] is compact). Since the same reasoning could be applied to any subsequence of (fn), wesee that (fn) converges uniformly. Since the pointwise limit of (fn) is f , it must be the casethat the uniform limit of (fn) is f . Since (fn) is a sequence of continuous function whichconverge uniformly, f must be continuous.

The same reasoning as above could be applied to (f ′n) to show that they converge uni-formly to some continuous function g on [0, 1].

Then since (fn) converges pointwise (in particular) and (f ′n) converges uniformly, weknow that f is differentiable and

f ′(x) = limn→∞

f ′n(x).

Problem S07.10. Suppose the functions (fn) on R satisfy

(i) 0 ≤ fn(x) ≤ 1, for all x ∈ R, n ∈ N,

(ii) fn(x) is an increasing function of x for all n ∈ N,

(iii) limn→∞

fn(x) = f(x) for x ∈ R where f(x) is continuous on R,

(iv) limx→−∞

f(x) = 0 and limx→∞ f(x) = 1.

Show that fn → f uniformly in R.

Solution. Let ε > 0.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 27

By (i) and (ii), f(x) is increasing and contained in [0, 1]. By (iv), there are m,M ∈ Rsuch that

x ≤ m =⇒ f(x) < ε/2

andx ≥M =⇒ f(x) > 1− ε.

We first prove that fn → f uniformly on [m,M ].Since [m,M ] is compact, f is uniformly continuous on [m,M ]. Thus there is δ > 0 such

that for any x, y ∈ [m,M ],

0 < |x− y| < δ =⇒ |f(x)− f(y)| < ε/5.

Cover [m,M ] with intervals [x0, x1], . . . , [xn−1, xn] such that x0 = m,xn = M and eachinterval has length less than δ. Since fn → f pointwise, for each xi, there is Ni ∈ N suchthat n ≥ Ni =⇒ |f(xi)− fn(xi)| < ε/5. Take N = max

0≤i≤nNi. Then for any x ∈ [m,M ],

there is i such that x ∈ [xi, xi+1], taking n ≥ N , we have

|fn(x)− f(x)| = |fn(x)− fn(xi) + fn(xi)− f(xi) + f(xi)− f(x)|= |fn(x)− fn(xi)|+ |fn(xi)− f(xi)|+ |f(xi)− f(x)|< (fn(x)− fn(xi)) + ε/5 + ε/5

< fn(xi+1)− fn(xi) + 2ε/5.

But fn(xi+1) is at most f(xi+1) + ε/5 and fn(xi) is at least f(xi)− ε/5. Then

|fn(x)− f(x)| < f(xi+1) + ε/5− (f(xi)− ε/5) + 2ε/5

= (f(xi+1)− f(xi)) + 4ε/5

< ε/5 + 4ε/5 = ε.

Since δ,N don’t depend on x, this implies that fn → f uniformly on [m,M ].For x ≤ m and n ≥ N , we have

|fn(x)− f(x)| ≤ fn(m)− 0 ≤ f(m)− ε/5 < ε/2 + ε/5 < ε

and for x ≥M and n ≥ N , we have

|fn(x)− f(x)| ≤ 1− f(M) < 1− (1− ε) < ε

so fn → f uniformly on all of R.

Problem S07.11.

(a) Consider the equations

u3 + xv − y = 0, v3 + yu− x = 0.

Can these equations be solved uniquely for u, v in terms of x, y in a neighborhood ofx = 0, y = 1, u = 1, v = −1?

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 28

(b) Give an example where the conclusion of the implicit function theorem is true but theconditions are not met.

Solution.

(a) Let f = (f1, f2) : R4 → R2 be defined

f(u, v, x, y) =

[f1(u, v, x, y)f2(u, v, x, y)

]=

[u3 + xv − yv3 + yu− x

], (u, v, x, y) ∈ R4.

We see f(1,−1, 0, 1) = (0, 0). Also at (1,−1, 0, 1), we have

det

([∂f1∂x

∂f1∂y

∂f2∂x

∂f2∂y

])= det

([−1 −1−1 1

])= −2 6= 0.

Thus the matrix is invertible. By the implicit function theorem, there is a open neigh-borhood U of (1,−1), an open neighborhood V of (0, 1) and a continuously differen-tiable map g = (g1, g2) : U → V such that

(u, v, g1(u, v), g2(u, v)) : (u, v) ∈ U = (u, v, x, y) : f(u, v, x, y) = (0, 0).

That is, we can uniquely for x, y in terms of u, v in a neighborhood of (1,−1, 0, 1). (Idid it backwards, but do that same thing but using the u, v derivatives to form thematrix to see the answer is yes.)

(b) Let F (x, y) = x3 − y and consider solving F (x, y) = 0 in a neighborhood of (0, 0). Weknow this has unique solution x = y1/3. However, ∂F

∂x= x2 = 0 at x = 0 which is not

invertible, so the conditions of the implicit function theorem are not met.

Problem S07.12. Let c0 be the normed space of real sequences x = (x1, x2, . . .) such thatlimk→∞

xk = 0 with the norm ||x|| = supk|xk|.

(a) Show that c0 is complete.

(b) Is the unit ball x ∈ c0 : ||x|| ≤ 1 compact?

(c) Is the set E =

x ∈ c0 :

∞∑k=1

k |xk| ≤ 1

compact?

Solution.

(a) Let (x(m))m∈N be a Cauchy sequence in c0. Let ε > 0, then for each ` ∈ N,∣∣∣x(m)` − x(n)

`

∣∣∣ ≤ ∣∣∣∣x(m) − x(n)∣∣∣∣ < ε

for sufficiently large m,n. This shows that for each ` ∈ N, the sequence (xm` )m∈N is aCauchy sequence in R and thus coverges to some x` ∈ R. Then (x(m)) converges term-by-term to x = (x1, x2, . . .) and so converges in norm. We need only check that x ∈ c0.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 29

For this, consider, for ε > 0, since lim`→∞ x(m)` = 0 for each m and limm→∞ x

(m) = x,there is N ∈ N so that `,m ≥ N implies that∣∣∣x` − x(m)

`

∣∣∣ < ε and∣∣∣x(m)`

∣∣∣ < ε.

For such m, `, we have

|x`| ≤∣∣∣x` − x(m)

`

∣∣∣+∣∣∣x(m)`

∣∣∣ < 2ε

which implies that lim`→∞ x` = 0 and so x ∈ c0.

(b) The unit ball is not compact. Consider a sequence (x(m)) where x(m)k = 1, k ≤ m and

x(m)k = 0, k ≥ m. Each member of the sequence is eventually constant at zero and so

has limit zero. Also each has∣∣∣∣x(m)

∣∣∣∣ = 1 so each member of the sequence is in theunit ball. However, the sequence has no limit in c0 since the limiting sequence is onethat does not go to zero.

(c) Put c∗0 = x ∈ c0 :∑

k k |xk| ≤ 1. We prove that c∗0 is sequentially compact and thuscompact. Suppose that

(x(n))n∈N be a sequence in c∗0. Then since

∞∑k=1

k∣∣∣x(n)k

∣∣∣ ≤ 1, for all n ∈ N,

we see that in particular, that∣∣∣x(n)k

∣∣∣ ≤ 1k

for all n, k ∈ N. Thus, by the Bolzano-

Weierstrass Theorem, for all k ∈ N, the sequence(x

(n)k

)n∈N

. has a convergent subse-

quence. Let(x

(n,1)1

)n∈N

be a convergent subsequence of(x

(n)1

)n∈N

with

limn→∞

x(n,1)1 = x1 ∈ R.

Then(x

(n,1)2

)n∈N

is a bounded sequence and so has a convergent subsequence(x

(n,2)2

)n∈N

withlimn→∞

x(n,2)2 = x2 ∈ R.

Then we still havelimn→∞

x(n,2)1 = x1 ∈ R.

Continuing this process inductively, for each m ∈ N, we get subsequence(x(n,m)

)n∈N

such that(x(n,m)

)n∈N is a subsequence of

(x(n,m−1)

)n∈N and for all k ≤ m, we get

limn→∞

x(n,m)k = xk.

Put y(n) = x(n,n), n ∈ N and x = (x1, x2, . . .). Then(y(n))n∈N is a subsequence of(

x(n))n∈N. We prove that y(n) → x and that x ∈ c∗0.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 30

First, it is clear thatlimn→∞

y(n)k = xk

or all k ∈ N. Since for each n ∈ N, we have

∞∑k=1

k∣∣∣y(n)k

∣∣∣ ≤ 1.

Thus

limn→∞

∞∑k=1

k∣∣∣y(n)k

∣∣∣ ≤ 1.

We can switch the sum and the limit since all terms in the sum are positive, so

∞∑k=1

k limn→∞

∣∣∣y(n)k

∣∣∣ ≤ 1 =⇒∞∑k=1

k |xk| ≤ 1 =⇒ x ∈ c∗0.

Finally, since x ∈ c∗0, we have |xk| ≤ 1k

for all k ∈ N (and similary∣∣∣y(n)k

∣∣∣ ≤ 1k

for all

k ∈ N). Let ε > 0 and choose N ∈ N with 1N< ε

4. Then for each k = 1, . . . , N there is

Mk ∈ N such that for n ≥Mk, we have∣∣∣y(n)k − xk

∣∣∣ < ε

2.

put M = maxM1, . . . ,MN.Take n ≥M . Then for k ≤ N , we have∣∣∣y(n)

k − xk∣∣∣ < ε/2

while for k > N , we have∣∣∣y(n)k − xk

∣∣∣ ≤ ∣∣∣y(n)k

∣∣∣+ |xk| ≤1

k+

1

k<

1

N+

1

N<ε

2.

Thus∣∣∣y(n)k − xk

∣∣∣ < ε/2 for all k ∈ N and so supk

∣∣∣y(n)k − xk

∣∣∣ ≤ ε/2 < ε. Since this holds

for all n ≥M , we see that y(n) → x in norm.

Hence c∗0 is sequentially compact and thus compact.

[Note: for those interested, switching the sum and the limit when all terms are positiveis an application of the Monotone Convergence Theorem when integration is takenwith respect to the counting measure on N and the sequences are viewed as functionsx : N→ R.]

Problem F08.1. For which a = 0, 1, 2 is f(t) = ta uniformly continuous on [0,∞)?

Solution. The function is uniformly continuous when a = 0, 1 but not when a = 2.For a = 0, this is trivial, since f(t)− f(s) = 0 for all t, s ∈ R.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 31

For a = 1, let ε > 0 and set δ = ε which does not depend on s, t ∈ R. Then

0 < |s− t| < δ =⇒ |s− t| < ε =⇒ |f(s)− f(t)| < ε,

which implies that f(t) = t is uniformly continuous.For a = 2, take ε = 1. Then for any δ > 0, we can take n ≥ 1/δ and take s = n, t =

n+ δ/2. Then

|f(s)− f(t)| =∣∣s2 − t2

∣∣ = |s+ t| |s− t| = (2n+ δ/2)(δ/2) = nδ + δ2/4 > nδ > 1.

Thus f(t) = t2 is not uniformly continuous.

Problem F08.2. Suppose A is a non-empty connected subset of R2.

(a) Prove that if A is open then A is path-connected.

(b) Is the same true if A is closed?

Solution.

(a) Since A is non-empty, choose x ∈ A and let B ⊆ A be the set of all elements of Awhich can be connected to x by a path lying entirely in A.

Clearly B is non-empty since x ∈ B. Take y ∈ B. Then y ∈ A and since A is open,there is an ε > 0 such that B(y, ε) ⊆ A. But for any z ∈ B(y, ε), we can connect z bya path to y which remains in A and then continue to x remaining in A. Thus z ∈ Band so B(y, ε) ⊆ B so B is open.

Now assume that A− B is non-empty. Taking y ∈ A− B,there is an r > 0 such thatthe ball B(y, r) ⊆ A−B because otherwise, there would be z ∈ B(y, r) which could beconnected to x by a path lying in A which could then be extended to a path connectinx to y. But y ∈ B − A procludes the existence of such path. Thus B(y, r) ⊂ A − Band A−B is open.

But now we see A = B ∪ (A − B) and clearly B ∩ (A − B) = ∅ and we have proventhat B,A − B are open. This implies that A is disconnected, a contradiction. Thusour assumption that A−B is non-empty must be incorrect. Hence A−B = ∅.

Thus A = B, so all points in A can be connected to x and so to each other. Thismeans that A is path-connected.

(b) No. It is well known that the set

A =(x, sin 1

x

): x ∈

(0, 1

π

]∪ (0, x) : x ∈ [−1, 1] ..= B ∪ C

is connected. If A was path-connected, then there is a continuous function f : [0, 1]→A such that f(0) = ( 1

π, 0) ∈ B and f(1) = (0, 0) ∈ C. Define

s = inft∈[0,1]

t : f(t) ∈ C .

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 32

Then f([0, s]) can contain only one point in C, otherwise the minimality of s is violated.However, f([0, s]) will contain all points in B and thus f([0, s]) contains all of C. Thusf([0, s]) is not closed (since it doesn’t equal its closure) and not compact. But [0, s]is compact and continuous functions map compact sets to compact sets so we have acontradiction. Thus A is not path-connected.

Problem F08.4. Suppose that K and F are closed subsets of R2 with K ∩ F = ∅.

(a) Prove that if K is bounded, then d(K,F ) > 0 where

d(K,F ) = inf d(x, y) : x ∈ K, y ∈ F .

(b) Is (a) necessarily true if K is not bounded?

Solution.

(a) Assume to the contrary that d(K,F ) = 0. Then for any ε > 0, there are x ∈ K, y ∈ Fsuch that d(x, y) < ε. In particular, for any n ∈ N, there are xn ∈ K, yn ∈ F suchthat d(xn, yn) < 1

n. Since K is bounded, by Bolzano-Weierstrass, the sequence (xn) has

a subsequence (xnm) which converges to some x ∈ R2 and since K is closed, x ∈ K.We show that (ynm) also converges to x. Let ε > 0. Then there is M ∈ N, such thatd(xnm , x) < ε/2. Taking m large enough so that nm > maxM, 2/ε. We see

d(x, ynm) ≤ d(x, xnm) + d(xnm , ynm) < ε/2 + 1nm

< ε/2 + ε/2 = ε.

Thus (ynm) converges to x as well and since F is closed, x ∈ F . But then K ∩ F 6= ∅as was assumed. The contradiction implies that d(K,F ) > 0.

(b) No. Let F = (x, 0) : x ∈ R, K =(x, 1

x

): x > 0

. Then it is clear that K,F are

closed and for any ε > 0, taking x > 1/ε gives existence of points p ∈ K, q ∈ F suchthat d(p, q) < ε. Letting ε→ 0 shows that d(K,F ) = 0.

Problem F08.10. Given v = (v1, . . . , vn) ∈ Rn, we let

||v|| =

√√√√ n∑i=1

v2i .

If f = (f1, . . . , fn) : [a, b]→ Rn, we define∫ b

a

f(t)dt =

(∫ b

a

f1(t)dt, . . . ,

∫ b

a

fn(t)dt

).

Prove that ∣∣∣∣∣∣∣∣∫ b

a

f(t)dt

∣∣∣∣∣∣∣∣ ≤ ∫ b

a

||f(t)|| dt.

Solution. If∣∣∣∣∣∣∫ ba f(t)dt

∣∣∣∣∣∣ = 0 then the inequality is trivial and we are done. Assume that∣∣∣∣∣∣∫ ba f(t)dt∣∣∣∣∣∣ > 0.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 33

For convenience, put xi =∫ bafi(t)dt and x = (x1, . . . , xn) so that x =

∫ baf(t)dt. We see

||x||2 =

∣∣∣∣∣∣∣∣∫ b

a

f(t)dt

∣∣∣∣∣∣∣∣2 =n∑i=1

(∫ b

a

fi(t)dt

)2

=n∑i=1

xi

∫ b

a

fi(t)dt

=

∫ b

a

(n∑i=1

xifi(t)

)dt

≤∫ b

a

||x|| ||f(t)|| dt

= ||x||∫ b

a

||f(t)|| dt

=

∣∣∣∣∣∣∣∣∫ b

a

f(t)dt

∣∣∣∣∣∣∣∣ ∫ b

a

||f(t)|| dt,

where the inequality incurred follows from Cauchy-Schwarz. Dividing by∣∣∣∣∣∣∫ ba f(t)dt

∣∣∣∣∣∣ gives∣∣∣∣∣∣∣∣∫ b

a

f(t)dt

∣∣∣∣∣∣∣∣ ≤ ∫ b

a

||f(t)|| dt.

Problem S08.2. Let (fn) be a sequence on functions on [0, 1] such that fn(x) ≥ 0 for alln ∈ N, x ∈ [0, 1] and for each x ∈ [0, 1],

limn→∞

fn(x) = 0.

Prove or give a counterexample to the assertion that

limn→∞

∫ 1

0

fn(x)dx = 0.

Solution. We give a counterexample. For each n ∈ N, n ≥ 3, define

fn(x) =

0, x < 1

n,

n2(x− 1

n

), 1

n≤ x < 2

n,

n2(

3n− x), 2

n≤ x < 3

n,

0, 3n≤ x.

Now fn(0) = 0, for all n ∈ N so limn→∞ fn(0) = 0 and for any x ∈ (0, 1], there is Nx ∈ N,such that 3/Nx < x. Hence for all n ≥ Nx, fn(x) = 0 and so limn→∞ fn(x) = 0. Thuslimn→∞ fn(x) = 0 for all x ∈ [0, 1].

However, each fn is a triangular spike of height n and base 2/n. This triangle has area1. Thus for each n ∈ N, we have∫ 1

0

fn(x)dx = 1 and so limn→∞

∫ 1

0

fn(x)dx = 1.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 34

Problem S08.3. Assuming f ∈ C4[a, b] is real, derive a formula for the error of approxi-mation E(h) when the second derivative is replaced by the finite difference formula

f ′′(x) ≈ f(x+ h)− 2f(x) + f(x− h)

h2

where h is the mesh size. (It is alright to assume that x− h, x, x+ h ∈ (a, b)).

Solution. By Taylor’s Theorem,

f(x± h) = f(x)± hf ′(x) + h2f′′(x)

2± h3f

′′′(x)

6+ h2f

(4)(ξ)

24

for some ξ between x and x ± h. Let K = supx∈[a,b]

∣∣f (4)(x)∣∣ . The supremum is finite (and

actually it is achieved for some x0 ∈ [a, b]) since f (4) is continuous and [a, b] is compact.Then

f(x+ h) + f(x− h) = 2f(x) + h2f ′′(x) + h4f(4)(ξ1) + f (4)(ξ2)

24≤ 2f(x) + h2f ′′(x) + h4K

12,

for some ξ1, ξ2 ∈ (x− h, x+ h). Then∣∣∣∣f(x+ h)− 2f(x) + f(x− h)

h2− f ′′(x)

∣∣∣∣ ≤ h2K

12=

supx∈[a,b]

∣∣f (4)(x)∣∣

12h2.

Thus the error E(h) is O(h2) [the exact formula can be taken to be the one above, thoughthis is actually a bound on the error].

Problem S08.6. Let Y be a complete and countable metric space. Prove there is y ∈ Ysuch that y is open.

Solution. Assume to the contrary that y is not open for all y ∈ Y . In any metric space(and more generally, any Hausdorff space), singleton sets are closed, so y = y and sincethe only subsets thereof are ∅ and y we conclude that the interior of y = y, which isthe largest open subset, is ∅. That is, if no singleton is open, then for each y ∈ Y , the sety has empty interior and thus y is nowhere dense. Then

Y =⋃y∈Y

y

is a countable union of closed no-where dense sets and since Y is complete, the Baire Cate-gory Theorem (metric space version) implies that Y is nowhere dense. But then since Y isboth open and closed we have Y = int(Y ) = ∅ which is an impossibility. Thus some y isopen.

Problem S08.7. Let a(x) be a function on R such that

(i) a(x) ≥ 0 for all x ∈ R, and

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 35

(ii) there exists M <∞ such that for all finite F ⊂ R,∑x∈F

a(x) ≤M.

Prove that A = x : a(x) > 0 is countable.

Solution. Without loss of generality, take M ∈ N. Define

An =x : a(x) > 1

n

, n = 1, 2, 3, . . .

so thatA =

⋃n

An.

Suppose that An has more that Mn elements. Then there is a finite subset F ⊂ An with atleast Mn+ 1 elements. This gives∑

x∈F

a(x) ≥∑x∈F

1

n≥ Mn+ 1

n= M +

1

n, which contradicts (ii).

Thus each An has less then Mn elements. Then A is countable since it is a countable unionof finite sets.

Problem F09.1.

(i) For each n ∈ N, let fn : N→ R be a function with |fn(m)| ≤ 1 for all m,n ∈ N. Provethat there is an infinite subsequence of positive integers (ni) such that for each m ∈ N,(fni(m))∞i=1 converges.

(ii) For ni as in (i), assume that in addition limm→∞

limi→∞

fni(m) = 0. Prove or disprove that

limi→∞

limm→∞

fni(m) = 0.

Solution.

(i) Let X = f : N→ [−1, 1] and equip X with the metric

d(f, g) = supm∈N|f(m)− g(m)| , f, g ∈ X.

Then d(f, g) ≤ 2 for all f, g ∈ X so X is totally bounded. Suppose (fk) is a Cauchysequence in X. Then for each m ∈ N, (fk(m)) is a Cauchy sequence in [−1, 1] and thusconverges to some value am ∈ [−1, 1]. Define f : N→ [−1, 1] by f(m) = am. Now forany ε > 0, m ∈ N, there is K ∈ N such that

|fk(m)− f(m)| < ε/2 for all k ≥ K.

Taking the supremum over all m gives d(fk, f) ≤ ε/2 < ε. Thus (fk) converges to f .Hence X is complete. Since X is complete and totally bounded, it is compact and thussequentially compact.

Let (fn) be a sequence in X. Then by sequential compactness, there is a subsequence(fni) which converges in the metric. It follows that (fni(m)) converges for each m ∈ N.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 36

(ii) No. Define fni(m) = 1,m > ni and fni(m) = 0,m ≤ ni. It’s clear that (fni) convergesto a function f which maps all natural numbers to zero, so f ∈ X. Then

limm→∞

limi→∞

fni(m) = 0 whereas limi→∞

limm→∞

fni(m) = 1.

Problem F09.2.

(i) Let X be a complete metric space with respect to the distance function d. We say thata map T : X → X is a contraction if for some λ ∈ (0, 1) and all x, y ∈ X:

d (T (x), T (y)) ≤ λd(x, y).

Prove that any contraction mapping on X has a unique fixed point.

(ii) Using (i), show that given a differentiable function f : R → R whose first derivativesatisfies f ′(x) = e−x

2 − e−x4 , x ∈ R, there exists α ∈ R such that f(α) = α.

Solution.

(i) Let ε > 0.

Let x0 ∈ X and define a sequence (xn) by xn = Txn−1, n = 1, 2, 3, . . . . We provethat (xn) is Cauchy. Indeed d(x2, x1) = d(Tx1, Tx0) ≤ λd(x1, x0). Further, d(x3, x2) =d(Tx2, Tx1) ≤ λd(x2, x1) ≤ λ2d(x1, x0). Indeed, by induction d(xn+1, xn) ≤ λnd(x1, x0).Let m,n ∈ N, n > m. Then

d(xn, xm) ≤ d(xn, xn−1) + d(xn−1, xn−2) + · · ·+ d(xm+1, xm)

≤ (λn−1 + λn−2 + · · ·+ λm)d(x1, x0)

≤ d(x1, x0)∞∑j=m

λj

= d(x1, x0)λm

1− λ.

But 0 < λ < 1 means that λm → 0 as m → ∞. Take N ∈ N such that d(x1,x0)λm

1−λ < εwhen m ≥ N . Then for n,m ≥ N , we have d(xn, xm) < ε. Thus (xn) is Cauchy andsince X is complete, (xn) converges to some x ∈ X. Then by continuity of T (whichfollows from the contraction property), we have

x = limn→∞

xn = limn→∞

Txn−1 = T(

limn→∞

xn−1

)= Tx.

To prove uniqueness, we consider x, y ∈ X such that Tx = x, Ty = y. Then

d(x, y) = d(Tx, Ty) ≤ λd(x, y) =⇒ (1− λ)d(x, y) ≤ 0 =⇒ d(x, y) ≤ 0

which implies that x = y.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 37

(ii) Let f : R→ R have derivative f ′(x) = e−x2 − e−x4 . Then for x ∈ [0, 1], we have

|f ′(x)| =∣∣∣e−x2 − e−x4∣∣∣ = e−x

4 − e−x2 ≤ e−x4 − e−1 ≤ e0 − e−1 = 1− 1

e< 1.

Also, for x ∈ [1,∞),

|f ′(x)| =∣∣∣e−x2 − e−x4∣∣∣ = e−x

2 − e−x4 ≤ e−x2 ≤ 1

e≤ 1− 1

e.

Thus for any x, y ∈ Rx 6= y, by the mean value theorem, there is c between x and ysuch that

|f(x)− f(y)| = |f ′(c)| |x− y| ≤(1− 1

e

)|x− y| .

Thus f is a contraction on R. Since R is complete with respect to the absolute value,by the above proof, f has a unique fixed point.

Problem F09.3. The purpose ofthis problem is to give a multi-variable calculus proof ofthe geometric mean-arithmetic mean inequality.

(i) Let R+n ⊂ Rn be the subset of vectors all of whom components are positive and let

f : R+n → R be defined by

f(x1, . . . , xn) = x1 + · · ·+ xn +1

x1 · · ·xnExplain why f attains a global (not necessarily unique) minimum as some p ∈ R+

n .

(ii) Find p.

(iii) Deduce that if all xi ∈ R are positive and∏xi = 1 then

∑xi ≥ n.

Solution.

(i) We see thatlimxi→0

f(x1, . . . , xn) = limxi→∞

f(x1, . . . , xn) =∞,

for any i = 1, . . . , n. Further f is continuous so it will attain a minimum on any set ofthe form [1/L, L]n, L > 1 since the set is compact. Fix M > 0 and let L > 1 be suchthat

f(x1, . . . , xn) > M

outside [1/L, L]. Then f attains a minimum (say m) inside [1/L, L] and is larger thanM outside [1/L, L]. Thus minm,M will be a global minimum for f .

(ii) Setting ∂f∂xj

= 0 for all i simultaneously, we see that

1− 1

x1 · · ·x2j · · ·xn

= 0 =⇒ xj =1∏n

j=1 xj.

Thenn∏j=1

xj =1(∏n

j=1 xj

)n =⇒

(n∏j=1

xj

)n+1

= 1.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 38

But since all xj are positive, we conclude that the minimum must occur along thehypersurface

n∏j=1

xj = 1.

But then xj = 1∏nj=1 xj

= 1. Thus the minimum occurs at (1, . . . , 1).

(iii) We see that for all x = (x1, . . . , xn) ∈ Rn+ such that

∏nj=1 xj = 1,

1 +n∑j=1

xj = f(x1, . . . , xn) ≥ f(1, . . . , 1) = n+ 1

which implies the desired inequality.

Problem F09.8. For a matrix A ∈Mn(R), define eA =∑∞

k=0Ak

k!. Let v0 ∈ Rn. Prove that

the function v : R→ Rn given by v(t) = eAtv0 solves the differential equation

v′(t) = Av(t)

v(0) = v0.

Solution. It is clear that v(0) = v0. Note that the matrices At and Ah commute for anyt, h ∈ R. Thus eA(t+h) = eAteAh. Then

v′(t) = limh→0

eA(t+h)v0 − eAtv0

h

= limh→0

eAh − Ih

eAtv0.

Further

eAh − I = hA+h2A2

2+h3A3

6+ · · · .

SoeAh − I

h= A+ o(h).

Hence

v′(t) = limh→0

eAh − Ih

eAtv0 = AeAtv0 = Av(t).

Problem F09.10.

(a) If f : I → R is a continuous function such that∫ 2

0f(x)dx = 36, prove that there is an

x ∈ [0, 2] such that f(x) = 18.

(b) Let g : [0, 2] × [0, 2] → R be a continuous function such that∫ 2

0

∫ 2

0g(x, y)dx dy = 36.

Show that there is an (x, y) ∈ I2 such that g(x, y) = 9.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 39

Solution. We can answer both these in one go. Suppose that there is no x ∈ [0, 2]and no (x, y) ∈ [0, 2] × [0, 2] such that f(x) = 18 or g(x, y) = 9. Then by continu-ity (and the intermediate value theorem), we have (wlog) f(x) > 18 for all x ∈ [0, 2]

and g(x, y) > 9 for all x ∈ [0, 2] × [0, 2]. But then∫ 2

0f(x)dx >

∫ 2

018 dx = 36 and∫ 2

0

∫ 2

0g(x, y)dx dy >

∫ 2

0

∫ 2

09 dx dy = 36. Both of these are contradictions with the assump-

tions, thus we do have such x and (x, y).

Problem S09.4. Let (X, d) be a metric space.

(a) Give a definition of compactness of X using open covers.

(b) Define completeness of X.

(c) Define connectedness of X.

(d) Is Q connected with the usual metric on R?

(e) Suppose X is complete. Prove that X is compact if and only if for any r > 0, X canbe covered by finitely many balls of radius r.

Solution.

(a) X is compact if every open cover of X admits and finite subcover.

(b) X is complete if every Cauchy sequence in X converges to a limit in X.

(c) X is connected if there is no decomposition X = A∪B with non-empty, open A,B ⊆ Xand A ∩B = ∅.

(d) No, Q is not connected. We find just such a decomposition. Let A = Q ∩ (−∞, π)and B = Q ∩ (π,∞). Then clearly A,B are nonempty, Q = A ∪ B and A ∩ B = ∅.Further, for x ∈ A, the rationals in the interval (x− π, π) is an open neighborhood ofx contained in A so A is open. B is also open by an identical arguement. Thus A,Bform a decomposition of Q which proves that Q is disconnected.

(e) Suppose X is complete and compact. Let r > 0. Then B(x, r)x∈X forms an opencover of X. By compactness, there is a finite subcover, B(x1, r), . . . , B(xk, r). ThenX is covered by finitely many balls of radius r.

Suppose X is complete and for any r > 0, X can be covered by finitely many balls ofradius r. We prove that X is sequentially compact which implies that X is compactin the sense of covers. Let (xn) be a sequence in X. Cover X with finitely many ballsof radius 1/2. Then one of the balls must contain infinitely many members of thesequence (xn). Let (x2,n) be a subsequence of (xn) which are all contained in a ballof radius 1/2. Cover this ball with finitely many balls of radius 1/3. Then infinitelymany members of (x2,n) will be contained in a single ball of radius 1/3. Cover this ballwith finitely many balls of radius 1/4 and repeat the process. By induction, we havesequences (xk,n)∞n=1 such that each (xk,n) is a subsequence of (xk−1,n) and each (xk,n)is contained in a ball of radius 1/k. Letting yn = xn,n, n ∈ N, we have a sequence (yn).

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 40

Then (yn)∞n=1 is a subsequence of (xn) and (yn)∞n=k is a subsequence of (xk,n) for eachk ∈ N. Let ε > 0 and choose k ∈ N so that 2/k < ε. Then by construction of (xk,n),for any m,n ≥ k,

d(ym, yn) < 2k< ε.

Thus (yn) is a Cauchy sequence and thus converges in X since X is complete. Thusevery sequence (xn) has a subsequence that converges and so X is sequentially compact.

Problem S09.6. Show that a continuous function f : [0, 1)→ R is uniformly continuous ifand only if there is a continuous function g : [0, 1]→ R such that f(x) = g(x) for all x ∈ [0, 1).

Solution. If such a g exists, then it is uniformly continuous, since it is continuous on acompact set. Thus f = g|[0,1) will be uniformly continuous as well.

Conversely, suppose f is uniformly continuous on [0, 1). Define xn = 1− 1n, n ∈ N. Then

|xn − xm| =∣∣∣∣ 1

m− 1

n

∣∣∣∣ ≤ max

1

m,

1

n

→ 0

as m,n → ∞. Thus (xn) is a Cauchy sequence. Let ε > 0. Then by uniform continuity,there is δ > 0 such that

0 < |x− y| < δ =⇒ |f(x)− f(y)| < ε.

Since (xn) is Cauchy, there is N ∈ N such that m,n ≥ N will give |xn − xm| < δ. For suchm,n, we have

|xn − xm| < δ =⇒ |f(xn)− f(xm)| < ε.

Thus (f(xn)) is a Cauchy sequence and has a limit in R. Say limn→∞

f(xn) = f1 ∈ R.

Define g : [0, 1] → R so that g(x) = f(x), x ∈ [0, 1) and g(1) = f1. We prove that g iscontinuous. It is clear that g is continuous on [0, 1) since it is equal to a continuous functionthere. Let (yn) be any sequence in [0, 1) converging to 1. Then

|g(yn)− f1| ≤ |g(yn)− f(xn)|+ |f(xn)− f1| ,

where xn is as defined above. However, since yn ∈ [0, 1), g(yn) = f(yn) and since xn and yngo to the same limit, they get arbitrarily close. That is, there is N ∈ N so that n ≥ N gives|yn − xn| < δ which then leads to |f(yn)− f(xn)| < ε. Also there is also N ∈ N such thatn ≥ N gives |f(xn)− f1| < ε. Taking the max of each N , we see

|g(xn)− f1| < 2ε

which proves that g(xn) → f1. Hence by the sequential criterion theorem, limx→1− g(x) =f1 = g(1) so g is continuous.

Problem S09.10.

(a) Rigorously justify the limit:∫ 1

0

1

1 + x2dx = lim

N→∞

N∑n=0

(−1)n

2n+ 1.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 41

(b) Deduce the value of 1− 13

+ 15− 1

7+ · · · .

Solution.

(a) Consider, for x ∈ [0, 1),

1

1 + x2=∞∑n=0

(−x2

)n=∞∑n=0

(−1)nx2n.

Moreover, the convergence is uniform on any interval [0, t], t < 1. Thus∫ t

0

1

1 + x2dx =

∫ t

0

(∞∑n=0

(−1)nx2n

)dx =

∞∑n=0

(−1)n∫ t

0

x2ndx =∞∑n=0

(−1)nt2n+1

2n+ 1

By the alternating series test, the series

∞∑n=0

(−1)n

2n+ 1

converges and thus by Abel’s Theorem,∫ 1

0

1

1 + x2dx = lim

t→1−

∫ t

0

1

1 + x2dx

= limt→1−

∞∑n=0

(−1)nt2n+1

2n+ 1

=∞∑n=0

limt→1−

(−1)nt2n+1

2n+ 1

=∞∑n=0

(−1)n

2n+ 1.

(b) Evaluating the integral, we see

∞∑n=0

(−1)n

2n+ 1=

∫ 1

0

1

1 + x2dx = arctan(x)

∣∣∣10

= arctan(1) =π

4,

which was the desired deduction.

Problem S09.12. Let F : R3 → R3 and ρ : R3 → R be smooth functions. Show that

div(F ) = ρ

for all (x, y, z) ∈ R3 if and only if∫∫∂Ω

F · dS =

∫∫∫Ω

ρ dx dy dz

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 42

for all open balls Ω ⊂ R3.[You may use without proof the various standard theorems of vector calculus.]

Solution. The Divergence theorem tells us that if F : R3 → R3 is smooth and D is a simplesolid region with smooth boundary ∂D, then∫∫

∂D

F · dS =

∫∫∫D

divF dx dy dz.

Assume ρ = divF . Then the divergence theorem immediately implies that∫∫∂Ω

F · dS =

∫∫∫Ω

ρ dx dy dz

for any open ball Ω ⊂ R3.

Conversely, assume that ∫∫∂Ω

F · dS =

∫∫∫Ω

ρ dx dy dz

for any open ball Ω ⊂ R3. Then∫∫∫Ω

divF dx dy dz =

∫∫∫Ω

ρ dx dy dz =⇒∫∫∫

Ω

Gdxdy dz = 0

for any open ball Ω ⊂ R3 where G ..= divF − ρ. Assume that G 6≡ 0 on R3. Then there isa point x ∈ R3 such that (wlog) G(x) = ε > 0. By countinuity (since G is smooth), there isδ > 0 such that

G(v) > ε/2 whenever ||x− v|| < δ.

Now B(x, δ) ⊂ R3 is certainly an open ball, but∫∫∫B(x,δ)

Gdxdy dz >

∫∫∫B(x,δ)

ε

2dx dy dz =

2πεδ3

3> 0.

This contradicts that G integrates to zero over any open ball in R3. Thus G ≡ 0 so ρ = divFon all of R3.

Problem F09.6. Consider the function f(x, y) = sin3(xy)+y2 |x| defined on S ⊂ R2 givenby

S = (x, y) ∈ R2 : x2010 + y2010 ≤ 1.

Define what it means for f to be uniformly continuous on S and prove that f is uniformlycontinuous on S.

Solution. f is uniformly continuous on S if for every ε > 0 there is δ > 0 such that for anyu, v ∈ S,

||u− v|| < δ =⇒ |f(u)− f(v)| < ε.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 43

Here ||·|| can be any norm on R2 restricted to S but is usually taken to be the Euclideannorm.

It is clear that f is continuous on S since f is composed of functions which are continuouseverywhere. We recall that continuous functions on compact spaces are uniforly continuous.Hence it is enough to show that S is compact. A subset of R2 is compact if and only if it isclosed and bounded. S is bounded, since for example, |x| , |y| ≤ 1. We need only show thatS is closed. Let (x, y) be a limit point of S. Then there is a sequence (xn, yn) in S whichconverges to (x, y). Then

0 ≤ x2010n + y2010

n ≤ 1

for all n ∈ N. Then x2010n + y2010

n → x2010 + y2010 (this follows since F (x, y) = x2010 + y2010

is continuous). But [0, 1] is compact, so it contains its limit points which means thatx2010 + y2010 ∈ [0, 1] and hence (x, y) ∈ S. Thus S is closed and so f is uniformly con-tinuous on S.

Problem F10.3. Suppose that f : R→ R and g : R2 → R have continuous derivatives upto order 3.

(a) State Taylor’s Theorem with remainder for f and g.

(b) Assuming the theorem for f , prove the theorem for g.

Solution.

(a) Let x, a ∈ R. Taylor’s Theorem for f says that there is ξ in between x and a such that

f(x) = f(a) + f ′(a)(x− a) +f ′′(a)

2!(x− a)2 +

f ′′′(ξ)

3!(x− a)3.

Let x, a ∈ R2. Define

Hg(x) =

[∂2g∂x2

(x) ∂2g∂x∂y

(x)∂2g∂y∂x

(x) ∂2g∂y2

(x)

]where x, y are the coordinates of x. Then Taylor’s Theorem for g says that

g(x) = g(a) +∇g(a) · (x− a) + a ·Hg(x)a + E(x)

where E(x) → 0 faster than ||x− a||22 as x → a. If x = (x, y) and a = (a, b) it issometimes convenient to write this in the form

g(x, y) = g(a, b) +∂g

∂x(a, b)(x− a) +

∂g

∂y(a, b)(y − b)

+1

2

(∂2g

∂x2(a, b)(x− a)2 + 2

∂2g

∂x∂y(a, b)(x− a)(y − b) +

∂2g

∂y2(a, b)(y − b)2

)+ E(x, y).

We note that this is equivalent because mixed partials of g are equal since g is threetimes continuous differentiable.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 44

(b) Let g : R2 → R be three times continuously differentiable. For fixed (x, y), (a, b) ∈ R2,define f : R→ R by

f(t) = g(a+ t(x− a), b+ t(y − b)).

Then f is also three times continuously differentiable, so

f(t) = f(0) + f ′(0)t+f ′′(0)

2!t2 +

f ′′′(ξ)

3!t3 (1)

for some ξ between 0 and t. Now f(0) = g(a, b). Using the chain rule, we see

f ′(t) =∂g

∂x(a+ t(x− a), b+ t(y − b))(x− a) +

∂g

∂y(a+ t(x− a), b+ t(y − b))(y − b)

so

f ′(t) =∂g

∂x(a, b)(x− a) +

∂g

∂y(a, b)(y − b).

Further (dropping the arguments), we see

f ′′(t) =∂2g

∂x2· (x− a)2 + 2

∂2g

∂x∂y· (x− a)(y − b) +

∂2g

∂y2· (y − b)2,

so

f ′′(0) =∂2g

∂x2(a, b)(x− a)2 + 2

∂2g

∂x∂y(a, b)(x− a)(y − b) +

∂2g

∂y2(a, b)(y − b)2.

Finally, it is clear that f ′′′(t) will contain terms multiplied by (x − a)3, (x − a)2(y −b), (x − a)(y − b)2 and (y − b)3. Thus f ′′′ → 0 faster than (x − a)2 + (y − b)2 as(x, y)→ (a, b). Evaluting (1) at t = 1, and noting that f(1) = g(x, y), we see

g(x, y) = g(a, b) +∂g

∂x(a, b)(x− a) +

∂g

∂y(a, b)(y − b)

+1

2

(∂2g

∂x2(a, b)(x− a)2 + 2

∂2g

∂x∂y(a, b)(x− a)(y − b) +

∂2g

∂y2(a, b)(y − b)2

)+ E(x, y),

where E(x, y)→ 0 fast as (x, y)→ (a, b). This proves Taylor’s Theorem for g.

Problem F10.4.

(a) Show that given a real-valued continuous function f on [0, 1]× [0, 1] and ε > 0, thereexist real valued continuous functions g1, . . . , gn, h1, . . . , hn on [0, 1] for some finite nsuch that ∣∣∣∣∣f(x, y)−

n∑i=1

gi(x)hi(y)

∣∣∣∣∣ ≤ ε, 0 ≤ x, y ≤ 1.

(b) If f(x, y) = f(y, x) for all x, y ∈ [0, 1], can this be done with gi = hi for all i?

Solution.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 45

(a) Let ε > 0. Since f is continuous on a compact set it is uniformly continuous. TakeN ∈ N so that |(x, y)− (z, w)| < 1

Ngives |f(x, y)− f(z, w)| < ε. Define a grid

xi = i/N, i = 0, 1, . . . , N and let

gi(x) =

0, x ≤ xi−1,

N(x− xi−1), xi−1 ≤ x ≤ xi,N(xi+1 − x), xi ≤ x ≤ xi+1,

0, x ≥ xi+1.

for i = 0, . . . , N . Then |gi(x)| ≤ 1 for all i, x and∣∣∣∣∣f(xi, y)−N∑i=0

gi(xi)f(xi, y)

∣∣∣∣∣ = 0 for all x = xi, i = 0, . . . , N, y ∈ [0, 1].

If x 6= xi for any i, then x ∈ (xj, xj+1) for some j. In this case, all gi(x) are zero exceptwhen i = j, j + 1 and we have gj(x) + gj+1(x) = 1. Thus∣∣∣∣∣f(x, y)−

N∑i=0

gi(x)f(xi, y)

∣∣∣∣∣ = |f(x, y)− gj(x)f(xj, y)− gj+1f(xj, y)|

= |(gj(x) + gj+1(x))f(x, y)− gj(x)f(xj, y)− gj+1f(xj+1, y)|≤ gj(x) |f(x, y)− f(xj, y)|+ gj+1(x) |f(x, y)− f(xj+1, y)|< gj(x)ε+ gj+1(x)ε = ε.

Hence the difference is small for all x ∈ [0, 1].

By the Stone-Weierstrass theorem, there are polynomials h0, · · · , hN such that for eachxi,

|f(xi, y)− hi(y)| < ε/(N + 1)

for all y ∈ [0, 1]. Then∣∣∣∣∣f(x, y)−N∑i=0

gi(x)hi(y)

∣∣∣∣∣ ≤∣∣∣∣∣f(x, y)−

N∑i=0

gi(x)f(xi, y)

∣∣∣∣∣+

∣∣∣∣∣N∑i=0

gi(x)f(xi, y)−N∑i=0

gi(x)hi(y)

∣∣∣∣∣≤ ε+

N∑i=0

|gi(x)| |f(xi, y)− hi(y)|

≤ ε+N∑i=0

ε/(N + 1) = 2ε.

Rescaling the epsilons and re-indexing the sums gives the solution. (For those inter-ested, the functions gi defined above are the functions used in the linear spline Galerkinfinite element method which is an important numerical method for solving PDE.)

(b) No. If f is a negative function then taking gi = hi, we will need to satisfy∣∣∣∣∣f(x, x)−n∑i=1

gi(x)2

∣∣∣∣∣ < ε

which isn’t possible for all ε since f is negative but the sum is nonnegative

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 46

Problem F10.10. Suppose f : R → R is bounded and Lipschitz continuous. For k ∈ N,define xk : [0, 1]→ R by xk(0) = 0 and

xk(t) = xk(n2−k) + (t− n2−k)f(xk(n2−k))

forn2−k < t ≤ (n+ 1)2−k, n ∈ N.

Explain why xk uniformly converges to a solution x : [0, 1]→ R of the ODE

x′(t) = f(x(t)), x(0) = 0,

as k →∞.

Solution. I’m not sure this problem is worth the time or effort required to complete it. Thefunction xk are the approximations to the solution of x′ = f(x) using Euler’s Method withstep size 2−k. It is well know that these will converge uniformly to x.

Problem F10.12. Define D(t) = (x, y) ∈ R2 : x2 + y2 ≤ r(t)2 where r(t) : R → Ris a continuously differentiable function. For a given smooth nonnegative function u(x, t) :R2 × R→ R, express the following in terms of a surface integral:

d

dt

(∫D(t)

u(x, t)dx

)−∫D(t)

∂u

∂t(x, t)dx.

Solution. Write the first integral above in polar coordinates:

d

dt

(∫D(t)

u(x, t)dx

)=

d

dt

∫ 2π

0

∫ r(t)

0

u(θ, ρ, t)ρ dρ dθ.

By Liebniz rule,

d

dt

∫ r(t)

0

u(θ, ρ, t)ρ dρ = u(θ, r(t), t)r(t)r′(t) +

∫ r(t)

0

∂u

∂t(θ, ρ, t)ρ dρ.

Thus

d

dt

∫ 2π

0

∫ r(t)

0

u(θ, ρ, t)ρ dρ dθ −∫ 2π

0

∫ r(t)

0

∂u

∂t(θ, ρ, t)ρ dρ dθ =

∫ 2π

0

u(θ, r(t), t)r(t)r′(t) dθ.

Hence

d

dt

(∫D(t)

u(x, t)dx

)−∫D(t)

∂u

∂t(x, t)dx

=

∫ 2π

0

u(θ, r(t), t)r(t)r′(t) dθ = r(t)r′(t)

∫∂D(t)

u(x, t)dx

is the desired representation as a surface integral.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 47

Problem S10.9. Assume that f(x, y, z) is a real valued, continuously differentiable func-tion such that f(x0, y0, z0) = 0. If ∇f(x0, y0, z0) 6= ~0, show that there is a differentiable sur-face given parametrically by (x(s, t), y(s, t), z(s, t)) with (x(0, 0), y(0, 0), z(0, 0)) = (x0, y0, z0)on which f = 0.

Solution. Since ∇f(x0, y0, z0) 6= ~0, assume without loss of generality that ∂f∂x

(x0, y0, z0) 6= 0.

Then the matrix [∂f∂x

(x0, y0, z0)] is invertible so by the implicit function theorem there is aneighborhood U of (y0, z0), a neighborhood V of x0 and a unique continuously differentiablefunction g : U → V with g(y0, z0) = x0 such that

(g(y, z), y, z) : y, z ∈ U = (x, y, z) : f(x, y, z) = 0.

Putting y(s, t) = y0 + s, z(s, t) = z0 + t, x(s, t) = g(y(s, t), z(s, t)) = g(y0 + s, z0 + t) for s, tsufficiently small, we see that

x(0, 0) = x0, y(0, 0) = y0, z(0, 0) = z0

and f = 0 on the surface (x(s, t), y(s, t), z(s, t)).

Problem S10.10. Let f : R2 → R be defined

f(x, y) =xy√x2 + y2

when (x, y) 6= (0, 0) and f(0, 0) = (0).

(a) Compute the directional derivatives of f at (0, 0) in all directions where they exist.

(b) Is f differentiable at (0, 0)?

Solution.

(a) Let v be a (unit) direction vector. The directional derivative of f at (0, 0) in thedirection of v is

limh→0

f(hv1, hv2)− f(0, 0)

h= lim

h→0

1

h

h2v1v2

||hv||= lim

h→0

hv1v2

|h|.

Since the limit of h/ |h| as h → 0 doesn’t exist, we need v1 = 0 or v2 = 0. Thusthe directional derivatives only exist in along the x or y axis and it is zero in thosedirections.

(b) No. If f was differentiable at (0, 0), the derivative would have to be the zero map sincethe directional derivatives along the x, y axes are zero there. Then

lim(x,y)→(0,0)

f(x, y)− f(0, 0)− 0

||(x, y)− (0, 0)||= lim

(x,y)→(0,0)

|f(x, y)|√x2 + y2

= lim(x,y)→(0,0)

|xy|x2 + y2

would need to be zero regardless of how we approach the point (0, 0). However, ap-proaching the origin along the line y = x (from either direction), we have

lim(x,y)→(0,0)

|xy|x2 + y2

= limx→0

x2

x2 + x2=

1

2.

Hence f is not differentiable at (0, 0).

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 48

Problem S10.12. Assume that (fn) is a sequence of nonnegative continuous function on

[0, 1] such that limn→∞

∫ 1

0

fn(x)dx = 0. Is it necessarily true that

(a) there is B ∈ R such that fn(x) ≤ B for x ∈ [0, 1] for all n ∈ N?

(b) there are points x0 in [0, 1] such that limn→∞

fn(x0) = 0?

Solution.

(a) No. For n ∈ N, n ≥ 2, define

fn(x) =

n3(x−

(12− 1

n2

)), 1

2− 1

n2 ≤ x < 12,

n3((

12

+ 1n2

)− x), 1

2≤ x < 1

2+ 1

n2 ,0, otherwise.

Then each fn is a triangular spike on(

12− 1

n2 ,12

+ 1n2

)of height n so clearly there is no

uniform bound B. However,∫ 1

0

fn(x)dx =2

n2· n =

2

n→ 0 as n→∞.

(b) No. Let (qn) be a denumeration of rationals in [0, 1]. Then define

fn(x) =

n(x−

(qn − 1

n

)), qn − 1

n≤ x < qn,

n((qn + 1

n

)− x), qn ≤ x < qn + 1

n,

0, otherwise.

Then fn is a spike of height 1 and base 2/n around the nth member of the sequence. Forany x ∈ [0, 1], there are infinitely many rationals in any neighborhood of x. Thus forinfinitely many n ∈ N , fn(x) > 1/2 (or any other number in (0, 1). Hence fn(x) 6→ 0for any x ∈ [0, 1].

Solution 2 (supplied by a classmate). Let χA denote the characteristic function of theset A; that is,

χA(x) =

1, x ∈ A0, x 6∈ A.

Then the sequence

χ[0,1/2], χ[1/2,1], χ[0,1/3], χ[1/3,2/3], χ[2/3,1], χ[0,1/4], χ[1/4,2/4], χ[2/4,3/4], χ[3/4,1], . . .

satisfies the property since the measure of the sets go to zero but each x ∈ [0, 1] iscontained in infinitely many sets of the form [k/n, (k+1)/n]. (Of course, these functionsaren’t continuous but you could mollify them).

Problem F11.3. Prove that the set of real numbers can be written as the union of un-countably many pairwise disjoint sets, each of which is uncountable.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 49

Solution. Since R× R has the same cardinality of R, there is a bijection F : R× R → R.Now

R× R =⋃x∈R

(x × R) .

ThenR = F (R× R) =

⋃x∈R

F (x × R) ..=⋃x∈R

Fx.

Each Fx is uncountable because it is an injective image of an uncountable set. Also theyare pairwise disjoint since the sets x × R are disjoint and F is injective. Thus we haveachieved the appropriate representation.

Problem F11.4. If you rearrange the order of terms in a sum∑an, sometime you can

change the value of the sum. Find all resulting limiting values of the following series andprove your assertions.

1.∞∑n=1

(−1)n−1

n

2.∞∑n=1

(−1)n−1

n2

Solution.

1. The sum can converge to any value x ∈ R. Indeed put an = (−1)n−1/n, bn = 1/(2n−1)and cn = −1/(2n), n = 1, 2, 3, . . .. Then clearly at least one of∑

n∈N

bn or∑n∈N

cn

must diverge because if both converged, then∑n∈N

1

n=∑n∈N

bn −∑n∈N

cn

would converge as well. However,

∞∑n=1

(−1)n−1

n=∑n∈N

bn +∑n∈N

cn

converges by the alternating series test, which would be impossible if one sum divergedwhile the other converged. Thus both sums diverge. We have that∑

n∈N

bn = +∞ and∑n∈N

cn = −∞.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 50

Let x > 0; the case that x < 0 is similar. Define BN =∑N

n=1 bn and Cn =∑N

n=1 cn.Then since BN →∞, there is N1 such that BN1 > x and the error will be no more than1/N1. Then there must be N2 so that BN1 + CN2 < x and the error will not exceedmax1/N1, 1/N2. Then there must be N3 so that BN3 + CN2 > x with error no morethan max1/N3, 1/N2. Repeating this process, we get a sequences so that

BNk−1+ CNk < x < BNk+1

+ CNk

where k is an even integer and the error from either side is at most

max1/Ni : i = k − 1, k, k + 1.

By construction, the error will go to zero and so limk→∞BNk−1+ CNk = x. Define a

rearrangement σ so that

aσ(1) = b1, . . . , aσ(N1) = bN1 , aσ(N1+1) = c1, . . . , aσ(N1+N2) = cN2 , . . . .

Then∞∑n=1

aσ(n) =∞∑n=1

(−1)σ(n)−1

σ(n)= x.

2. The sum converges absolutely, so any rearrangement of the sum will have the samevalue. We prove this here in a couple steps. Assume that (an)n∈N is a sequence suchthat

∞∑n=1

|an|

converges. We first prove that∞∑n=1

an

converges. Indeed, we see that

∞∑n=1

an =∞∑n=1

(|an|+ an)−∞∑n=1

|an| .

Further0 ≤ an + |an| ≤ 2 |an| .

Then

sN =N∑n=1

an + |an|

is a monotone sequence of terms which is bounded by 2∑∞

n=1 |an| and thus converges.Hence we have expressed

∞∑n=1

an

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 51

as a difference of convergent sequence and thus∑∞

n=1 an converges. Put

A =∞∑n=1

an.

Let σ : N → N be any bijection of N to itself. Let ε > 0. Then there is N ≥ 0 suchthat for k ≥ N , first ∣∣∣∣∣A−

k∑n=1

an

∣∣∣∣∣ < ε/2

and second, for k ≥ N , ∣∣∣∣∣∞∑n=k

an

∣∣∣∣∣ < ε/5.

Since σ is a bijection, there is a finite N∗ ∈ N so that 1, . . . , N ⊆ σ(1), . . . , σ(N∗).Let N = σ(1), . . . , σ(N∗) \ 1, . . . N. Then N ⊆ k+ 1, k+ 2, . . .. Now we see, forany ` ≥ N∗, ∣∣∣∣∣A− ∑

n=1

aσ(n)

∣∣∣∣∣ ≤∣∣∣∣∣A− ∑

n=1

an

∣∣∣∣∣+

∣∣∣∣∣∑n=1

an −∑n=1

aσ(n)

∣∣∣∣∣≤ ε

2+

∣∣∣∣∣ ∑n=N+1

an −∑n∈N

aσ(n)

∣∣∣∣∣≤ ε

2+

∣∣∣∣∣ ∑n=N+1

an

∣∣∣∣∣+

∣∣∣∣∣∑n∈N

aσ(n)

∣∣∣∣∣≤ ε

2+

∣∣∣∣∣∞∑

n=N+1

an

∣∣∣∣∣+

∣∣∣∣∣∞∑

n=N+1

an

∣∣∣∣∣≤ ε

2+ε

5+ε

5< ε.

Hence by definition,∞∑n=1

aσ(n) = A.

Now we have established that any rearrangement of the sum will yield the same value.We simply need to find the value. Starting from

∞∑n=1

1

n2=π2

6,

we see∞∑n=1

1

(2n)2=π2

24.

Then∞∑n=1

(−1)n−1

n2=∞∑n=1

1

n2− 2

∞∑n=1

1

(2n)2=π2

6− π2

12=π2

12.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 52

Problem S11.7. Prove there is a real number x such that

x5 − 3x+ 1 = 0.

Solution. Let p(x) = x5 − 3x + 1. Then p(1) = −1 < 0 and p(2) = 27 > 0. Sincepolynomials are continuous, by the intermediate value theorem, we can conclude there is anx ∈ [1, 2] such that p(x) = 0.

Problem S11.8. Give examples of:

1. A function f(x) on [0, 1] which is not Riemann integrable for which |f(x)| is Riemannintegrable.

2. Continuous functions fn and f on [0, 1] such that fn(t) → f(t) for all t ∈ [0, 1] but∫ 1

0fn(t)dt does not converge to

∫ 1

0f(t)dt.

Solution.

1. Define

f(x) =

1, x ∈ Q ∩ [0, 1],−1, x ∈ (R−Q) ∩ [0, 1]

Then since both the rational and irrationals are dense, any piecewise constant majoriz-ing function of f must remain greater than or equal to 1 and any piecewise constantminorizing function of f must be less than or equal to −1. Hence∫ 1

0

(g(x)− h(x))dx ≥ 2

for any piecewise constant majorizer g of f and any piecewise constant minorizer hof f . Since this value cannot become arbitrarily small, f is not Riemann integrable.However, |f(x)| = 1, x ∈ [0, 1] which is clearly Riemann integrable since it is a constantfunction.

2. Use triangles of height n and width 1/n (see F05.3, S08.2).

Problem S11.9. Prove that if f is a continuous function on [a, b] such that f(x) ≥ 0 forall x ∈ [a, b], then ∫ b

a

f(x)dx = 0

implies f(x) = 0 for all x ∈ [a, b].

Solution. Suppose to the contrary,that f 6≡ 0. Then there is t ∈ [a, b] such that f(t) 6= 0.Without loss of generality, let f(t) = ε > 0. By continuity, there is δ > 0 such that f(x) >

ε/2 when x ∈ (t−δ, t+δ). Since f is non-negative, we know that∫ t−δa

f(x)dx,∫ bt+δ

f(x)dx ≥ 0.Then ∫ b

a

f(x)dx =

∫ t−δ

a

f(x)dx+

∫ t+δ

t−δf(x)dx+

∫ b

t+δ

f(x)dx

≥∫ t+δ

t−δf(x)dx ≥

∫ t+δ

t−δ

ε

2dx = δε > 0.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 53

This contradicts that∫ baf(x)dx = 0. Hence we conclude f(x) = 0 for all x ∈ [a, b].

A small note: we implicitly assumed that t ∈ (a, b). However, if t = a, we can bound faway from zero when x ∈ [a, a + δ) and if t = b, we do the same when x ∈ (b − δ, b] so theproof still works.

Problem S11.11.

1. Show that a connected subset A ⊂ R is path connected.

2. Give an example of a subset of R2 which is connected but not path connected.

Solution.

1. First we claim that a connected subset of R is an interval. Indeed, if A ⊂ R is notan interval, then there is x 6∈ A such that for some y, z ∈ A, y < x < z. ThenA = (A ∩ (−∞, x)) ∪ (A ∩ (x,∞)) shows that A is disconnected.

Thus A, being connected, is an interval: A = [a, b] (the interval could just as well beopen, half-open or infinite; the proof is the same in all cases). Then for x, y ∈ A, theline f(t) = x+ t(y − x) will stay entirely inside A and connect x to y. Thus A is pathconnected.

2. The topologist’s sine curve works (see F08.1).

Problem S11.12. Give a metric space (M,d) and a constant 0 < r < 1, a continuousfunction T : M →M is called and r-contraction if

d(Tx, Ty) ≤ rd(x, y)

for all x, y ∈ M . A well know fixed point theorem states that if M is complete and T is anr-contraction on M , then T has a unique fixed point (don’t prove this). This result is oftenused to prove existence and uniqueness of solutions to differential equations.

1. Illustrate this technique for the equation

f ′(t) = f(t), f(0) = 1 (DE)

by letting M = C[0, c] for sufficiently small c with the uniform distance

d(f, g) = sup0≤t≤c

|f(t)− g(t)| , f, g ∈M

and defining

(Tf)(t) = 1 +

∫ t

0

f(s)ds. (2)

2. What approximations do you obtain from the sequence T (0), T 2(0), T 3(0), . . . ?

Solution.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 54

1. By the fundamental theorem of calculus, we see that (DE) implies

f(t) = 1 +

∫ t

0

f(s)ds.

Further, if f is continuous, the step can be reversed. Thus for continuous f , we havethat f satisfies (DE) if and only if f satisfies the above integral equation. Define T asin (2). Then f satisfies the integral equation if and only if f is a fixed point of T . Thatis, if we can prove that T has a unique fixed point, then (DE) has a unique solution.Consider for continuous x, y and t ∈ [0, c],

|(Tx)(t)− (Ty)(t)| =∣∣∣∣∫ t

(x(s)− y(s))ds

∣∣∣∣≤∫ t

0

|x(s)− y(s)| ds ≤ ||x− y||∞∫ t

0

ds ≤ c ||x− y||∞ .

Taking the supremum over all t on the left hand side gives

||Tx− Ty||∞ ≤ c ||x− y||∞ .

Thus if we take 0 < c < 1, T is a contraction and thus has a unique fixed point. Hence(DE) has a unique solution on [0, c].

2. Let f0(t) = 0 and define fn = Tfn−1, n = 1, 2, 3, . . .. Then

f1(t) = 1 +

∫ t

0

0ds = 1,

f2(t) = 1 +

∫ t

0

1ds = 1 + t,

f3(t) = 1 +

∫ t

0

(1 + s)ds = 1 + t+ t2

2,

...

fn(t) =n−1∑k=0

tk

k!.

It ie easy to see that [T n(0)](t) = fn(t). Thus our nth estimate gives us the Maclaurinexpansion of et truncated at term n − 1. From this it is clear that [T n(0)](t) → et

uniformly as n → ∞ which we expect since we know that (DE) has unique solutionf(t) = et.

Problem F12.1. Let bn∞n=1 be a sequence of real numbers and let M > 0 be such that∣∣∣∣∣N∑n=1

bn

∣∣∣∣∣ ≤M

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 55

for all N ∈ N. Let an∞n=1 be a sequence of positive real numbers which decreases to 0 asn→∞. Show that the series

∑∞n=1 anbn converges.

Solution. Let Bn =∑n

k=1 bk and sn =∑n

k=1 akbk, n ∈ N. We know that Bn is a boundedsequence and we need to prove that sn has a limit as n → ∞. Consider, for m,n ∈ N,n > m,

|sn − sm| =

∣∣∣∣∣n∑

k=m+1

akbk

∣∣∣∣∣ .Using summation by parts, we see

n∑k=m+1

akbk = (Bnan+1 −Bmam+1)−n∑

k=m+1

Bk(ak+1 − ak).

Then

|sn − sm| =

∣∣∣∣∣(Bnan+1 −Bmam+1)−n∑

k=m+1

Bk(ak+1 − ak)

∣∣∣∣∣≤ |Bn| |an+1|+ |Bm| |am+1|+

∣∣∣∣∣n∑

k=m+1

Bk(ak+1 − ak)

∣∣∣∣∣≤M(an+1 + am+1) +

n∑k=m+1

M(ak − ak+1),

where we have used the fact that (an) is a positive decreasing sequence. Then the sumtelescopes so

|sn − sm| ≤M(an+1 + am+1) +M(am+1 − an+1) = 2Mam+1.

Since the sequence (an) goes to zero as n → ∞, this shows that (sn) is a Cauchy sequenceand hence converges. Thus

∑∞n=1 anbn converges.

Problem F12.3. Let fn be a sequence of non-negative continuous functions on a com-pact metric space X. Assume that fn(x) ≥ fn+1(x) for all n and x so that lim

n→∞fn(x) = f(x)

exists for every x ∈ X. Show that f is continuous if and only if fn → f uniformly in X.

Solution. ⇐=. Assume that fn → f uniformly. Let ε > 0. Then, by uniform convergence,there is N ≥ 0, such that for any x ∈ X,

n ≥ N =⇒ |fn(x)− f(x)| < ε/3.

Further, for x, y ∈ X, and fixed n ≥ N , by continuity of fn, there is δ > 0, such that

d(x, y) < δ =⇒ |fn(x)− fn(y)| < ε/3.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 56

By repeated use of the triangle inequality,

|f(x)− f(y)| ≤ |f(x)− fn(x)|+ |fn(x)− f(y)|≤ |f(x)− fn(x)|+ |fn(x)− fn(y)|+ |fn(y)− f(y)|< ε/3 + ε/3 + ε/3 = ε, whenever d(x, y) < δ.

Thus f is continuous.=⇒ . Assume f is continuous and that fn → f pointwise. Let ε > 0. Define An = x ∈

X : |fn(x)− f(x)| < ε. Since f, fn are continuous, An is open since it is the pullback of theopen set [0, ε) under the continuous function |fn − f | (we consider [0, ε) open because it isopen in the relative topology on [0,∞) which is the largest potential co-domain of |f − fn|;alternatively, we could replace it with (−∞, ε) if we consider the range of |f − fn| to be R).Since fn → f pointwise, eventually |fn(x)− f(x)| < ε for every x. Hence (An) form an opencover of X. Since X is compact, there is a finite subcover:

X =N⋃n=1

An.

By construction, we have A1 ⊂ A2 ⊂ A3 ⊂ · · · . Thus X = AN . But this implies that fn → funiformly, because for all x ∈ X, |fN(x)− f(x)| < ε (and the same will hold for any n > N).

Problem F12.4. A subset K of a metric space (X, d) is called nowhere dense is K hasempty interior. Prove the Baire Category Theorem that if (X, d) is a complete metric space,then X is not a countable union of closed nowhere dense sets.

Solution. First, we need to get the correct definition. A subset K of a metric space (X, d)is nowhere dense if the closure of K has empty interior. The closure part is key because,e.g., the rationals should not be considered nowhere dense. [Note: in the case of closed K,we have that K equals its closure so the correct definition reduces to the definition given.]

First, it is clear that X is not nowhere dense since the interior of the closure of X is X.Assume to the contrary that X = ∪nKn where each Kn is closed and nowhere dense. Since Xis not nowhere dense, X 6= K1 and since K1 is closed, X−K1 is open. Take x1 ∈ X−K1. Byopenness, there is δ1 > 0 such that B(x1, δ1) ⊂ X−K1; wlog take δ1 < 1/2 (we can do this bysimply reducing δ1 if necessary). Since K2 is nowhere dense, it cannot contain B(x1, δ1/2).Take x2 ∈ (X−K2)∩B(x1, δ1/2) (which is an open set). Then there is δ2 < δ1/2 < 1/4 suchthat B(x2, δ2) ⊂ (X −K2) ∩ B(x1, δ1/2). Continue this procedure inductively to constructa sequence (xn) in X such that xn 6∈ K1, . . . , Kn and d(xn, xn−1) ≤ 1

2n−1 . Then for n > m,d(xn, xm) ≤ d(xn, xn−1) + · · ·+ d(xm+1, xm) <

∑∞j=m

12m

. Taking m large enough shows thatthis is a Cauchy sequence. Since X is complete, the sequence converges to some x ∈ X.However, for any m ∈ N, the sequence (xn) eventualy resides in a ball of distance at least1/2m−1 away from X − Km. Thus x 6∈ Km for all m ∈ N. Hence x ∈ X − ∪nKn which isa contradiction to our assumption that X = ∪mKm. Thus X is not the union of countablymany closed, nowhere dense sets.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 57

Problem S12.1. Let Ω denote the set of all closed subsets of [0, 1] and let ρ : Ω×Ω→ [0, 1]be defined by

ρ(A,B) ..= max

supx∈A

infy∈B|x− y| , sup

y∈Binfx∈A|x− y|

,

for A,B ∈ Ω. Prove that (Ω, ρ) is a metric space.

Solution. Symmetry is obvious, as if the fact that ρ(A,A) = 0. We simply have to provethat ρ is positive on unequal sets and that the triangle inequality is satisfied. Suppose thatA,B ∈ Ω, A 6= B. Then (without loss of generality) we can take x ∈ A such that x 6∈ B.Then, since B is closed, there is ε > 0 such that (x− ε, x+ ε)∩B = ∅. This means that forany y ∈ B, |x− y| > ε. Then

infy∈B|x− y| ≥ ε > 0

which leads toρ(A,B) ≥ sup

x∈Ainfy∈B|x− y| > 0.

Next, let A,B,C ∈ Ω. Then for all a ∈ A, b ∈ B, c ∈ C,

|a− b| ≤ |a− c|+ |c− b| .

Taking the infimum over all b ∈ B gives

infb∈B|a− b| ≤ |a− c|+ inf

b∈B|c− b| .

But infb∈B|c− b| ≤ sup

c∈Cinfb∈B|c− b|, so

infb∈B|a− b| ≤ |a− c|+ sup

c∈Cinfb∈B|c− b| .

Next taking the infimum over all c ∈ C and then the supremum over all a ∈ A (in thatorder), gives

supa∈A

infb∈B|a− b| ≤ sup

a∈Ainfc∈C|a− c|+ sup

c∈Cinfb∈B|c− b| ≤ ρ(A,C) + ρ(B,C).

A very similar sequence of operations gives supb∈B

infa∈A|a− b| ≤ ρ(A,C) + ρ(B,C) so we can

conclude that ρ(A,B) ≤ ρ(A,C) + ρ(B,C), which completes the proof.

Problem S12.2. Recall that f : [a, b]→ R is convex if for all x, y ∈ [a, b],

f(αx+ (1− a)y) ≤ αf(x) + (1− α)f(y), α ∈ [0, 1].

Let fn : [a, b]→ R be convex functions and suppose that f(x) = limn→∞ fn(x) exists for allx ∈ [a, b]. Prove that if f(x) is continuous on [a, b], then fn → f uniformly.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 58

Solution. Fix ε > 0. Since f is continuous on a compact set, it is uniformly continuous.Thus we can take a uniform partition a = a0 < a1 < · · · < aN = b fine enough that for allx ∈ [a, b],

|f(x)− f(ai)| , |f(x)− f(ai+1)| ≤ ε

where [ai, ai+1] is the subinterval containing x. Thus

f(x) ≥ f(ai)− ε and f(x) ≥ f(ai+1)− ε =⇒ f(x) ≥ maxf(ai), f(ai+1) − ε.

Now choose M large enough that for n ≥ M , we have |fn(ai)− f(ai)| < ε for all ai. Thenfor any x ∈ [a, b], using the convexity and the above inequality, we see that for n ≥M ,

fn(x) ≤ maxfn(ai), fn(ai+1) ≤ maxf(ai), f(ai+1)+ ε ≤ f(x) + 2ε.

This gives the necessary uniform upper bound. I could not figure out how to achieve thecorresponding lower bound.

Problem S12.3. Prove the Bolzano-Weierstrass Theorem in the following form: Everysequence (an)n∈N of numbers in [0, 1] has a convergent subsequence.

Solution. Let (an)n∈N be a sequence in [0, 1]. Divide the interval into [0, 1/2] and [1/2, 1]At least one of these subintervals must contain infinitely many of (an). Let (a1,n) be a subse-quence that is entirely in one of the half intervals. Divide that interval into two subintervalswhich cover the set and are of length 1/4. Then there is an infinite subsequence of (a1,n)which lies entirely in one interval of length 1/4. Repeat this process indefinitely. Then wehave inductively created sequences (ak,n)∞n=1 such that each (ak,n)∞n=1 is a subsequence of(ak−1,n)∞n=1 and each sequence (ak,n)∞n=1 is contained in an interval of length 1/2k. Definexn = an,n, n ∈ N. Then (xn)∞n=1 is a subsequence of (an)∞n=1 and for each k ∈ N, (xn)∞n=k

is a subsequence of (ak,n)∞n=1. Let ε > 0. Choosing k ∈ N so that 1/2k−1 < ε, we see thatm,n ≥ k implies

|xm − xn| ≤ 12k−1 < ε.

Thus (xn) is a Cauchy sequence and thus converges since [0, 1] is complete (which is becauseit is a closed subset of a complete space). Thus every sequence (an) in [0, 1] has a convergentsubsequence.

Problem S12.4. For a sequence (an) of non-negative numbers, let sn ..=∑n

i=1 ai. Supposethat (sn) tends to a number s ∈ R in the Cesaro sense:

limn→∞

s1 + s2 + · · ·+ snn

= s.

Show that∑∞

n=1 an exists and is equal to s.

Solution. To show the infinite sum exists, we use a proof by contradiction. Assuming thesum diverges, since all terms are non-negative, the sum must diverge to +∞. Then for anyL > 0, there is N ∈ N, such that sn =

∑ni=1 ai > L for all n > N . Choosing n > N , we see

s1 + s2 + · · ·+ snn

=s1 + · · ·+ sN

n+sN+1 + · · ·+ sn

n>

(n−N)L

n.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 59

Letting L→∞ implies thats1 + s2 + · · ·+ sn

n

can be made arbitrarily large so that (sn) cannot converge in the Cesaro sense which con-tradicts our assumption. We conclude that

∑∞n=1 an exists.

To prove that∑∞

n=1 an = s, we need to prove that sn → s. Suppose that sn → s′

for some s′ ∈ R. Let ε > 0. There is N ∈ N so that |sn − s′| < ε/2, n > N . Then|s1 − s′|+ · · · |sN − s′| is some finite number. Choose N∗ > N so that

|s1 − s′|+ · · · |sN − s′|N∗

< ε/2.

Then for n > N∗, we have∣∣∣∣s1 + · · ·+ snn

− s′∣∣∣∣ =

∣∣∣∣(s1 − s′) + · · · (sn − s′)n

∣∣∣∣≤∣∣∣∣(s1 − s′) + · · · (sN − s′)

n

∣∣∣∣+

∣∣∣∣(sN+1 − s′) + · · · (sn − s′)n

∣∣∣∣≤ |s1 − s′|+ · · · |sN − s′|

n+|sN+1 − s′|+ · · · |sn − s′|

n

<|s1 − s′|+ · · · |sN − s′|

N∗+

(n−N)

n· ε

2< ε/2 + ε/2 = ε.

Hence

limn→∞

s1 + · · ·+ snn

= s′.

Since limits in R are unique, we must have s′ = s and so limn→∞ sn = s as desired.

Problem S12.5. Prove there is a unique continuous function y : [0, 1]→ R such that

y(x) = ex +y(x2)

2, x ∈ [0, 1].

Solution. Define the operator T on C[0, 1] by

(Ty)(x) = ex +y(x2)

2, x ∈ [0, 1].

Then clearly, Ty is continuous whenever y is continuous so T : C[0, 1] → C[0, 1]. Consider,for any y, z ∈ C[0, 1],

|(Ty)(x)− (Tz)(x)| =∣∣∣∣z(x2)

2− y(x2)

2

∣∣∣∣ =1

2

∣∣z(x2)− y(x2)∣∣ ≤ 1

2||z − y||∞ , x ∈ [0, 1].

Thus taking the supremum over all x, we have

||Ty − Tz||∞ ≤1

2||y − z||∞ .

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 60

Hence T is a contraction mapping and since C[0, 1] is complete with respect to the sup norm,there is a unique y ∈ C[0, 1] such that Ty = y. This y is the unique continuous functionsatisfying the desired equation.

Problem S12.6. Let γ be a smooth curve in R2−(0, 0) which begins and ends at (1, 0)and winds once around the origin counterclockwise. Compute the integral

I(γ) =

∮γ

y dx− x dyx2 + y2

.

Solution. We can parameterize the curve in any way we want since the problem doesn’tspecify. The most natural parameterization is

x = cos θ, y = sin θ, θ ∈ [0, 2π].

Then

I(γ) =

∫ 2π

0

sin θ(− sin θ)− cos θ cos θ

cos2 θ + sin2 θdθ

= −∫ 2π

0

dθ = −2π.

Problem F13.1. For a sequence an of positive real numbers, define Pn =∏n

j=1(1 + aj).Prove that limn→∞ Pn exists if and only if limn→∞

∑nj=1 aj exists.

Solution. Since all terms are positive Pn and Sn =∑n

j=1 aj are increasing sequences. If wecan show they are bounded above then they converge.

Suppose that Sn converges. Then Sn is a bounded sequence; let M = supn Sn. For allx ≥ 0, we have 1 + x ≤ ex (this is easily verified by looking at the Taylor Expansion of ex).Then

Pn =n∏j=1

(1 + aj) ≤n∏j=1

eaj = exp

(n∑j=1

aj

)= eSn ≤ eM .

Thus Pn converges.Suppose Pn converges. We see that

Pn =n∏j=1

(1 + aj) = 1 +

(n∑j=1

aj

)+

(∑i 6=j

aiaj

)+ · · ·+

n∏j=1

aj = Sn + (positive terms).

Thus Sn ≤ Pn, so Sn converges since Pn converges.

Problem F13.2. Let f : R→ R be a nondecreasing function.

(a) Prove that x ∈ R : f is not continuous at x is countable.

(b) Let S ⊂ R be a countable set. Prove there is a nondecreasing function f : R→ R suchthat f is discontinuous at x if and only if x ∈ S.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 61

Solution.

(a) Let Ω = x ∈ R : f is not continuous at x. Then for y ∈ Ω, we must have

limx→y−

f(x) 6= limx→y+

f(x),

because otherwse f would be continuous at y. In particular, since f is nondecreasing,

limx→y−

f(x) < limx→y+

f(x).

Since the rationals are dense in the reals, for each y ∈ Ω, we can find q(y) ∈ Q so that

limx→y−

f(x) < q(y) < limx→y+

f(x).

Consider, for y 6= z ∈ Ω (wlog we take y < z), we have

limx→y−

f(x) < q(y) < limx→y+

f(x) ≤ limx→z−

f(x) < q(z) < limx→z+

f(x),

and thus q(y) 6= q(z). This means that Ω injects into Q and thus must be countable.

(b) Let S ⊂ R be countable. Then we can say S = sn : n ∈ N. For any x define Ax ⊂ Nsuch that Ax = n ∈ N : sn < x. Then

f(x) =∑n∈Ax

1

2n

has the desired property. Indeed for x < y, if there is any n ∈ N such that x < sn < y,then clearly f(y) is greater than f(x) by at least 2−n; if there is no such n, thenf(x) = f(y). Thus f is nondecreasing. Further, for any n ∈ N,

limx→s−n

f(x) < limx→s+n

f(x) = limx→s−n

f(x) +1

2n

so f is discontinuous at all points in S. Also, for y 6∈ S, take δ > 0 small enough sothat s1, . . . , sN are not in (y − δ, y + δ) [this is possible for any N since none of thesevalues can equal y]. Then for |x− y| < δ, we have

|f(x)− f(y)| ≤∞∑n=N

1

2n.

Choosing N large enough, this bound becomes arbitrarily small. Thus f is continuousat y when y 6∈ S.

Problem F13.3. Let γ : [0, 1]→ R2 be a continuous injective function. By definition, thelength of the range γ([0, 1]) is

L(γ) = sup

n−1∑j=0

||γ(tj+1)− γ(tj)|| : 0 ≤ t0 < t1 < · · · < tn ≤ 1, n <∞

.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 62

(a) Suppose that f is continuous and non-decreasing on [0, 1] and let γ(t) = (t, f(t)) (sothat the range of γ is the graph of f). Prove

L(γ) ≤ 1 + f(1)− f(0).

(b) Show there exists a continuous nondecreasing f(t) on [0, 1] such that f(0) = 0, f(1) = 1and L(γ) = 2 when γ(t) = (t, f(t)).

Solution.

(a) Let 0 = t0 < t1 < · · · < tn = 1 be an arbitrary partition of [0, 1] Then

L ≤n−1∑j=0

||γ(tj+1)− γ(tj)||

=n−1∑j=0

||(tj+1, f(tj+1))− (tj, f(tj))||

=n−1∑j=0

||(tj+1, f(tj+1))− (tj+1, f(tj)) + (tj+1, f(tj))− (tj, f(tj))||

≤n−1∑j=0

||(tj+1, f(tj+1))− (tj+1, f(tj))||+ ||(tj+1, f(tj))− (tj, f(tj))||

=n−1∑j=0

|f(tj+1)− f(tj)|+ |tj+1 − tj|

= f(tn)− f(t0) + tn − t0 = 1 + f(1)− f(0).

Note we use |f(tj+1)− f(tj)| = f(tj+1)−f(tj) which is valid since f is non-decreasing.

(b) This question seems unreasonably difficult for the basic exam but here we go.

Let f0(x) = x, x ∈ [0, 1] and iteratively define

fn+1(x) =

fn(3x)/2, 0 ≤ x ≤ 1/3,

1/2, 1/3 ≤ x ≤ 2/3,1/2 + fn(3x− 2)/2, 2/3 ≤ x ≤ 1.

We argue that these are continuous functions. If fn is continuous, then fn+1 is con-tinuous on [0, 1/3) and (2/3, 1]. It is also clear that fn+1 is continuous on (1/3, 2/3).By induction, we see that fn(0) = 0, fn(1) = 1 for all n ∈ N and thus fn+1(1/3) =fn(1)/2 = 1/2 and fn+1(2/3) = 1/2 + fn(0)/2 = 1/2 so fn+1 is continuous at 1/3, 2/3.Thus fn+1 is continuous whenever fn is continuous. Thus, since f0 is continuous, wehave fn continuous for all n. Next, we see that if x ∈ (1/3, 2/3), then fn(x) = fn+1(x),n ≥ 1. Next, we see

sup0≤x≤1/3

|fn+1(x)− fn(x)| = sup0≤x≤1/3

∣∣12fn(3x)− 1

2fn−1(3x)

∣∣= 1

2sup

0≤x≤1|fn(x)− fn−1(x)|

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 63

and similarly

sup2/3≤x≤1

|fn+1(x)− fn(x)| = sup2/3≤x≤1

∣∣12− 1

2fn(3x− 2)− 1

2+ 1

2fn−1(3x− 1)

∣∣= 1

2sup

0≤x≤1|fn(x)− fn−1(x)| .

Thus we have

sup0≤x≤1

|fn+1(x)− fn(x)| = 12

sup0≤x≤1

|fn(x)− fn−1(x)| .

Hence by induction,

sup0≤x≤1

|fn+1(x)− fn(x)| = 12n

sup0≤x≤1

|f1(x)− f0(x)| .

This shows that fn is uniformly Cauchy and thus converges uniformly to some functionf . Since f is the uniform limit of continuous functions, it is continuous.

By constuction, f is constant on the intervals [1/3, 2/3], [1/9, 2/9], [7/9, 8/9], [1/27, 2/27],etc. The sum of the lengths of the intervals is

13

+ 2 · 19

+ 4 · 127

+ · · · = 13

∞∑k=0

(23

)k= 1

3

(1

1−1/3

)= 1.

However, f(0) = 0 and f(1) = 1, thus on the set where f is non-constant, the arclength must be at least 1 since the vertical change is 1. This means that the total arclength is greater than or equal to 2. Using (a), we see that the arc length must be 2.

[Note: The function f we constructed is called the Cantor step function and has afew remarkable properties. It is constant on the complement of the Cantor set. Asa consequence, it is differentiable almost everywhere and has derivative zero almosteverywhere. However, it still increases from 0 to 1. It is uniformly continuous (since itis continuous on a compact set) but it is not absolutely continuous since it cannot bewritten as the derivative of its (Lebesgue) integral. Finally, the function g(x) = x +f(x), x ∈ [0, 1] maps the Cantor set (a set of measure zero) homeomorphically to a setof measure 1. This is a bizarre result because, while we know continuous functions canstretch and enlarge volume, it seems counterintuitive that they can “create” volume.]

Problem F13.5. A function f : Rd → R is said to be convex if

f(tx+ (1− t)y) ≤ tf(x) + (1− t)f(y), for all x, y ∈ Rd, 0 ≤ t ≤ 1.

Assume that f is continuously differentiable and

(∇f(x)−∇f(y)) · (x− y) ≥ 0, x, y ∈ Rd.

Prove that f is convex.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 64

Solution. For simplicity, for fixed x, y ∈ Rd, define F,G : [0, 1]→ R by

F (t) = f(tx+ (1− t)y), G(t) = tf(x) + (1− t)f(y), t ∈ [0, 1].

We need to prove that F (t) ≤ G(t) or rather G(t)− F (t) ≥ 0, for all t ∈ [0, 1]. We see

G′(t)− F ′(t) = f(x)− f(y)−∇f(tx+ (1− t)y) · (x− y).

For all r > s, we have

(G′(r)− F ′(r))− (G′(s)− F ′(s)) = [∇f(sx+ (1− s)y)−∇f(rx+ (1− r)y)] · (x− y).

But sx+ (1− s)y − (rx+ (1− r)y) = (s− r)(x− y), and s− r < 0 so by our assumption

(s− r)((G′(r)− F ′(r))− (G′(s)− F ′(s))) ≥ 0

so (G′(r)− F ′(r))− (G′(s)− F ′(s)) ≤ 0 or

G′(r)− F ′(r) ≤ G′(s)− F ′(s), r, s ∈ [0, 1], r > s.

That is, the derivative of G− F is decreasing.It is clear that G(1)−F (1) = 0 = G(0)−F (0). Now assume that there is t ∈ (0, 1) such

that G(t) − F (t) < 0. Then by the mean value theorem, there are c ∈ (0, t), d ∈ (t, 1) suchthat

0 < G(t)− F (t)− (G(0)− F (0)) = (G′(c)− F ′(c))t

and0 < G(t)− F (t)− (G(1)− F (1)) = (G′(d)− F ′(d))(t− 1).

ThenG′(d)− F ′(d) < 0 < G′(c)− F ′(c)

which contradicts that the derivative of G − F is decreasing. Hence there is no such t, soG(t) ≥ F (t) as required.

Problem F13.6. Let X be a metric space and let (xn) be a sequence in X such that if(yn) is a subsequence of (xn) then there is a subsequence (zn) of (yn) converging to x ∈ X.Prove that (xn) converges to x ∈ X.

Solution. Suppose that xn 6→ x. Then there is ε > 0 such that for any k ∈ N there isn ≥ k such that d(xn, x) ≥ ε. For each k ∈ N, let nk ≥ k be a natural number such thatd(xn, x) ≥ ε. Then (xnk) is certainly a subsequence of (xn) but any member of (xnk) is atleast ε away from x. Hence no subsequence of (xnk) converges to x which is a contradiction.Thus x→ x.

Problem S13.1.

(a) Define what it means for a function on [0, 1] to be Riemann integrable.

(b) Show that every monotone increasing function on [0, 1] is Riemann integrable.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 65

(c) Use part (b) to construct a Riemann integrable function on [0, 1] which has infinitelymany discontinuities.

Solution.

(a) Let f : [0, 1]→ R. Then by definition, f is Riemann integrable if and only if∫ 1

0

f(x)dx =

∫ 1

0

f(x)dx,

where∫ 1

0

f(x)dx = inf

∫ 1

0

g(x)dx : g is a piecewise constant majorizing function of f on [0, 1]

,

and∫ 1

0

f(x)dx = sup

∫ 1

0

h(x)dx : h is a piecewise constant minorizing function of f on [0, 1]

.

(b) Left ε > 0, and n ∈ N such that f(1)−f(0)n

< ε. Let 0 = x0 < x1 < · · · < xn = 1 be thepartition of [0, 1]; i.e., xi = i/n, i = 0, . . . , n. Define

g(x) = f(xi+1), x ∈ [0, 1], xi ≤ x < xi+1

where i = 0, . . . , n − 1 and by convention g(1) = f(1). Then g is piecewise constantand majorizes f . Likewise, let

h(x) = f(xi), x ∈ [0, 1], xi ≤ x < xi+1,

for i = 0, . . . , n − 1, and h(1) = f(xn−1). Then h is piecewise constant and minorizesf . Further ∫ 1

0

f(x)dx−∫ 1

0

f(x)dx ≤∫ 1

0

g(x)dx−∫ 1

0

h(x)dx

=n−1∑i=0

∫ xi+1

xi

(g(x)− h(x))dx

=n−1∑i=1

∫ xi+1

xi

(f(xi+1)− f(xi))dx

=n−1∑i=1

(xi+1 − xi)(f(xi+1)− f(xi))

=1

n

n−1∑i=1

(f(xi+1)− f(xi))

=1

n(f(1)− f(0)) < ε.

Since ε was arbitrary, this implies that∫ 1

0f(x)dx =

∫ 1

0f(x)dx so f is Riemann inte-

grable.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 66

(c) Define

f(x) =

0, x = 0,1

n+1, x ∈

[1

n+1, 1n

), n = 1, 2, 3, . . . ,

1, x = 1.

Then f is nondecreasing, and hence Riemann integrable, but f is discontinuous atx = 1

nfor all n = 1, 2, 3, . . . .

Problem S13.2. The approximation from Simpson’s Rule for∫ baf(x)dx is given by

S(f) =

[2

3f

(a+ b

2

)+

1

3

(f(a) + f(b)

2

)](b− a).

Suppose that f has continuous derivatives up to order 3. Show that∣∣∣∣∫ b

a

f(x)dx− S(f)

∣∣∣∣ ≤ C(b− a)4 supa≤x≤b

∣∣f (3)(x)∣∣

where C is a constant which does not depend on f .

Solution. We construct a second-degree polynomial p that agrees with f at a, b and a+b2.

Put c = a+b2

. The correct polynomial is

p(x) = f(c)(x− a)(x− b)(c− a)(c− b)

+ f(b)(x− a)(x− c)(b− a)(b− c)

+ f(a)(x− b)(x− c)(a− b)(a− c)

.

Put g(x) = f(x)− p(x). Next, choose d ∈ [a, b], d 6= a, d 6= b, d 6= c. Then

q(x) = g(d)(x− a)(x− b)(x− c)(d− a)(d− b)(d− c)

is a third degree polynomial which agrees with g at 4 points. Hence by higher order Rolle’stheorem, there is y ∈ (a, b) such that g′′′(y)− q′′′(y) = 0. But g′′′ = f ′′′ so

f ′′′(y) = q′′′(y) =6g(d)

(d− a)(d− b)(d− c)

and thus

|g(d)| = |f′′′(x)| |d− a| |d− b| |d− c|

6≤ (b− a)3

6supa≤t≤b

|f ′′′(t)| .

Since d was arbitrary, we conclude that

|g(x)| = |f(x)− p(x)| ≤ (b− a)3

6supa≤t≤b

|f ′′′(t)| , for all x ∈ [a, b].

Finally, let h = c− a = b− c, then

p(x) = −f(c)

h2(x− a)(x− b) +

f(b)

2h2(x− a)(x− c) +

f(a)

2h2(x− b)(x− c).

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 67

So ∫ b

a

p(x)dx =

∫ h

−hp(x+ c)dx

=1

h2

∫ h

−h

1

2(f(b)x(x+ h) + f(a)x(x− h))− f(c)(x− h)(x+ h)dx

=1

h2

(1

2· 2h3(f(b) + f(a))

3+

4h3f(c)

3

)=h

3(f(b) + f(a)) + 4f(c))

= (b− a)

[2

3f

(a+ b

2

)+

1

3

(f(a) + f(b)

2

)]= S(f).

Thus ∣∣∣∣∫ b

a

f(x)dx− S(f)

∣∣∣∣ =

∣∣∣∣∫ b

a

f(x)− p(x)dx

∣∣∣∣≤∫ b

a

|f(x)− p(x)| dx

≤ (b− a)3

6supa≤t≤b

|f ′′′(t)|∫ b

a

dx

=(b− a)4

6supa≤t≤b

|f ′′′(t)| ,

so C = 1/6.

Problem S13.3. Prove that a metric space is sequentially compact if and only if it iscomplete and totally bounded.

Solution. Let X be a metric space which is complete and totally bounded. Let (xn) bea sequence in X. Since we will be inductively choosing more and more subsequences, it isconvenient to say (xn) = (x1,n). Since X is totally bounded, it can be covered with finitelymany balls of radius 1/2. One of these balls must contain infinitely many of (x1,n). Let(x2,n) be a subsequence of (x1,n) which is contained in a ball of radius 1/2. Cover this ballwith finitely many balls of radius 1/3. Then infinitely many of (x2,n) must be contained ina single ball. Let (x3,n) be a subsequence of (x2,n) which is completely contained in a ballof radius 1/3. Continuing this process we get sequences (xk,n) each contained in a ball ofradius 1/k and each a subsequence of (xk−1,n). Let (yn) be defined yn = xn,n for all n ∈ N.Then (yn) is a subsequence of (xn) and by construction,

d(yn, ym) < max

2n, 2m

.

From this we see that (yn) is a Cauchy sequence and is hence convergent since X is complete.Thus (xn) has a convergent subsequence and so X is sequentially compact.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 68

Now suppose that X is sequentially compact. Let (xn) be a Cauchy sequence in X. Thenthere is a subsequence (xnk) which converges to some x ∈ X. Let ε > 0, then there is N ∈ Nsuch that d(xn, xnk) < ε/2 and d(xnk , x) < ε/2 when n, k > N . For such n, k, we have

d(xn, x) ≤ d(xn, xnk) + d(xnk , x) < ε

so (xn) converges to x as well. Thus Cauchy sequences in X converge and so X is complete.Again, suppose X is sequentially compact. Let ε > 0. For contradiction, assume that

X cannot be covered by finitely many balls of radius ε. Let x1 ∈ X. Then by assump-tion B(x1, ε) does not cover X, so there is x2 outside B(x1, ε). Again, by assumption,B(x1, ε), B(x2, ε) do not cover X so there is x3 outside of both B(x1, ε), B(x2, ε). Repeatingthis process inductively, we construct a sequence (xn) such that d(xn, xm) ≥ ε for all m 6= n.This sequence has no Cauchy subsequence and hence no convergent subsequence. This is acontradiction since we assumed our space is sequentially compact. Hence X can be coveredby finitely many balls of radius ε and so X is totally bounded.

Problem S13.4. Denote by hn the nth harmonic number:

hn = 1 +1

2+

1

3+ · · ·+ 1

n.

Prove that there is a limitγ = lim

n→∞(hn − lnn).

Solution. Let xn = hn − lnn. Define fn : [1,∞)→ R by

fn(x) =

1/j, j ≤ x < j + 1, j = 1, 2, . . . n− 1,0, x > n.

Then each fn is piecewise constant an majorizes f(x) = 1/x for x ∈ [1, n]. Then

hn = 1 +1

2+ · · ·+ 1

n=

∫ n

1

fn(x)dx ≥∫ n

1

1

xdx = lnn.

Thus xn = hn − lnn ≥ 0. Also

xn+1 − xn = hn+1 − hn + lnn− ln(n+ 1) =1

n+ 1+ lnn− ln(n+ 1).

By the mean value theorem, there is c ∈ (n, n+ 1) such that

ln(n)− ln(n+ 1) =1

c(n− (n+ 1)) = −1

c.

Then

xn+1 − xn =1

n+ 1− 1

c≤ 0.

Thus xn is a decreasing sequence which is bounded below so it converges to its infimum.

Problem S13.5. Define polynomials Un(x), n = 0, 1, 2, . . . as follows:

U0(x) = 1, U1(x) = 2x, Un(x) = 2xUn−1(x)− Un−2(x), n = 2, 3, 4, . . . .

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 69

(a) Prove that Un(cos θ) =sin((n+ 1)θ)

sin θfor all n.

(b) Prove that the polynomials satisfy∫ 1

−1

Um(x)Un(x)√

1− x2dx =π

2δm,n,

where δm,n is the Kronecker delta (i.e., δm,n = 1 if m = n and 0 otherwise).

Solution.

(a) The claim is most easily proven by induction. Note U0(cos θ) = 1 = sin((0+1)θ)sin θ

so theclaim clearly holds for n = 0. Also

U1(cos θ) = 2 cos θ =2 cos θ sin θ

sin θ=

sin(2θ)

sin θ=

sin((1 + 1))θ

sin θ.

Hence the claim also holds for n = 1.

Set n ≥ 1 and assume the claim holds for all k ≤ n. Then

Un+1(cos θ) = 2 cos θUn(cos θ)− Un−1(cos θ)

= 2 cos θsin((n+ 1)θ)

sin θ− sin(nθ)

sin θ

=2 cos θ sin((n+ 1)θ)− sin((n+ 1)θ − θ)

sin θ

=2 cos θ sin((n+ 1)θ)− (cos θ sin((n+ 1)θ)− cos((n+ 1)θ) sin θ)

sin θ

=cos θ sin((n+ 1)θ) + cos((n+ 1)θ) sin θ)

sin θ

=sin((n+ 1)θ + θ)

sin θ=

sin((n+ 2)θ)

sin θ.

This completes the induction so the claim holds for all n.

(b) Use the substitution x = cos θ. Then∫ 1

−1

Um(x)Un(x)√

1− x2dx =

∫ 0

π

Um(cos θ)Un(cos θ)√

1− cos2 θ(− sin θ)dθ

=

∫ π

0

sin((m+ 1)θ)

sin θ

sin((n+ 1)θ)

sin θsin2 θdθ

=

∫ π

0

sin((m+ 1)θ) sin((n+ 1)θ)dθ.

From here, the using the identity sinA sinB = 12(cos(A − B) − cos(A + B)) and

performing the integration gives the answer.

Problem S13.11. Define the Fibonacci sequence Fn by F0 = 0, F1 = 1 and Fn =Fn−1 + Fn−2, for n = 2, 3, 4, . . ..

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 70

(a) Show that the limit as n→∞ of FnFn−1

exists and find its value.

(b) Prove that F2n+1F2n−1 − F 22n = 1, n ≥ 1.

Solution.

(a) To find the limit, we simply solve the recurrence relation and take the limit explicitly.Assume Fn = C · xn for some real number x. Then from the recurrence relation, wesee

Cxn = Cxn−1 + Cxn−2 =⇒ x2 − x− 1 = 0.

Solving gives

x± =1±√

5

2.

In particular, |x−| < 1 andFn = C1x

n+ + C2x

n−.

We could solve for C1, C2 using F0, F1, but we will see it doesn’t matter what they are.We notice

FnFn−1

=C1x

n+ + C2x

n−

C1xn−1+ + C2x

n−1−

and since limn→∞

xn− = 0, we have

limn→∞

FnFn−1

= limn→∞

C1xn+

C1xn−1+

= x+ =1 +√

5

2≈ 1.618.

(b) We prove the claim by induction, noticing that F3F1−F 22 = 2 · 1− 12 = 1. Assume for

some n ≥ 1, we have F2n+1F2n−1 − F 22n = 1. Then

F2n+3F2n+1 − F 22n+2 = (F2n+2 + F2n+1)F2n+1 − F 2

2n+2

= F2n+2F2n+1 + F 22n+1 − F 2

2n+2

= F2n+2F2n+1 + F 22n+1 − (F2n+1 + F2n)2

= F2n+2F2n+1 − 2F2n+1F2n − F 22n

= F2n+1(F2n+2 − 2F2n)− F 22n

= F2n+1(F2n+1 − F2n)− F 22n

= F2n+1F2n−1 − F 22n = 1 by our inductive hypothesis.

Thus the claim is proven for all n ≥ 1.

Problem F14.1. Show that the function

H(x, y) = x2 + y2 + |x− y|−1

achieves its global minimum somewhere in the set (x, y) ∈ R2 : x 6= y.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 71

Solution. By symmetry, it is enough to check x > y. In that region,

∂H

∂x= 2x− 1

(x− y)2,

∂H

∂y= 2y +

1

(x− y)2.

Setting these both equal to zero, we see

2x(x− y)2 = 1, (3)

−2y(x− y)2 = 1. (4)

Adding (1),(2) gives(x− y)3 = 1 =⇒ x = y + 1.

Plugging this into (2) yields y = −1/2 and so x = 1/2. This point must be a minimum sincethere are no local/global maxima for this function (by the very nature of the square and abso-lute value functions). By symmetry, the function has global minima at (x, y) = (1/2,−1/2)and (x, y) = (−1/2, 1/2).

Problem F14.2. Let A,B be closed subsets of Rn such that A ∪ B and A ∩ B are con-nected. Prove that A is connected.

Solution. Suppose to the contrary that A is disconnected. Then there are open, disjoint,nonempty C,D such that A = C ∪D. By shrinking each of C,D and taking the closures, itis actually alright to take C,D to be closed. Then since A∩B is connected, it must be con-tained in either C orD.Without loss of generality, say A∩B ⊂ C. Then A∪B = D∪(B∪C).However, since C,D are disjoint, and since A ∩ B,D are disjoint, this shows that A ∪ B isnot connected; a contradiction. Thus A is connected.

Problem F14.3. Let fn : [0, 1]→ R be a sequence of monotonically increasing continuousfunctions. Assume that fn converge pointwise to a continuous function f : [0, 1]→ R. Showthat the convergence is uniform.

Solution. Let ε > 0.Since each of fn is increasing, by passing to the limit, we see that f is increasing. Also,

since f is continuous and [0, 1] is compact, f is uniformly continuous. Take δ > 0 such thatfor all x, y ∈ [0, 1], |x− y| < δ implies that |f(x)− f(y)| < ε/5.

Choose N ∈ N such that 1/N < δ and let 0 = x0 < x1 < . . . < xN = 1 be the uniformpartition of [0, 1]; i.e., xi = i/N, i = 0, 1, . . . , N .

Since fn → f pointwise, for each i = 0, 1, . . . , N there is Mi ∈ N such that n ≥ Mi

implies that |fn(xi)− f(xi)| < ε/5 for each i. Take M = maxM0,M1, . . . ,MN (so that Mis a uniform constant that guarantees convergence at the grid points).

For any x ∈ [0, 1], there is i = 0, 1 . . . , N − 1 such that x ∈ [xi, xi+1]. Then for n ≥ M ,

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we have

|fn(x)− f(x)| = |fn(x)− fn(xi) + fn(xi)− f(xi) + f(xi)− f(x)|≤ |fn(x)− fn(xi)|+ |fn(xi)− f(xi)|+ |f(xi)− f(x)|< fn(xi+1)− fn(xi) + ε/5 + ε/5

< f(xi+1) + ε/5− (f(xi)− ε/5) + 2ε/5

< f(xi+1)− f(xi) + 4ε/5 < ε.

Notice that M does not depend on x, thus the convergence is uniform.

Problem F14.4. Let fn : [−2, 2] → [0, 1] be a sequence of convex functions. Show thatthere is a subsequence which converges uniformly on [−1, 1].

Solution. From the range of f , we see |fn(x)| ≤ 1 so the sequence is uniformly bounded.Also, since each fn is convex, we know that

fn(s)− fn(t)

s− t

is a non-decreasing function of s and t. Then for s, t ∈ [−1, 1], we have

fn(s)− fn(t)

s− t≤ fn(2)− fn(t)

2− t≤ fn(2)− fn(1)

2− 1≤ 1.

Also,fn(s)− fn(t)

s− t≥ fn(−1)− fn(t)

−1− t≥ fn(−1)− fn(−2)

−1− (−2)≥ −1.

Thus ∣∣∣∣fn(s)− fn(t)

s− t

∣∣∣∣ ≤ 1 =⇒ |fn(s)− fn(t)| ≤ |s− t| .

Thus each fn is Lipschitz with the same Lipschitz constant. Hence the sequence is equicontin-uous. Thus by the Arzela-Ascoli Theorem, there is a subsequence that converges uniformly.

Problem F14.5. Consider the sequence

a1 =√

2 and an+1 =√

2 + an, n ≥ 1.

Prove that the sequence converges and find the limit.

Solution. We prove convergence by the monotone convergence theorem. It is clear that allterms in the sequence are non-negative. Consider, a1 ≤ 2. Also, if an ≤ 2 then

2 + an ≤ 2 =⇒√

2 + an ≤ 2 =⇒ an+1 ≤ 2.

Thus by induction, the sequence is bounded above by 2. Also a1 =√

2 ≤√

2 +√

2 = a2

and if an ≤ an+1, we have

2 + an ≤ 2 + an+1 =⇒√

2 + an ≤√

2 + an+1 =⇒ an+1 ≤ an+2.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 73

Thus by induction the sequence is increasing. Thus the sequence converges.Let limn→∞ an = a. Then by continuity, a must satisfy

a =√

2 + a =⇒ a2 − a− 2 = 0 =⇒ a = −1 or a = 2.

But all an are nonnegative, so we cannot have a = −1. Thus limn→∞ an = 2.

Problem F14.6. Let f : [0, 1]→ R be a C1 function. Prove that

limn→∞

n−1∑k=0

∣∣f (k+1n

)− f

(kn

)∣∣ =

∫ 1

0

|f ′(t)| dt.

Solution. Since f ′ is continuous so is |f ′| and since [0,1] is compact, |f ′| is uniformlycontinuous. Let ε > 0. then there is δ > 0 such that |x− y| < δ =⇒ ||f ′(x)| − |f ′(y)|| <ε. Choose N ∈ N such that 1/N < δ. For n > N , let 0 = x0 < x1 < · · · < xn = 1 bethe uniform partition of [0, 1]. By the mean value theorem, in each interval (xi, xi+1), i =0, 1, . . . , n− 1, there is yi such that

|f(xi+1)− f(xi)| = |f ′(yi)| |xi+1 − xi| = 1n|f ′(yi)| .

Putting this all together:∣∣∣∣∣n−1∑k=0

∣∣f (k+1n

)− f

(kn

)∣∣− ∫ 1

0

|f ′(x)| dx

∣∣∣∣∣ =

∣∣∣∣∣n−1∑k=0

(1

n|f ′(yi)| −

∫ xk+1

xk

|f ′(x)| dx)∣∣∣∣∣

=

∣∣∣∣∣n−1∑k=0

∫ xk+1

xk

[|f ′(yi)| − |f ′(x)|]dx

∣∣∣∣∣≤

n−1∑k=0

∫ xk+1

xk

||f ′(yi)| − |f ′(x)|| dx

<n−1∑k=0

∫ xk+1

xk

εdx =n−1∑k=0

ε

n= ε.

Thus

limn→∞

n−1∑k=0

∣∣f (k+1n

)− f

(kn

)∣∣ =

∫ 1

0

|f ′(t)| dt

as desired.

Problem S14.7. Find a doubly infinite sequence an,m, n,m ∈ Z such that for all m,∑n∈Z

an,m = 0

and for all n, ∑m∈Z

an,m = 0

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 74

with all of these series converging absolutely but such that∑n∈Z

∑m∈Z

|an,m| =∞.

Solution. Define

an,m =

1, m = n,−1, m = n+ 1, 0, m 6= n, n+ 1.

Then an,m is an “infinite matrix” with 1 along the diagonal and -1 on the first superdiagonal.Fixing m and summing down any “column,” we will get∑

n∈Z

an,m = am,m−1 + am,m = −1 + 1 = 0 and∑n∈Z

|an,m| = |am−1,m|+ |am,m| = 2.

Similarly, fixing n and summing across any “row,” we have∑m∈Z

an,m = an,n + an,n+1 = 1 + (−1) = 0 and∑m∈Z

|an,m| = |an,n|+ |an,n+1| = 2.

Thus all these sums are zero and all of them converge absolutely. However,∑n∈Z

∑m∈Z

|an,m| =∞

because we are adding 1 infinitely many times.

Problem S14.8.

(a) Prove that the function f : R→ R defined by

f(t) =

e−

1t , t > 0

0, t ≤ 0

is infinitely differentiable.

(b) Find a function ϕ : Rn → R such that ϕ(x) ≥ 0 for all x ∈ Rn, ϕ(x) = 0 if |x| > 1 and∫Rnϕ(x)dx = 1.

Solution.

(a) It is clear that f is infinitely differentiable on each of its pieces. We simply need toto prove that it is infinitely differentiable at 0. We see that, f (n)(t) = 0 for all t < 0,n ∈ N. Thus limt→0− f

(n)(t) = 0. Thus the goal is to show that limt→0+ f(n)(t) = 0 for

all n ∈ N. For t > 0,

f ′(t) =1

t2e−

1t , f ′′(t) =

(1

t4− 2

t3

)e−

1t , . . . .

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 75

In general, f (n)(t) = pn(

1t

)e−

1t , for t > 0, where pn is a polynomial of degree 2n. Thus

if we can prove that p(

1t

)e−

1t → 0 as t → 0+ for all polynomials p, then we will be

done. This is easy by a simple change of variables. Put s = 1t. This is a bijective map

for t > 0 and t→ 0+ =⇒ s→ +∞. Then

limt→0+

p(

1t

)e−

1t = lim

s→+∞p(s)e−s = 0, by L’Hopital’s rule.

Thus f is infinitely differentiable.

(b) Define ψ : Rn → R by

ψ(x) =

e− 1

1−|x|2 , |x| < 10, |x| ≥ 1.

Then ψ is nonnegative, zero outside the unit ball and infinitly differentiable by thesame reasoning as in (a). By continuity, ψ has a finite integral on the unit ball in Rn

and thus on all of Rn since it is zero outside the unit ball. Define ϕ : Rn → R by

ϕ(x) =ψ(x)∫

Rn ψ(y)dy, x ∈ Rn.

Then ϕ has all the desired properties.

Problem S14.9. Find a function that minimizes∫ 1

0|f ′(x)|2 dx among all f ∈ C1[0, 1] such

that f(0) = 0, f(1) = 1. Is the minimizing function unique?

Solution. By the Cauchy- Schwarz inequality∫ 1

0

f ′(x)dx ≤(∫ 1

0

12dx

)(∫ 1

0

f ′(x)2dx

)=

∫ 1

0

|f ′(x)|2 dx

with equality if and only if 1 and f ′(x) are linearly dependent. However, using our conditionsf(0) = 0 and f(1) = 1, we can evaluate the right hand side to see

1 ≤∫ 1

0

|f ′(x)|2 dx.

Thus we have an upper bound for all such quantities. We can achieve equality by ensuringthat 1 and f ′(x) are linearly dependent; that is f ′(x) = α, a constant. Then f(x) = αx+ βfor some constant β. But using the conditions f(0) = 0, f(1) = 1 again, we see that f(x) = x.Checking, we see that the solution does indeed meet the lower bound and it is unique sincethe Cauchy-Schwarz inequality gives necessary and sufficient conditions for equality.

Problem S14.10. Let F be a set of continuous real valued functions on [0, 1]. Assumethat

(i) F is uniformly bounded; i.e., there is M < ∞ such that |f(x)| ≤ M for all f ∈ F ,x ∈ [0, 1].

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 76

(ii) F is equicontinuous; i.e., for every ε > 0 there is δ > 0 such that for all x, y ∈ [0, 1],

|x− y| < δ =⇒ |f(x)− f(y)| < ε.

Prove that every sequence in F has a convergent subsequence.

Solution. This is the forward direction of the Arzela-Ascoli Theorem.Let ε > 0. Take δ > 0 be such that |x− y| < δ =⇒ |f(x)− f(y)| < ε/3 for all

x, y ∈ [0, 1], f ∈ F .Let (fn) be a sequence in F . Since we will be inductively creating more and more se-

quences, it is convenient to write (fn) = (f0,n). Let (xk) be a denumeration of the rationalsin [0, 1]. By uniform boundedness, (f0,n(x1)) is a bounded sequence in R. Thus by Bolzano-Weierstrass, there is a convergent subsequence (f1,n(x1)). Again, (f1,n(x2)) is bounded sothere is a convergent subsequence (f2,n(x2)). Since this is a subsequence, we still have con-vergence of (f2,n(x1)) as well.

Continuing this by induction, we create subsequences (fm,n) such that (fm,n) is a subse-quence of (fm−1,n) for all m and (fm,n(xj)) converges for all j ≤ m. Taking gn = fn,n forall n ∈ N, we see that (gn) is a subsequence of (fn) and (gn(x)) converges for all rationalx ∈ [0, 1] by construction. We prove that (gn) is uniformly Cauchy and thus convergesuniformly.

The open intervals (xk−δ, xk+δ) create an open cover of [0, 1] since the rationals are dense.By compactness, there is a finite subcover: (xk1−δ, xk1 +δ), . . . , (xkN −δ, xkN +δ). Since (gn)converges at each rational (and is thus Cauchy at each rational), there are K1, . . . , Kj ∈ Nsuch that n,m ≥ Kj implies

|gn(xk)− gm(xk)| < ε/3,

for each j = 1, . . . , N . Take K = maxjKj.Let x ∈ [0, 1]. Then x ∈ (xkj − δ, xkj + δ) for some j = 1, . . . , N . Then

|gn(x)− gm(x)| =∣∣gn(x)− gn(xkj) + gn(xkj)− gm(xkj) + gm(xkj)− gm(x)

∣∣≤∣∣gn(x)− gn(xkj)

∣∣︸ ︷︷ ︸<ε/3 by equicontinuity

+∣∣gn(xkj)− gm(xkj)

∣∣︸ ︷︷ ︸<ε/3 by convergence at xkj

+∣∣gm(xkj)− gm(x)

∣∣︸ ︷︷ ︸<ε/3 by equicontinuity

< ε.

Thus (gn) is uniformly Cauchy and so (fn) has a uniformly convergent subsequence.

Problem S14.11. Let F be a set of continuous real values functions on [0, 1]. Assumethat every sequence in F has a uniformly convergent subsequence. Prove that

(i) F is uniformly bounded; i.e., there is M < ∞ such that |f(x)| ≤ M for all f ∈ F ,x ∈ [0, 1].

(ii) F is equicontinuous; i.e., for every ε > 0 there is δ > 0 such that for all x, y ∈ [0, 1],

|x− y| < δ =⇒ |f(x)− f(y)| < ε.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 77

Solution. This is the backwards direction of the Arzela-Ascoli Theorem. We prove both(i) and (ii) by contradiction. We denote

||f ||∞ = sup0≤x≤1

|f(x)| , f ∈ F .

(i) Assume that F is not uniformly bounded. Then for each n ∈ N there is a function fn ∈F such that ||fn||∞ > n. By assumption, this sequence (fn) must have a subsequence(gn) converging uniformly to some f [it may not be the case that f ∈ F ]. Then f mustbe continuous as a uniform limit of continuous function. Then

n ≤ ||gn||∞ ≤ ||gn − f ||∞ + ||f ||∞ .

Taking the limit as n → ∞, we see that ||gn − f ||∞ → 0 by uniform convergence andso ||f ||∞ is unbounded. But this is impossible since f is a continuous function on acompact set. This contradiction implies that F is uniformly bounded.

(ii) Assume that F is not equicontinuous. Then there exists an ε > 0 such that for anyδ > 0, there are x, y ∈ [0, 1], f ∈ F such that |x− y| < δ but |f(x)− f(y)| ≥ ε.Choosing δ = 1

nfor each n ∈ N, we get sequences (xn), (yn) in [0, 1] and (fn) in F such

that

|xn − yn| <1

nbut |fn(xn)− fn(yn)| ≥ ε.

This sequence (fn) must have a subsequence (gn) converging uniformly to some f[again, not necessarly f ∈ F but f will be continuous since each gn is continuous].Since f is continuous, there is N1 sufficiently large so that |x− y| < 1/N1 =⇒|f(x)− f(y)| < ε/4. Further there isN2 such that n ≥ N2 implies that |gn(x)− f(x)| <ε/4 for all x ∈ [0, 1] (by uniform convergence). Then for n ≥ maxN1, N2,

ε ≤ |gn(xn)− gn(yn)| = |gn(xn)− f(xn) + f(xn)− f(yn) + f(yn)− gn(y)|= |gn(xn)− f(xn)|+ |f(xn)− f(yn)|+ |f(yn)− gn(yn)|< ε/4 + ε/4 + ε/4 = 3ε/4.

This is a contradiction. Thus F must be equicontinuous.

Problem S14.12. Assume [0, 1] =∞⋃n=1

In where In = [an, bn] 6= ∅ and

In ∩ Im = ∅

when n 6= m.

(a) Let E = an : n ≥ 1 ∪ bn : n ≥ 1. Prove that E is closed.

(b) Prove no such family of intervals In exists.

Solution.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 78

(a) Let x ∈ [0, 1]. Then since the given intervals are disjoint, there is a unique n ∈ N suchthat x ∈ [an, bn]. Then x 6∈ E implies x ∈ (an, bn). Conversely, if x ∈ (an, bn) for somen ∈ N then x 6∈ E. Hence

[0, 1]− E =∞⋃n=1

(an, bn)

and so [0, 1]− E is open since it is a union of open sets. Thus E is closed.

(b) No idea.

Problem S15.1. Let f : [0,∞)→ [0,∞) be continuous with f(0) = 0. Show that if

f(t) ≤ 1 +1

10f(t)2, for all t ≥ 0

then f is bounded on [0,∞).

Solution. Assume there is t ∈ [0,∞) such that f(t) ≥ 2. Then by continuity, there mustbe s ∈ [0,∞) such that f(s) = 2. But then

2 = f(s) ≤ 1 +1

10f(s)2 ≤ 1 +

4

10= 1.4;

a contradiction. Hence f(t) < 2 for all t ∈ [0,∞). By assumption f is a non-negativefunction so this implies f is bounded.

Problem S15.2. Let f : [0, 1]→ R, α ∈ (0, 1). We say that f is α-Holder continuous (andwrite f ∈ Cα[0, 1]) if

||f ||Cα = supx∈[0,1]

|f(x)|+ supx,y∈[0,1],x 6=y

|f(x)− f(y)||x− y|α

<∞.

This defines a norm on Cα[0, 1] (no need to prove this).Prove that any bounded sequence in C1/2[0, 1] admits a convergent subsequence in C1/3[0, 1].

Solution. Let (fn) be a bounded sequence in C1/2[0, 1]; say M is the bound. Then in

particular, for all x, y ∈ [0, 1], |x− y| ≤ 1 so |x− y|1/2 ≤ |x− y|1/3 . Then 1

|x−y|1/3≤ 1

|x−y|1/2so

||fn||C1/3 = supx∈[0,1]

|fn(x)|+ supx,y∈[0,1],x 6=y

|fn(x)− fn(y)||x− y|1/3

≤ supx∈[0,1]

|fn(x)|+ supx,y∈[0,1],x 6=y

|fn(x)− fn(y)||x− y|1/2

= ||fn||C1/2 ≤M.

Thus (fn) is uniformly bounded in C1/3[0, 1]. Further, it is clear from ||fn||C1/3 ≤M that

|fn(x)− fn(y)| ≤M |x− y|1/3 .

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 79

Let ε > 0. Then putting δ = (ε/M)3, we see that

|x− y| < δ =⇒ |fn(x)− fn(y)| < M((ε/M)3)1/3 = ε

for all n ∈ N and x, y ∈ [0, 1]. Thus (fn) is equicontinuous. Hence by Arzela-Ascoli, there isa subsequence of (fn) which converges uniformly in C1/3[0, 1].

Problem S15.3. Let f : R→ R be a Lipschitz function. Suppose that for every x ∈ R,

limn→∞

n[f(x+ 1

n

)− f(x)

]= 0.

Prove that f is differentiable.

Solution. Since f is Lipschitz, it is differentiable Lebesgue almost everywhere and

f(x) = f(0) +

∫ x

0

f ′(t)dλ(t)

for all x ∈ R where λ is the Lebesgue measure on R (actually absolute continuity is enoughto guarantee this; Lipschitz continuity is a stronger condition). Wherever the derivative isdefined, the given assumption implies that the derivative is zero by the sequential criteriontheorem. Thus f ′(t) exists and is zero for Lebesgue almost every t ∈ R. But then∫ x

0

f ′(t)dλ(t) = 0, for all x ∈ R.

Hence f(x) = f(0) for all x ∈ R. Thus f is constant which certainly implies that it isdifferentiable.

[Note: I’m not sure this was the intended solution since the concepts of absolute conti-nuity and the Lebesgue measure aren’t covered in boot camp but I can’t find an elementarysolution to this problem which uses only the definition of the derivative and related concepts.]

Problem S15.4. Let f : [0, 1] → R be a function satisfying the intermediate value prop-erty; that is, whenever 0 ≤ a < b ≤ 1 and y lies between f(a) and f(b), there is x ∈ (a, b)such that f(x) = y. Assume that for any y ∈ R, the preimage f−1 (y) is closed. Provethat f is continuous.

Solution. Suppose f is discontinuous at some point x ∈ R. Then there is an ε > 0 suchthat for any δ > 0 there is x0 such that |x− x0| < δ but |f(x)− f(x0)| ≥ ε. Specifically, foreach n ∈ N, there is xn ∈ R such that |x− xn| < 1

nbut

|f(xn)− f(xn)| ≥ ε.

Each f(xn) is either greater than or less than f(x). Thus there are either infinitely many nsuch that f(xn) ≥ f(x) + ε or infinitely many n such that f(xn) ≤ f(x) − ε. Without lossof generality, we can assume there are infinitely many n such that f(xn) ≥ f(x) + ε. Weconsider this subsequence of (xn) but for notational convenience, we still call it (xn).

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 80

Take c ∈ (f(x), f(xn)) for each n. Note, c need not depend on n because |f(x)− fn(x)| ≥ε for all n. Then by the intermediate value property, there is yn ∈ (x, xn) such that f(yn) = c.Clearly the sequence yn converges to x because for each n, we have |x− yn| < |x− xn| < 1

n.

But each yn ∈ f−1 (c) which is closed by assumption and so x ∈ f−1 (c) which impliesf(x) = c. This is a contradiction because we assumed that f(x) < c.

Hence f is continuous at all x ∈ R.

Problem S15.5. Let f : [1,∞) → [0,∞) be a bounded decreasing function such thatlimx→∞ f(x) = 0. Prove that ∫ n+1

1

f(x)dx−n∑k=1

f(k)

tends to a finite limit as n→∞.

Solution. Write

xn =

∫ n+1

1

f(x)dx−n∑k=1

f(k), n ∈ N.

We prove that xn is a Cauchy sequence and hence converges.Consider, for n > m, we have

xn − xm =

∫ n+1

1

f(x)dx−n∑k=1

f(k)−∫ m+1

1

f(x)dx+m∑k=1

f(k)

=

∫ n+1

m+1

f(x)−n∑

k=m+1

f(k)

=n∑

k=m+1

(∫ k+1

k

f(x)dx− f(k)

).

Since f is decreasing, we have f(k + 1) ≤∫ k+1

kf(x)dx ≤ f(k) for all k. Then

−f(m+1) ≤ f(n+1)−f(m+1) =n∑

k=m+1

(f(k + 1)− f(k)) ≤ xn−xm ≤n∑

k=m+1

(f(k)− f(k)) = 0.

Let ε > 0. Since f(x) → 0 as x → ∞, there is N ∈ N such that x ≥ N =⇒ f(x) < ε.Taking n,m ≥ N , we see |xn − xm| < ε. Thus (xn) is Cauchy and hence convergent.

Problem S15.6. Prove that the integral equation

f(x) = ex2

+ 12

∫ 1

0

cos(y)f(y)dy

admits a unique continuous solution f : [0, 1]→ R.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 81

Solution. Define the operator T on C[0, 1] by

(Tf)(x) = ex2

+ 12

∫ 1

0

cos(y)f(y)dy, x ∈ [0, 1], f ∈ C[0, 1].

Since C[0, 1] is complete with respect to the supremum norm, it will suffice to prove thatT : C[0, 1] → C[0, 1] and that T is a contraction; then the Banach Fixed Point Theoremassures the existence of a unique fixed point for T which will be a unique solution to theintegral equation in C[0, 1].

If f ∈ C[0, 1], then∫ 1

0cos(y)f(y)dy converges, so it is clear that Tf is a well-defined

and continuous function since it is a constant added to the continuous function ex2. Thus

T : C[0, 1]→ C[0, 1].Take f, g ∈ C[0, 1]. Then for x ∈ [0, 1],

|(Tf)(x)− (Tg)(x)| =∣∣∣∣ex2 + 1

2

∫ 1

0

cos(y)f(y)dy − ex2 − 12

∫ 1

0

cos(y)g(y)dy

∣∣∣∣≤ 1

2

∫ 1

0

cos(y) |f(y)− g(y)| dy

≤ 12

∫ 1

0

|f(y)− g(y)| dy

≤ 12||f − g||∞ .

Since this holds for all x ∈ [0, 1], we see

||Tf − Tg||∞ ≤12||f − g||∞

so T is a contraction. Thus there is a unique function f such that Tf = f . This function isthe unique solution to the integral equation.

Note: it is likely that the integral equation was supposed to read

f(x) = ex2

+ 12

∫ x

0

cos(y)f(y)dy.

In this case, the property still holds but we need a bit more work to prove continuity of T.Let ε > 0, f ∈ C[0, 1]. Then since ex

2is continuous (and thus uniformly continuous on

[0, 1]), there is δ > 0 such that for all x, y ∈ [0, 1],

|x− y| < δ =⇒∣∣∣ex2 − ey2∣∣∣ < ε

2.

Also, since f is continuous on a compact set, it is bounded by some M ∈ R. We can takeδ < ε/2M by making it smaller if necessary. Then for |x− y| < δ (wlog x > y), we see

|(Tf)(x)− (Tf)(y)| =∣∣∣∣ex2 − ey2 +

∫ x

y

cos(z)f(z)dz

∣∣∣∣≤∣∣∣ex2 − ey2∣∣∣+

∫ x

y

|f(z)| dz

< ε/2 +M(x− y) < ε/2 +M(ε/2M) = ε.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 82

Thus Tf is continuous. The proof that T is a contraction follows exactly as above.

Problem F15.1. Let (an)∞n=1 be a sequence of positive real numbers such that

an+m ≤ an + am, n,m ≥ 1.

Prove that limn→∞

ann

exists.

Solution. Using the “subadditivity,” it is not difficult to see that for any n,m ≥ 1, we have

anm = an+···+n ≤ man.

Clearly the set a1/1, a2/2, . . . is bounded below (by zero) and so it has a finite infimum

L. We show that limn→∞

ann

= L.

Let ε > 0. Since L is the infimum of the sequence, there is k ∈ N such that ak/k ≤ L+ε/2.There is also N ∈ N such that N ≥ k and

maxa1, . . . , ak−1N

≤ ε/2.

Then for n ≥ N , write n = dk + r for some d, r ∈ N, r ≤ k. Then

L ≤ ann

=adk+r

dk + r≤ adk + ar

dk + r=

adkdk + r

+arn≤ dak

dk+

maxa1, . . . , ak−1N

≤ akk

+ε/2 ≤ L+ε.

Thus by definition, limn→∞

ann

= L.

Problem F15.2. Let a, b ∈ R, a < b. Show that if g, h : [a, b] → R are continuous withh ≥ 0, then there is c ∈ [a, b] such that∫ b

a

g(x)h(x)dx = g(c)

∫ b

a

h(x)dx.

Solution. If h ≡ 0 and/or g is constant, the conclusion is trivial and actually holds for allc ∈ [a, b].

Assume h 6≡ 0 and g is not constant. Then a standard result tells us that∫ bah(x)dx > 0.

We seek to prove that there is c ∈ [a, b] such that g(c) =∫ ba g(x)h(x)dx∫ ba h(x)dx

.

Assume no such c exists. If there are values c+, c− ∈ [a, b] such that

g(c+) >

∫ bag(x)h(x)dx∫ bah(x)dx

and g(c−) <

∫ bag(x)h(x)dx∫ bah(x)dx

then we have violated the intermediate value theorem and we are done. Otherwise one ofc+, c− doesn’t exists, and then

g(c) is always greater than or always less than

∫ bag(x)h(x)dx∫ bah(x)dx

.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 83

Assume wlog that the latter is true. Then∫ b

a

(g(x)− g(c))h(x)dx > 0

for all c ∈ [a, b].Since [a, b] is compact and g is continuous, there is z ∈ [a, b] such that g(z) ≥ g(x) for

all x ∈ [a, b]. Then g(x) − g(z) ≤ 0, for all x ∈ [a, b]. Then (g(x) − g(z))h(x) ≤ 0 for allx ∈ [a, b]. But this implies that ∫ b

a

(g(x)− g(z))h(x)dx ≤ 0

which contradicts the above statement. This implies the existence of some such c.

Problem F15.3. Let fn be a sequence of continuous functions fn : [−1, 1]→ [0, 1] suchthat for every x ∈ [−1, 1]

(a) the sequence of numbers fn(x) is non-increasing, and

(b) limn→∞

fn(x) = 0.

Define

gn(x) =n∑

m=1

(−1)mfm(x).

Prove that gn converges pointwise to some function g on [−1, 1] and that g is continuous.

Solution. The series

limn→∞

gn(x) = limn→∞

n∑m=1

(−1)mfm(x)

converges for each x ∈ [−1, 1] by the alternating series test. Define

g(x) = limn→∞

n∑m=1

(−1)mfm(x), x ∈ [−1, 1].

It remains to prove that g is continuous.We first show that fn → 0 uniformly (this is actually called Dini’s Theorem and proven

a bit more generally in F12.3). Indeed, let ε > 0 and define An = x ∈ [−1, 1] : fn(x) < ε.Each An is open since it is the pullback of an open set under a continuous function. Sincefn(x)→ 0 for all x ∈ [−1, 1], we see that

[−1, 1] =∞⋃n=1

An.

Then since [−1, 1] is compact there is a finite subcover An1 , . . . , Ank . Then

[−1, 1] =k⋃`=1

An` .

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 84

However, by construction An ⊂ Am when n ≤ m. Thus [−1, 1] = Ank . Then for all n ≥ nk,we have fn(x) < ε for all x ∈ [−1, 1]. Thus fn → 0 uniformly.

Take n,m ∈ N (wlog n ≥ m). Then

|gn(x)− gm(x)| = |fm(x)− fm+1(x)± · · · ± fn(x)| .

However, since fn(x) is non-increasing for each x, we see that fk+1(x)− fk(x) ≤ 0 for all x.Then

|gn(x)− gm(x)| ≤ fm(x).

Since fm → 0 uniformly, this shows that gn is uniformly Cauchy and hence converges uni-formly. Clearly each gn is continuous since it is a finite sum of continuous functions. So g isthe uniform limit of continuous functions and is thus continuous.

Problem F15.4. Let fn : [0,∞) → R be functions defined recursively by f0(x) = 0, x ∈[0,∞) and

fn+1(x) = e−2x +

∫ x

0

fn(t)e−2tdt, n ≥ 1, x ∈ [0,∞).

Show that f(x) = limn→∞

fn(x) exists and explicitly find f .

Solution. Let V = g ∈ C[0,∞) : ||f ||∞ ≤ 2. Assume that gn is a sequence in V whichconverges to some function g ∈ C[0,∞). Then by the continuity of the norm, we see

||g||∞ =∣∣∣∣∣∣ limn→∞

gn

∣∣∣∣∣∣∞

= limn→∞

||gn||∞ ≤ 2.

Thus V is a closed subspace of a complete space and is thus complete.For f ∈ V , define

(Tf)(x) = e−2x +

∫ x

0

f(t)e−2tdt, x ∈ [0,∞).

Then f0 = T 0f0, f1 = Tf0, f2 = T 2f0, . . . . Thus it will suffice to show that limn→∞ Tnf0

exists.It is clear that T is a linear operator; we must prove that T : V → V and that T is

a contraction. Then by the Banach Fixed Point theorem, T has a unique fixed point f .Moreover, to find the fixed point, we can begin from an arbitrary member of V and iteratewith T . That is limn→∞ fn = f where Tf = f .

Let f ∈ V . Then for x ∈ [0,∞),

|(Tf)(x)| =∣∣∣∣e−2x +

∫ x

0

f(t)e−2tdx

∣∣∣∣≤∣∣e−2x

∣∣+

∫ x

0

|f(t)| e−2tdt

≤ 1 + 2

∫ x

0

e−2tdt

≤ 1 + 2

∫ ∞0

e−2tdt = 2.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 85

Thus ||Tf ||∞ ≤ 2 when ||f ||∞ ≤ 2 and so T : V → V .Let f, g ∈ V .Then

|(Tf)(x)− (Tg)(x)| =∣∣∣∣e−2x +

∫ x

0

f(t)e−2tdt− e−2x −∫ x

0

g(t)e−2tdt

∣∣∣∣=

∣∣∣∣∫ x

0

(f(t)− g(t))e−2tdt

∣∣∣∣≤∫ x

0

|f(t)− g(t)| e−2tdt

≤ ||f − g||∞∫ x

0

e−2tdt

≤ ||f − g||∞∫ ∞

0

e−2tdt = 12||f − g||∞ .

Thus||Tf − Tg||∞ ≤

12||f − g||∞

and so T is a contraction.Thus f = lim

n→∞fn = lim

n→∞T nf0 exists. Further, f(x) is continuous and

Tf = f =⇒ f(x) = e−2x +

∫ x

0

f(t)e−2tdt, x ∈ [0,∞).

However, if f is continuous, then∫ x

0f(t)e−2tdt is differentiable, so f is differentiable. Differ-

entiating the above equation, we see f satisfies the differential equation

f ′(x)− e−2xf(x) = −2e−2x

f(0) = 1.

Using an integrating factor, we see

d

dx

(f(x) exp

(12e−2x

))= −2e−2x exp

(12e−2x

)which implies that

f(x) exp(

12e−2x

)− e1/2 = −2

∫ x

0

e−2t exp(

12e−2t

)dt = 2

(exp

(12e−2t

)− e1/2

).

Then

f(x) = exp(

12− 1

2e−2x

)+ 2

(1− exp

(12− 1

2e−2x

))= 2− exp

(12− 1

2e−2x

).

Problem F15.5. Let F : R3 → R be differentiable. Suppose F (0, 0, 0) = 0 and that nocomponent of ∇F is 0 at (0, 0, 0). Show that if x = x(y, z), y = y(x, z), z = z(x, y) define thesurface F (x, y, z) = 0 in a neighborhood of (0, 0, 0), then

∂x

∂y

∂y

∂z

∂z

∂x= −1.

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 86

Solution. Alternately using the implicit value theorem in each coordinate, we get differen-tiable functions x(y, z), y(x, z), z(x, y) such that

F (x(y, z), y, z) = 0,

F (x, y(x, z), z) = 0,

F (x, y, z(x, y)) = 0,

in some neighborhood of (0, 0, 0). Differentiate the first equation with respect to y, thesecond with respect to z and the third with respect to x to get

∂F

∂x

∂x

∂y+∂F

∂y= 0,

∂F

∂y

∂y

∂z+∂F

∂z= 0,

∂F

∂z

∂z

∂x+∂F

∂x= 0.

Rearranging and using that no component ∇F is 0 in a neighborhood of (0, 0, 0) [which,incidentally, we also needed to call on the implicit function theorem] yields the result.

Problem F15.6. Let X = R− 0. Find a metric ρ on X such that

(i) (X, ρ) is a complete metric space, and

(ii) for any sequence (xn) in X and x ∈ X,

|xn − x| → 0 if and only if ρ(xn, x)→ 0 (as n→∞).

Solution. We immediately see that any that any sequence in R which goes to zero inabsolute value cannot be a Cauchy sequence with respect to ρ. A first thought may be

ρ(x, y) =|x− y||xy|

=

∣∣∣∣1x − 1

y

∣∣∣∣ , x, y ∈ X

since this will blow up for sequences that go to zero. However, we also need to require thatany sequence that does not converge in absolute value is not a Cauchy sequence in ρ andthis doesn’t hold for ρ defined above because that choice of ρ makes the natural numbers aCauchy sequence. To make sure this doesn’t happen, we modify ρ a bit by setting

ρ(x, y) = max

|x− y| ,

∣∣∣∣1x − 1

y

∣∣∣∣ , x, y ∈ X.

If (xn) is a sequence in X and x ∈ X, then certainly ρ(xn, x)→ 0 implies that |xn − x| → 0.Conversely, if xn → x 6= 0, we know that 1

xn→ 1

xand so |xn − x| → 0 implies that

ρ(xn, x) → 0. Finally, if (xn) is a Cauchy sequence in (X, ρ) then it is also a Cauchysequence in R with the absolute value so there is limit x ∈ R. We simply need to prove that

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Christian Parkinson UCLA Basic Exam Solutions: Analysis 87

x 6= 0 (i.e., x ∈ X) to conclude that X is complete. Since (xn) is Cauchy in (X, ρ), there isN ∈ N, so that m,n ≥ N give

|xn − xm||xnxm|

< 1 =⇒ |xn − xm| < |xnxm| .

If x = 0 [and thus xn → 0], then there is M ∈ N so that m ≥ M gives |xm| < 1 and so forn,m ≥ maxM,N we have

|xn| > |xn − xm| .

Plugging in m = M and using the reverse triangle inequality gives that

|xn| ≥ |xM | − |xn| =⇒ |xn| ≥ 12|xM | > 0, n ≥ maxM,N.

But this directly contradicts that xn → 0 since it gives a positive lower bound for sufficientlylarge n. Hence we cannot have x = 0 and we conclude that X is complete.