HW #1 SOLUTIONS

1. Let f : [a, b] → R and let v : I → R, where I is an interval containing f([a, b]). Suppose that f is con-tinuous at x ∈ [a, b]. Suppose that lims→f(x) v(s) = v(f(x)). Using the ε − δ definition of limit, prove thatlimt→x v(f(t)) = v(f(x)).

Solution.

Take an ε > 0. Since lims→f(x) v(s) = v(f(x)), there exists ε1 > 0 such that if s ∈ I and |s − f(x)| < ε1then |v(s) − v(f(x))| < ε. Now, f is continuous at x, so there exists δ > 0 such that if t ∈ [a, b] is suchthat |t − x| < δ then |f(t) − f(x)| < ε1. Since I contains f([a, b]), we automatically have f(t) ∈ I, and also|f(t) − f(x)| < ε1 from above, so |v(f(t)) − v(f(x))| < ε as desired. This shows that limt→x v(f(t)) = v(f(x)),i.e. that the composition v ◦ f is continuous at x.

2. Let g : I → R be a function, where I is an interval containing 0 in its interior. Suppose limx→0g(x)x = 0.

Define

f(x) =

{g(x) sin 1

x if x 6= 0,0 if x = 0

Prove that f ′(0) = 0.

Solution.

By definition, f ′(0) = limx→0f(x)−f(0)

x−0 = limx→0g(x) sin 1

x

x . We have the inequalities 0 ≤∣∣∣ g(x) sin 1

x

x

∣∣∣ ≤ ∣∣∣ g(x)x ∣∣∣for x 6= 0, since | sin y| ≤ 1 for all y ∈ R, in particular for y = 1

x . Since we are given that limx→0g(x)x = 0, we

also have limx→0

∣∣∣ g(x)x ∣∣∣ = 0. Of course, limx→0 0 = 0, so by the Squeeze Theorem, limx→0

∣∣∣ g(x) sin 1x

x

∣∣∣ = 0, or

limx→0g(x) sin 1

x

x = 0.

Remark. We’re using the easy-to-check fact that for a function h : (a, b)→ R and t ∈ (a, b), limx→t h(x) = 0 if andonly if limx→t |h(x)| = 0. To see this, we need only unravel the definition of limit in this context: limx→t h(x) = 0says that for all ε > 0, there exists δ > 0 such that if |x − t| < δ then |h(x) − 0| < ε, whereas limx→t |h(x)| = 0says that for all ε > 0, there exists δ > 0 such that if |x − t| < δ then ||h(x)| − 0| < ε. Note however that||h(x)| − 0| = |h(x)| = |h(x)− 0| so the two statements are saying the same thing.

3. Let f, g : R→ R be differentiable functions satisfying f ′(x) > g′(x) for all x ∈ R and f(0) = g(0). Prove thatf(x) > g(x) for x > 0 and f(x) < g(x) for x < 0.

Solution.

Let h : R → R be defined by h(x) = f(x) − g(x) for x ∈ R. Then h(0) = f(0) − g(0) = 0 and h′(x) =1

2 HW #1 SOLUTIONS

f ′(x) − g′(x) > 0 if x ∈ R . If x > 0, then by the Mean Value Theorem, h(x) = h(x) − h(0) = h′(c)(x − 0) forsome c between 0 and x. But then h′(c) = f ′(c)− g′(c) > 0, so h(x) = h′(c)x > 0.

If x < 0, then the same computation h(x) = h′(c)x for some c between 0 and x shows that h(x) < 0 be-cause h′(c) > 0 and x < 0.

4. Rudin Ch 5. Exercise #7. Suppose f ′(x), g′(x) exist, g′(x) 6= 0, and f(x) = g(x) = 0. Prove that

limt→x

f(t)

g(t)=f ′(x)

g′(x).

This holds also for complex functions.

Solution.

limt→x

f(t)

g(t)= lim

t→x

f(t)−f(x)t−x

g(t)−g(x)t−x

, and we are given that limt→xf(t)−f(x)

t−x = f ′(x) and limt→xg(t)−g(x)t−x = g′(x) 6= 0

both exist, hence the ratio limt→xf(t)g(t) = f ′(x)

g′(x) (note that we need g′(x) = 0).

5. Rudin Ch 5. Exercise #13. Suppose a and c are real numbers, c > 0, and f is defined on [−1, 1] by

f(x) =

{|x|a sin(|x|−c) (if x 6= 0),0 (if x = 0).

Prove the following statements:(a) f is continuous if and only if a > 0.(b) f ′(0) exists if and only if a > 1.(c) f ′ is bounded if and only if a ≥ 1 + c.(d) f ′ is continuous if and only if a > 1 + c.(e) f ′′(0) exists if and only if a > 2 + c.(f) f ′′ is bounded if and only if a ≥ 2 + 2c.(g) f ′′ is continuous if and only if a > 2 + 2c.

Solution.

(a) Suppose a ≤ 0. Then for each positive integer k, consider a real number xk =(

1π2 +2πk

)1/c∈ (0, 1] ⊂ [−1, 1].

Then, f(xk) = |xk|a sin(|xk|−c) =(

1π2 +2πk

)a/csin(π2 + 2πk) = (π2 + 2πk)−a/c. If a = 0, then limk→∞ f(xk) = 1,

whereas if a < 0 then limk→∞ f(xk) =∞. In either case, we have limk→∞ f(xk) 6= f(0), whereas limk→∞ xk = 0(c > 0). Therefore f is not continuous at 0 (recall that a function f defined on a neighborhood of 0 is continuousat 0 if and only if limk→∞ f(xk) = f(0) for all sequences {xk}k such that limk→∞ xk = 0). In fact, we have shownthat f has neither the left nor the right limit at 0.

Suppose now that a > 0. Then if x 6= 0, we have |f(x)| = |x|a| sin(|x|−c) ≤ |x|a, and for x = 0 of coursewe have the same inequality (since a > 0, |0|a = 0 is defined also of course). Since limx→0 |x|a = 0, we havelimx→0 |f(x)| = 0 by the Squeeze Theorem.

HW #1 SOLUTIONS 3

(b) f ′(0) = limx→0f(x)−f(0)

x−0 = |x|a sin(|x|−c)x if the limit exists.

If a ≤ 1, then limx→0+ xa−1 sin(|x|−c) does not exist, as in part (a) by considering the sequence xk =

(1

π2 +2πk

)1/c.

Therefore we need a > 1. In this case,∣∣∣ |x|a sin(|x|−c)

x

∣∣∣ = |x|a−1 sin(|x|−c) ≤ |x|a−1 and the latter tends to 0

as x→ 0, so f ′(0) exists by the Squeeze Theorem and f ′(0) = 0 in fact in this case.

(c) Note first that f(x) = f(−x) if x ∈ [−1, 1], i.e. f is an even function. For x 6= 0, f is certainly differ-entiable at x since f is given by the expression xa sin(x−c) if x > 0 on a neighborhood of x not containing 0 andby (−x)a sin((−x)c) on a neighborhood of x not containing 0 and both expressions are products of compositionsof differentiable functions, hence differentiable. We assume that a > 1 from part (b) to discuss the existence off ′ on [−1, 1] in the first place.

For x > 0, f ′(x) = axa−1 sin(x−c) + xa cos(x−c)(−cx−c−1) = axa−1 sin(x−c) − cxa−c−1 cos(x−c) (recall theremark above that f is given by x 7→ xa sin(x−c) in a neighborhood of x in [−1, 1] not containing 0). Also fora > 1, f ′(0) exists and f ′(0) = 0.

For x < 0, f is again differentiable at x because f(x) = f(−x) expresses f as a composition of the differen-tiable function f |(0,1] (this is the restriction of f on (0, 1]) together with the differentiable function x 7→ −x on aneighborhood of x not containing 0. Using the chain rule now, we have f ′(x) = −f ′(−x), so that

f ′(x) =

axa−1 sin(x−c)− cxa−c−1 cos(x−c) if x > 00 if x = 0c|x|a−c−1 cos(|x|−c)− a|x|a−1 sin(|x|−c) if x < 0

At any rate, we then need only analyze the boundedness of the expression axa−1 sin(x−c) − cxa−c−1 cos(x−c)on (0, 1] (for 0 < x ≤ 1). Since a > 1, the first term has absolute value |axa−1 sin(x−c)| ≤ a|x|a−1, which tends to0 as x→ 0+ because a > 1. It is thus bounded. For the latter term, |cxa−c−1 cos(x−c)| ≤ cxa−c−1, and this tends

to 0 as x→ 0+ if a ≥ 1+ c. In the case a < 1+ c, we take the sequence {xk}k≥1 defined by xk =(

12πk

)1/c ∈ (0, 1],

which tends to 0 as k →∞, and yet f(xk) = −c(

12πk

)(a−c−1)/c → −∞ as k →∞, showing that f is not boundedon (0, 1] if a < 1 + c.

(d) We assume f ′ exists throughout [−1, 1], i.e. a > 1.

Assume first that a > 1+ c, then from f ′(x) = −f ′(−x) again, we need only show that limx→0+ f′(x) = f ′(0) = 0

to check contiuity of f ′ (it is continuous on (0, 1]) because it is expressed on a neighborhood of a point in(0, 1] by a combination of products, addition, and composition of differentiable functions). limx→0+ f

′(x) =limx→0+(axa−1 sin(x−c)− cxa−c−1 cos(x−c)), and the first term has limit limx→0+ ax

a−1 sin(x−c) = 0 since a > 1(Squeeze Theorem yet again here). For the latter term, note that limx→0+ −cxa−c−1 cos(x−c) = 0 again by theSqueeze Theorem because a > 1 + c.

Assume now that f ′ is continuous on [−1, 1]. Then f ′([−1, 1]) is compact hence bounded, so part (c) implies thata ≥ 1 + c. We need only rule out the case a = 1 + c then to show that a > 1 + c. So suppose a = 1 + c. Thenfor x > 0, f ′(x) = axc sin(x−c)− c cos(x−c). Now take the sequence xk = (2πk)−1/c, which has the properly thatxk → 0+ as k →∞, and f ′(xk) = −c for all k. That is, limk→∞ f ′(xk) = −c 6= 0 = f ′(0), so f ′ is not continuousat 0.

(e) We assume here that f ′ is continuous throughout [−1, 1], i.e. a > 1 + c by part (d).

f ′′(0) = limx→0f ′(x)−f ′(0)

x−0 . For x < 0, f ′(x)−f ′(0)x−0 = −f ′(−x)−f ′(0)

x−0 = f ′(−x)−x , and for x > 0, f ′(x)−f ′(0)

x−0 =

4 HW #1 SOLUTIONS

f ′(x)x . Therefore f ′′(0) exists if and only if the right hand limit limx→0+

f ′(x)x = limx→0+(axa−2 sin(x−c) −

cxa−c−2 cos(x−c)) exists and if this right hand limit exists then its limit is f ′′(0). Write g(x) = axa−2 sin(x−c)−cxa−c−2 cos(x−c) then.

Suppose a < 2 + c. Then take the sequence xk = (2πk)−1/c. Then xk → 0 as k → ∞ whereas g(xk) =−c(2πk)(c+2−a)/c, so that limk→∞ g(xk) = −∞, so f ′′(0) does not exist.

Suppose a = 2 + c. Then take xk = (πk)−1/c, so g(xk) = −c(−1)k, and limk→∞ g(xk) does not exist.

Suppose a > 2 + c. Then the first term |axa−2 sin(x−c)| ≤ axa−2 and this tends to 0 as x → 0. For thesecond term, |cxa−c−2 cos(x−c)| ≤ cxa−c−2 and again this tends to 0 as x → 0. So by the Squeeze Theorem,f ′′(0) exists.

(f) Assume now that a > 2 + c. We know from the above computation in part (e) that f ′′(0) = 0. f ′ is expressedby differentiable function on [−1, 0) ∪ (0, 1], so f ′ is differentiable on it. Therefore with the assumption thtaa > 2+c, f ′ is differentiable on [−1, 1]. We have f ′(x) = −f ′(−x), so by the Chain Rule, f ′′(x) = f ′′(−x), i.e. f ′′

is an even function and it suffices to compute f ′(x) for x > 0, i.e. ddx (axa−1 sin(x−c) − cxa−c−1 cos(x−c)) =

(a(a − 1)xa−2 − c2xa−2c−2) sin(x−c) − c(−2a + c + 1)xa−c−2 cos(x−c). Note that both a(a − 1) > 0 and(−2a + c + 1) > 0 from a > 2 + c and c > 0. All the terms appearing here except for xa−2c−2 sin(x−c)tends to 0 as x→ 0, so f ′′ is bounded on [−1, 1] if and only if xa−2c−2 sin(x−c) is bounded on (0, 1]. By the sameargument as in the previous parts, this happens exactly when a− 2c− 2 ≥ 0, or when a ≥ 2 + 2c.

(g) Suppose f ′′ is continuous on [−1, 1]. Then f ′′([−1, 1]) is compact since [−1, 1] is compact (closed and bounded,Heine-Borel). So by part (f), f ′′ is bounded on [−1, 1] and thus a ≥ 2 + 2c. We rule out the case a = 2 + 2c. Ifa = 2 + 2c, we have for x > 0, f ′′(x) = (a(a− 1)xa−2 − c2xa−2c−2) sin(x−c)− c(−2a+ c+ 1)xa−c−2 cos(x−c). Set

xk =(π2 + 2πk

)−1/cas usual, then f ′′(xk) = a(a− 1)(π2 + 2πk)−(a−2)/c− c2, which does not tend to 0 as k →∞.

Since f ′′(0) = 0 and limk→∞ xk = 0, f ′′ is not continuous at 0.

Suppose a > 2 + 2c. Then all the terms in the expression above for f ′′(x) for x > 0 tends to 0 as x → 0by the Squeeze Theorem, so f ′′ is continuous at 0. It’s continuous everywhere else so f ′′ is continuous on [−1, 1].

6. Rudin Ch 5. Exercise #14. Let f be a differentiable real function defined in (a, b). Prove that f is con-vex if and only if f ′ is monotonically increasing. Assume next that f ′′(x) exists for every x ∈ (a, b), and provethat f is convex if and only if f ′′(x) ≥ 0 for all x ∈ (a, b).

Solution.

Recall that a real function f defined on (a, b) is convex if and only if for all x < y < z in (a, b), f(y)−f(x)y−x ≤ f(z)−f(y)z−y .

We will prove this equivalence at the end of this solution.

Suppose that f is convex. Take x < y in (a, b). Take sequences {xn}n and {yn}n such that x < xn < yn < y for alln, limn→∞ xn = x and limn→∞ yn = y. Since f is assumed to be differentiable, we then have f ′(x) = limn→∞ anand f ′(y) = limn→∞ bn if an = f(xn)−f(x)

xn−x and bn = f(yn)−f(y)yn−y . By the convexity, an = f(xn)−f(x)

xn−x ≤ f(yn)−f(xn)yn−xn ≤

f(y)−f(yn)y−yn = bn for all n ≥ 1. Therefore f ′(x) = limn→∞ an ≤ limn→∞ bn = f ′(y). That is, f ′ is monotonically

increasing.

Suppose now that f ′ is monotonically increasing. Take x < y < z in (a, b). By the Mean Value Theorem,

HW #1 SOLUTIONS 5

f(y)−f(x)y−x = f ′(c) for some c ∈ (x, y) and f(z)−f(y)

z−y = f ′(d) for some d ∈ (y, z). Then x < c < y < d < z, and in

particular c < d. Therefore f ′(c) ≤ f ′(d), and so f is convex.

Assuming that f ′′(x) exists for all x ∈ (a, b), note only that f ′ is monotonically increasing if and only if f ′′(x) ≥ 0for all x ∈ (a, b) by the Mean Value Theorem. One direction follows from Theorem 5.11. Suppose that f ′ ismonotonically increasing. Then fix y ∈ (a, b) and let h : [y, b) → R be defined by h(x) = f ′(x) − f ′(y). Thenh(y) = 0, h is differentiable since f ′ is differentiable, and h(x) = h(x)− 0 = h(x)− h(y) = h′(c)(x− y) for somec ∈ (y, x). But h′(c) = f ′′(c) ≥ 0 (we’re fixing y), so that h(x) ≥ 0. That is, f ′(x)− f ′(y) ≥ 0, or f ′(x) ≥ f ′(y).So f ′ is monotonically increasing.

Proof of equivalent definitions of convexity. We compare the two conditions on a function f : (a, b) → R: i)For all t ∈ (0, 1) and x, y ∈ (a, b), f(tx + (1 − t)y) ≤ tf(x) + (1 − t)f(y) and ii) For all x, y, z ∈ (a, b) such that

x < y < z, f(y)−f(x)y−x ≤ f(z)−f(y)

z−y .

i) implies ii). Take x < y < z in (a, b). We take here t = z−yz−x , so that t ∈ (0, 1). Then y = tx+ (1− t)z, and so

f(y) ≤ tf(x)+(1−t)f(z) = z−yz−xf(x)+ y−x

z−xf(z). Rearranging this inequality gives precisely f(y)−f(x)y−x ≤ f(z)−f(y)

z−y .

ii) implies i). If t ∈ (0, 1) and x < y are in (a, b), then x < tx + (1 − t)y < y. So using the condition ii),

we have f(tx+(1−t)y)−f(x)tx+(1−t)y−x ≤ f(y)−f(tx+(1−t)y)

y−(tx+(1−t)y) . Rearranging this inequality gives precisely that f(tx+ (1− t)y) ≤tf(x) + (1− t)f(y).

7. Rudin Ch 5, Exercise #15. Suppose a ∈ R1, f is twice-differentiable real function on (a,∞), and M0,M1,M2

are the least upper bounds of |f(x)|, |f ′(x)|, |f ′′(x)|, respectively, on (a,∞). Prove that M21 ≤ 4M0M2. Does this

hold for vector valued functions too?

Solution.

If x ∈ (a,∞) and h > 0, then by the theorem of Taylor, there is ζ ∈ (x, x + h) such that f(x + h) =

f(x) + f ′(x)h + f ′′(ζ)2 h2, or f ′(x)h = f(x + h) − f(x) − f ′′(ζ)

2 h2. Therefore |f ′(x)|h ≤ M0 + M0 + M2

2 h2.

Or |f ′(x)| ≤ 2M0

h + M2

2 h. Fixing h > 0, we see here then that the set {|f ′(x)| |x ∈ (a,∞)} has 2M0

h + M2

2 h as an

upper bound, or M1 ≤ 2M0

h + M2

2 h, or M1h ≤ 2M0

+M2

2 h2. This inequality certainly holds for h ≤ 0 as well since

the RHS is nonnegative for h ∈ R.

We have so far that M2

2 h2 − M1h + 2M0 ≥ 0 for all h ∈ R. Considering the real vlaued function g(h) =

M2

2 h2 −M1h + 2M0 defined on R, a degree 2 polynomial, the fact that it is always nonnegative implies that it

has at most 1 root, i.e. that its discriminant M21 − 4(M2

2 )(2M0) ≤ 0, or M21 ≤ 4M0M2.

Let’s consider now a vector valued, twice differentiable function f : (a,∞)→ Rn. We reduce the problem to theone-variable case by considering various projections of the curve f in Rn onto the 1-dimensional subspaces of Rn.To that end, take a direction u ∈ Sn−1 = {(a1, a2, . . . , an) |

∑ni=1 a

2i = 1}. We then consider the projection of the

curve f onto the line spanned by u, i.e. consider the function g : (a,∞) → R by g(t) = u · f(t) = |u||f(t)| cos θwhere θ is the “angle” between u and f(t).

Using the one-variable case, we have (supt∈(a,∞) |g′(t)|)2 ≤ 4 supt∈(a,∞) |g(t)| supt∈(a,∞) |g′′(t)|. Note that g′(t) =

u · f ′(t) and g′′(t) = u · f ′′(t), so by the Cauchy-Schwartz inequality we have |g′(t)| ≤ |u||f ′(t)| = |f ′(t)| andlikewise |g′′(t)| ≤ |f ′′(t)|.

6 HW #1 SOLUTIONS

We have thus shown that for all u ∈ Sn−1, (supt∈(a,∞) |u · f ′(t)|)2 ≤ 4M0M2. Note that (supt∈(a,∞) |f ′(t)|)2 =

supt∈(a,∞) |f ′(t)|2. Now take any ε > 0. Then there exists ζ ∈ (a,∞) such that supt∈(a,∞) |f ′(t)| − ε < |f ′(ζ)|.We now then choose u ∈ Sn−1 to maximize the magnitude of the projection f ′(ζ) onto u, i.e. take u = f ′(ζ)

|f ′(ζ)| in

the direction parallel to f ′(ζ). Then we have the inequalities

( supt∈(a,∞)

|f ′(t)| − ε)2 ≤ |f ′(ζ)|2 = |u · f ′(ζ)|2 ≤ ( supt∈(a,∞)

|u · f ′(t)|)2 ≤ 4M0M2

.

Since ε > 0 was arbitrary, we have M21 ≤ 4M0M2.

Now take

f(x) =

{2x2 − 1 if − 1 < x < 0x2−1x2+1 if x ≥ 0

We show that M0 = 1, M1 = 4, and M2 = 4 for the above f defined on (−1,∞). It’s easy to verify thefollowing:

f ′(x) =

{4x if − 1 < x < 0

4x(x2+1)2 if x ≥ 0

f ′′(x) =

{4 if − 1 < x < 0−12x2+4(x2+1)3 if x ≥ 0

The only care one needs to take in differentiating the above f is at 0, but this is handled in a straightfor-

ward manner by comparing the right-hand and the left-hand limits at 0. Writing x2−1x2+1 = 1− 2

x2+1 , it’s clear thatM0 = 1. Likewise for M1 = 4 and M2 = 4.

8. Rudin Ch 5, Exercise #18. Suppose f is a real function on [a, b], n is a positive integer, and f (n−1) ex-ists for every t ∈ [a, b]. Let α, β, P be as in Taylor’s theorem (5.15). Define

Q(t) =f(t)− f(β)

t− βfor t ∈ [a, b], t 6= β, differentiate

f(t)− f(β) = (t− β)Q(t)

n-1 times at t = α, and derive the following version of Taylor’s theorem:

f(β) = P (β) +Q(n−1)(α)

(n− 1)!(β − α)n.

Solution.

We’ll verify the conclusion by induction on n. Note that P (β) depends on n, as P (β) = f(α) + f ′(α)(β −α) + f ′′(α)

2! (β − α)2 + · · ·+ f(n−1)(α)(n−1)! (β − α)n−1.

If n = 1, then the conclusion reads f(β) = P (β) + Q(α)(β − α) = f(α) + Q(α)(β − α), which just comesfrom the definition of Q(t) (and evaluating this at t = α).

HW #1 SOLUTIONS 7

Before we proceed to the inductive step, we develop a formula for derivatives of f(t)−f(β) = (t−β)Q(t). Differen-tiating this once, we have f ′(t) = Q(t)+(t−β)Q′(t), differentiating yet again, we have f ′′(t) = 2Q′(t)+(t−β)Q′′(t).It’s easy to see by (yet another) induction that f (n)(t) = nQ(n−1)(t)+(t−β)Q(n)(t) for n ≥ 1. Plugging in t = α,we have f (n)(α) = nQ(n−1)(α) + (α− β)Q(n)(α).

Now we resume the induction. Assuming that the conclusion holds for n, we have f(β) = P (β)+Q(n−1)(α)(n−1)! (β−α)n.

From here, one simply substitutes Q(n−1)(α) = f(n)(α)+(β−α)Q(n)(α)n from the equation we derived in the previous

paragraph.

9. Rudin Ch 5, Exercise #26. Suppose f is differentiable on [a, b], f(a) = 0, and there is a real number Asuch that |f ′(x)| ≤ A|f(x)| on [a, b]. Prove that f(x) = 0 for all x ∈ [a, b].

Solution.

Note that we may as well assume that A > 0, for if A ≤ 0 then 0 ≥ |f ′(x)| ≤ 0, or f ′(x) = 0 for all x ∈ [a, b], inwhich case we have that for t ∈ [a, b], f(t) = f(t) − 0 = f(t) − f(a) = f ′(c)(t − a) for some c ∈ [a, t], and sincef ′(c) = 0 we have f(t) = 0, i.e. f(x) = 0 for all x ∈ [a, b].

Assuming now that A > 0, set ∆ = b−an , where n ∈ Z+ is chosen such that b−a

n < 1A (note that we can ar-

range for this by the Archimedean property). For i ∈ {0, 1, 2, . . . , n}, set ti = a+ i∆, so that [a, b] = ∪ni=1[ti−1, ti].

We show by induction on i that f is 0 on [ti−1, ti].

Base case is when i = 1, or showing that f is 0 on [a, a+ ∆]. Set M be the sup of the set {|f(x)| |x ∈ [a, a+ ∆]}.Taking x ∈ [a, a+∆], we have by the Mean Value Theorem there exists c ∈ [a, x] such that |f(x)| = |f(x)−f(a)| =|f ′(c)||x−a| ≤ A|f(c)||x−a| ≤ A ·∆ ·M . This being true for all x ∈ [a, a+∆], we have M ≤ A ·∆ ·M . If M > 0,then dividing the inequality by M , we have 1 ≤ A ·∆, contradicting ∆ < 1

A . Therefore M = 0 and f(x) = 0 forall x ∈ [a, a+ ∆].

Assume now that f is 0 on [t0, t1] ∪ [t1, t2] ∪ · · · ∪ [ti−1, ti]. We now show that f is 0 on [ti, ti+1]. We knowthat f(ti) = 0. The same argument applies: Let M be the sup of the set {|f(x)| |x ∈ [ti, ti+1]}. For anyx ∈ [ti, ti+1], we have |f(x)| = |f(x)− f(ti)| = |f ′(c)||x− ti| ≤ A|f(c)| ·∆ ≤ A ·∆ ·M , for some c ∈ [ti, x]. Thisbeing true for all x ∈ [ti, ti+1], we have M ≤ A ·∆ ·M . Same argument as before, we must then have M = 0.Therefore f = 0 on [ti, ti+1].

10. Rudin Ch 5, Exercise #27. Let φ be a real function defined on a rectangle R in the plane, given bya ≤ x ≤ b, α ≤ y ≤ β. A solution of the initial-value problem

y′ = φ(x, y), y(a) = c (α ≤ c ≤ β)

is, by definition, a differentiable function f on [a, b] such that f(a) = c, α ≤ f(x) ≤ β, and

f ′(x) = φ(x, f(x)) (a ≤ x ≤ b).

8 HW #1 SOLUTIONS

Prove that such a problem has at most one solution if there is a constant A such that

φ(x, y2)− φ(x, y1)| ≤ A|y2 − y1|whenever (x, y1) ∈ R and (x, y2) ∈ R.

Solution.

Suppose g : [a, b]→ R is yet another solution of the above initial-value problem. Consider a function h : [a, b]→ Rdefined by h(x) = f(x)−g(x) for x ∈ [a, b]. Then for x ∈ [a, b], |h′(x)| = |f ′(x)−g′(x)| = |φ(x, f(x))−φ(x, g(x))| ≤A|f(x)− g(x)| = A|h(x)|. Also, h(a) = f(a)− g(a) = c− c = 0. By problem #26 then we have h = 0 on [a, b].

We now classify all solutions to y′ = y1/2, y(0) = 0. Certainly zero function is a solution to this differentialequation. Suppose now that f is a solution to the differential equation on an interval I containing 0. Certainlyf(x) ≥ 0 for all x ∈ I, so that f ′(x) =

√f(x) implies that f is nondecreasing on I. For x ∈ I and x ≤ 0, we must

thus have 0 ≤ f(x) ≤ f(0) = 0, or f(x) = 0. We can then just (uniquely by monotonicity of f) extend f to all of(−∞, 0] ∪ I by defining it to be 0 on (−∞, 0], therefore we assume that our function is defined f : (−∞, a]→ R,where a ≥ 0. We can assume that a > 0 (if a = 0 the monotonicity argument above already implies that zerofunction is the only possible solution on (−∞, 0]). Zero function defined on all of (−∞,∞) is of course a solutionto the differential equation. Since f is nondecreasing, if f is not the zero function then f(a) = b > 0.

Now let c = sup{x | f(x) = 0} = S. Note that c exists because S has a as an upper bound and is nonemptybecause it contains 0. Then there is a sequence {xn}n≥1 in S such that limn→∞ xn = c, and by continuity of fwe must have f(c) = 0. We have c < a. Take a point l such that c < l < a, and note that f(l) = s > 0. Now letKn = [c+ l−c

n , l+ a−ln ] for n ≥ 2. Then Kn is compact, f is defined on Kn and f is never zero on Kn. Therefore

f(Kn) is compact and does not contain 0. Considering the function φ(y) = y1/2 defined on Kn, we see thenthat φ′(y) = 1

2y−1/2 is defined and continuous on f(Kn), so φ′(f(Kn)) is compact, i.e. there is a real number

A (may depend on n) such that for all x ∈ f(Kn), |φ′(x)| ≤ A. Therefore if q, w ∈ f(Kn) then by the MeanValue Theorem, |φ(q) − φ(w)| = |φ′(c)||q − w| ≤ A|q − w|, where c is between q and w. Also f(Kn) is compactand connected in R1, so it is a closed interval (Theorem 2.47, say). If we use the rectangle R as in problem #27as Kn × f(Kn), then f ′(x) = φ(f(x)) and |φ(y1) − φ(y2)| ≤ A|y1 − y2| for all y1, y2 ∈ f(Kn), so the conditionsof problem #27 are satisfied (note that φ(y) is independent of x here, an “autonomous” case). Therefore thesolution to this differential equation with the initial condition f(l) = s is unique, and it’s readily checked that

the mapping x 7→ (x+2√s−l)2

4 is a solution to this initial differential equation, so we must have f(x) = (x+2√s−l)2

4on Kn.

Now we take xn = c+ l−cn for n ≥ 2, so that xn ∈ Kn and f(xn) = (xn+2

√s−l)2

4 . Note that limn→∞ xn = c, and

so by continuity of f we must then have 0 = f(c) = (c+2√s−l)2

4 . s = f(l), so we must have c+ 2√f(l)− l = 0, or

f(l) = (l−c)24 . Therefore

f(x) =

{0 if x ≤ c(l−c)2

4 if x ≥ c