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  • 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 continuous function 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 we have an open cover of K. Then by compactness of K, there is a finite subcover (f−1(Uαi))

    N i=1.

    Take y ∈ f(K). Then there is x ∈ K such that f(x) = y. Then x ∈ f−1(Uαi) for some i = 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 that f(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 as such. 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 → R under 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.

  • 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, we see

    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 find δ > 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∗−δ

    F (x, y)dx dy >

    ∫ y∗+δ y∗−δ

    ∫ x∗+δ x∗−δ

    ε

    2 dx dy = 2εδ2 > 0.

    This contradicts that F integrates to zero over any rectangle, hence F is identically zero which 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 that the 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’t be bothered to write down the proof.

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

  • Christian Parkinson UCLA Basic Exam Solutions: Analysis 3

    Prove the following stronger statement: there is an integer M such that for each element x ∈ [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 ∈ N since it is the pullback of an open set under a continuous function. Thus En is open as a union of open sets. Thus En, n ∈ N forms and open cover of [0, 1]. By compactness, there is 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. Assume that S is a countable intersection of open sets Dn, n = 1, 2, 3, . . .. Then Q+ is also a countable intersection 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) denumerate the 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 be

    open and dense in R+. This is impossible because⋂ n

    D+n = Q+ whereas ⋂ n

    An = R+ −Q+.

    The contradiction means that (2) is satisfied.

  • Christian Parkinson UCLA Basic Exam Solutions: Analysis 4

    Assume that S is a countable union of closed sets:

    S = ⋃ n

    Cn.

    Then R− S =

    ⋂ n

    (R− Cn)

    which is a countable intersection of open sets. This is impossible by the same reasoning as above 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→ R are 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 some polynomials 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 (

    1 2 (x+ y)2 + 1

    2 (x− y)2

    ) + b (

    1 2 (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) = 1 2 (a+ b)x2 and g(x) = 1

    2 (a− b)x2 works.

  • 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 when n ≥ N . By definition of the Riemann integral, there are some piecewise constant functions gn, 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 is piecewise 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 p