mathematical methods for statistics homework...

Mathematical Methods for StatisticsHomework Assignments

Luke Tierney

Spring 2016

STAT:5120 (22S:190), Spring 2016 Tierney

Assignment 1

Due on Friday, January 29, 2016.

1. Let f : R → R be a non-decreasing function, and let D be the set of points atwhich f is not continuous. Show that D is at most countable.

2. Let (S,B, P ) be a probability space and assume that s ∈ B for all s ∈ S.Let A = s ∈ S : P (s) > 0. The elements of A are sometimes called atomsof the probability P . Show that A is at most countable. [Hint: for a positiveinteger n, how many atoms can there be with probabilities greater than or equalto 1/n?]

3. Let A′ denote the set of limit points of a set A.

(a) Show that A′ is closed.

(b) Show that A and its closure A have the same limit points.

(c) Do A and A′ always have the same limit points?

4. Suppose A1, A2, . . . are subsets of a metric space.

(a) If Bn =⋃ni=1Ai show that Bn =

⋃ni=1Ai.

(b) If B =⋃∞i=1Ai show that B ⊃

⋃∞i=1Ai. Show by example that the inclu-

sion may be proper.

5. (a) Prove that the complement of the interior A of a set A is the closure ofthe complement Ac.

(b) Do A and the closure A always have the same interiors?

(c) Do A and A always have the same closures?

1


Solutions

1. For each x ∈ R let L(x) = supf(y) : y < x and U(x) = inff(y) : y > x.Since f is non-decreasing L(x) ≤ U(x) for all x, and for x ≤ y we have U(x) ≤L(y). Furthermore, a point x is a discontinuity if and only if L(x) < U(x).Suppose we can define a function on D such that for each x in D the valueh(x) is a rational number in the open interval Ix = (L(x), U(x)). Since theseintervals are disjoint, h(x) = h(y) for x, y ∈ D is and only if x = y. So thefunction h is a one to one mapping from D onto a subset of the rationals. Sucha subset is at most countable, so D is at most countable.

To complete the proof we need to show that such a function h exists. Thiscan be done by an appeal to the Axiom of Choice, but this can be avoidedby an explicit construction such as this: For each positive integer k let Qn bethe set of rationals of the form k

nfor some integer k. For each x ∈ D let nx

be the smallest integer n such that the intersection Qn ∩ (L(x), U(x)) is notempty. Since (L(x), U(x)) is bounded the set Qn ∩ (L(x), U(x)) is a finite setof rationals; let h(x) be the smallest rational in this set.

2. For positive integers n let An = s : P (s) ≥ 1/n. Then A =⋃∞n=1An.

Since P (An) ≤ 1 the set An is a finite set with at most n elements. So A is acountably infinite union of finite sets and is therefore at most countable.

3. (a) This uses the topological definition of neighborhood of x as any open setcontaining x and the fact that a point x is a limit point of A if and only ifevery open neighborhood of x contains infinitely many points of A:

Let x be a point in (A′)c, i.e. a point that is not a limit point of A. Thenthere exists an open neighborhood G of x such that G ∩ A is at mostfinite. For every point y in G the set G is an open set neighborhood of ycontaining at most finitely many points from A, so no point in G is a limitpoint. So the set (A′)c is open, and therefore A′ is closed.

An alternative using only metric space concepts: Suppose x is not a limitpoint. Then there is an open ball Nr(x) centered at s containing no pointof A other than possibly x itself. For any y ∈ Nr(x) we can find an openball Nr′(y) centered at y such that Nr′(y) ⊂ Nr(x) and x 6∈ Nr′(y). Theball Nr′(y thus contains no points of A, and therefore y is not a limit point.So no point in Nr(x) is a limit point and thus x is an interior point of thecomplement of A′. Since this holds for all x that are not limit points, thecomplement of A‘ is open and thus A′ is closed.

(b) Since A ⊂ A any limit point of A is also a limit point of A. To show theconverse, let x be a limit point of A. To show that x is also a limit pointof A we need to show that any open ball Nr(x) centered at x contains apoint in A different from x. Since x is a limit point of A the neighborhoodcontains a point y ∈ A with y 6= x. If y ∈ A we are done. Otherwise, y

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is a limit point of A. Let r′ = min(d(x, y), r − d(x, y)). Then since y is alimit point of A the neighborhood Nr′(y) contains a point z ∈ A. Sinced(y, z) < d(x, y) we have z 6=; since d(y, z) < r − d(x, y) we have

d(x, z) ≤ d(x, y) + d(y, z) < r

and so y ∈ Nr(x), which completes the proof.

(c) Consider A = 1, 1/2.1/3, . . . . The set of limit points of A is is A′ = 0,which has no limit points. So A and A′ need not have the same limitpoints.

4. If x is a limit point of Ai for some i then x is also a limit point of B and of Bn

for i ≤ n. So Bn ⊃⋃ni=1Ai and B ⊃

⋃∞i=1Ai.

(a) Suppose x is a limit point of Bn but not a limit point of any of A1, . . . , An.Then for each i there exists a radius ri > 0 such that Nri(x) does notcontain a point of Ai different from x. Let r = min(r1, . . . , rn), which ispositive since it is a minimum of a finite set of positive values. Then Nr(x)contains no point of the union B other than possibly x, contradicting theassumption that x is a limit point of B. So x must be a limit point of oneof A1, . . . , An, and therefore x ∈

⋃ni=1Ai. So Bn ⊂

⋃ni=1Ai as well, and

thus Bn =⋃ni=1Ai.

A version of the argument not using metric properties: Suppose x is a limitpoint of Bn but not a limit point of any of A1, . . . , An. Then for each ithere exists an open set Gi with x ∈ Gi that contains no point in Ai exceptpossibly x. The finite intersection G =

⋂ni=1Gi is then an open set with

x ∈ G that contains no point in the union Bn other than possibly x. Thiscontradicts the assumption that x is a limit point of Bn.

(b) Let An = [1/n, 1]. Then the Ai are closed and B =⋃∞i=1Ai =

⋃∞i=1Ai =

(0, 1], but B = [0, 1] 6= B.

5. (a) The complement of the interior of A is a closed set containing the com-plement of A. So (A)c ⊃ (Ac). No suppose x ∈ (A)c. Since x is not inthe interior of A, every open neighborhood of x must contain a point inAc. So x is a contact point of Ac, i.e. x ∈ (Ac). So (A)c ⊂ (Ac). Thus(A)c = (Ac).

(b) No: Let A = Q, the rationals. Then the interior of A is A = ∅, and theclosure of A is A = R, with interior (A) = R.

(c) No: The closure of the rationals Q is R; the interior of the rationals is theempty set, with closure the empty set.

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Assignment 2

Due on Friday, February 5, 2016.

1. Give an example of an open cover of the interval (0, 1) that does not have afinite subcover.

2. Let (X, τ) be a Hausdorff space and K a compact subset of X. Show that K isclosed.

3. Are closures and interiors of connected sets always connected? (Look at subsetsof R2.)

4. Let (X, d) be a compact metric space. Show that X contains an at most count-able dense set, i.e. that (X, d) is separable. [Hint: For every positive integer nthere are finitely many open balls of radius 1/n whose union covers the metricspace.]

5. Show that a sequence xn in a metric space (X, d) converges to a point x ∈ X ifand only if every subsequence xnk has a further subsequence xnki that convergesto x. [Hint: for the “if” direction suppose xn does not converge to x and arguethat there must exist a subsequence that is uniformly bounded away from x.]

6. Show that convergence of a sequence of real numbers xn implies convergence ofthe sequence |xn| converges as well. Is the converse true as well? converse

4


Solutions

1. Let Gn = (1/n, 1). Then (0, 1) =⋃ni=1Gn, but no finite sub-collection of the

Gn covers (0, 1).

2. Let x be a point in Kc and for each y ∈ K let Ay and By be open sets suchthat x ∈ By, y ∈ Ay, and Ay ∩ By = ∅. The union

⋃y∈LAy covers K. Since

K is compact there is a finite subcover Ay1 , . . . Ayn . Let By1 , . . . , Byn be thecorresponding open sets containing x and let B =

⋂ni=1Byi . Then x ∈ B,

B is open since it is the intersection of a finite collection of open sets, andB ⊂

⋂ni=1A

cyi⊂ (⋃ni=1Ayi)

c ⊂ Kc. So every x ∈ Kc is an interior point of Kc.Thus Kc is open and K is closed.

3. Suppose E is not connected. Then there exist nonempty sets A,B with E =A ∪ B such that A ∩ B = A ∩ B = ∅. Let A1 = A ∩ E and B1 = B ∩ E. ThenE = A1 ∪ B1, A1 ∩ B1 = ∅, and A1 ∩ B1 = ∅. Suppose A1 is empty. ThenE ⊂ B1 ⊂ B. Since A is non-empty it must contain a limit point of E. ButE ⊂ B implies that all limit points of E are in B, and A ∩ B = ∅. This is acontradiction, so A1 is not empty. Similarly, B1 is not empty, and therefore E isthe union of two separated sets and is not connected. So closures of connectedsets are connected.

The interior of a connected set need not be connected. Consider two closeddiscs in R2 that touch at one point.

4. Since X is compact and N1/n(x) : x ∈ X is an open cover of X there is

a subcover Cn = N1/n(x(n)1 ), . . . , N1/n(x

(n)kn

) corresponding to points Dn =

x(n)1 , . . . , x(n)kn. Let D =

⋃∞i=nDn. To show that D is dense we need to show

that every point in X is a contact point of D. Let x be a point in X and ε > 0.Chose n such that 1/n < ε. Since Cn is a cover of X there is a point y ∈ Dn

such that x ∈ N1/n(y) and thus y ∈ Nε(x). Thus x is a contact point of D.

5. If xn converges to x then all subsequences converge to x, so all subsequencesof subsequences converge to x. If xn does not converge to x then there existsan ε > 0 such that d(xn, x) ≥ varepsilon for infinitely many indices n. Letn1, N2, . . . be the sequence of these indices. Then every further subsequencexnki satisfies d(xnki , x) ≥ varepsilon for all i and so cannot converge to x. Soif every subsequence has a further subsequence that converges to x then thesequence xn must converge to x.

6. For any normed linear space xn → x implies ‖xn‖ → ‖x‖: By the triangleinequality, for any x, y

‖x‖ = ‖x− y + y‖ ≤ ‖x− y‖+ ‖y‖

and therefore‖x‖ − ‖y‖ ≤ ‖x− y‖.

5


Reversing the roles of x and y produces

‖y‖ − ‖x‖ ≤ ‖x− y‖,

and together these imply

|‖x‖ − ‖y‖| ≤ ‖x− y‖.

Thus ‖xn − x‖ → 0 implies ‖xn‖ − ‖x‖ → 0, or ‖xn‖ → ‖x‖.The converse is not true: Let xn = (−1)n. Then |xn| = 1 for all n and thusconverges, but xn does not.

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Assignment 3

Due on Friday, February 12, 2016.

1. Let X be the set of all bounded infinite sequences of real numbers,

X =

x = (a1, a2, . . . ) : ai ∈ R for all i and sup

1≤i<∞|ai| <∞

with the norm ‖x‖ = sup1≤i<∞ |ai|. Show that the metric space (X, d) withd(x, y) = ‖x− y‖ is complete.

2. Show that the metric space (X, d) with X the real line and the distance

d(x, y) = | tan−1(x)− tan−1(y)|

is not a complete metric space.

3. Let f be a continuously differentiable real valued function on an interval [a, b]with f(a) < 0 < f(b) and 0 < m ≤ f ′(x) ≤M <∞ for all x ∈ [a, b]. Let

A(x) = x− f(x)

M

Show f(x) has a unique root x∗ in [a, b] and that for any initial value x0 ∈ [a, b]the sequence xn defined by xn = A(xn−1) converges to x∗. [Hint: Show that Ais a contraction on [a, b]; you can show that A([a, b]) ⊂ [a, b] by showing thatA(x) is increasing and a ≤ A(a) and A(b) ≤ b; you may also want to use themean value theorem.]

4. Let C[0,1] be the normed linear space of continuous funcitons on [0, 1] with norm‖x‖ = sup0≤t≤1 |x(t)| and define A : C[0,1] → C[0,1] by

A(x)(t) =

∫ t

0

(t− u)x(u)du.

Show that A is a contraction.

5. Show that for any sequence xn

lim supn→∞

xn = limn→∞

[supxm : m ≥ n] .

6. Findlimn→∞

(√n2 + n− n).

[Hint: Multiply and divide by√n2 + n+n. Alternatively, factor out n and use

a Taylor series expansion of the square root.]

7


Solutions

1. Let xn = (a1,n, a,n, . . . ) be a Cauchy sequence. Then for each i and any n,m

|ai,n − ai.m| ≤ ‖xn − xm‖.

So the sequence ai,n is a Cauchy sequence in R and therefore has a limit ai. Letx = (a1, a2, . . . ) be the sequence of these coordinate-wise limits. Fix an ε > 0and find N such that ‖xn − xm‖ < ε/2 foa all n,m ≥ N . Then for any integerM and any n,m ≥ N

M∑i=1

|ai,n − ai,m| ≤ ‖xn − xm‖ < ε/2.

Taking the limit as m → ∞ gives∑M

i=1(ai,n − ai) ≤ ε/2, and then taking thelimit as M →∞ gives

∞∑i=1

|ai,n − ai| = ‖xn − x‖ ≤ ε/2 < ε.

So for n ≥ N‖x‖ ≤ ‖xn‖+ ‖xn − x‖ ≤ ‖xn‖+ ε/2 <∞;

therefore x ∈ X and xn → x. Since ‖xn − x‖ ≤ ε for n ≥ N the sequence xnconverges to x in (X, d). Thus every Cauchy sequence has a limit and (X, d) iscomplete.

2. Let xn = n. Then yn = tan−1(xn) → π/2, so yn is a Cauchy sequence withrespect to the usual distance, and therefore xn is a Cauchy sequence for thedistance d. For any finite real number x we have tan−1(x) < π/2, so

limn→∞

d(x, xn) = π/2− tan−1(x) > 0.

So xn does not have a limit in R under distance d and thus the metric space isnot complete. R with tan−1(+∞) = π/2 and tan−1(−∞) = −π/2 is completeand compact under this distance.

3. The function A is increasing since

A′(x) = 1− f ′(x)

M≥ 1− M

M= 0.

Since f(a) < 0 and f(b) > 0 this gives

a < a− f(a)

M= A(a) ≤ A(x) ≤ A(b) = b− f(b)

M< b.

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So A([a, b]) ⊂ [a, b]. The function f is increasing on [a, b], so for any x, y ∈ [a, b]

A(x)− A(y) = x− y − f(x)− f(y)

M

By the mean value theorem there is a value ξ between x and y such that

f(x)− f(y) = f ′(ξ)(x− y).

Since m ≤ f ′(ξ) ≤M

|A(x)−A(y)| =∣∣∣∣x− y − f ′(ξ)(x− y)

M

∣∣∣∣ = |x−y|(

1− f(ξ)

M

)≤ |x−y|

(1− m

M

).

So A is a contraction mapping with α = 1 − mM< 1, x∗ is a fixed point of A,

and the claims follow from the fixed point theorem.

4. The fundamental theorem of calculus implies that A(x) is a continuous functionif x is continuous, so A is a mapping from C[0,1] into itself. For any x, y ∈ C[0,1]

and any t ∈ [0, 1]

|A(x)(t)− A(y)(t)| =∣∣∣∣∫ t

0

(t− u)(x(u)− y(u))du

∣∣∣∣≤∫ t

0

(t− u)‖x− y‖du

≤ ‖x− y‖∫ 1

0

(1− u)du =1

2‖x− y‖.

So A is a contraction mapping with α = 12. The unique fixed point is x(t) ≡ 0.

5. The sequence yn = supxm : m ≥ n is monotone decreasing and xn ≤ yn forall n. So yn converges, possibly to −∞, and

lim supn→∞

xn ≤ limn→∞

yn.

Suppose z > lim supn→∞ xn. Then by Rudin’s Theorem 3.17 (b) there existsan integer N such that xn < z for all n ≥ n, and therefore yn ≤ z for n ≥ N aswell. So

limn→∞

yn ≤ z

for all z > lim supn→∞ xn and therefore

limn→∞

yn ≤ lim supn→∞

xn.

Solimn→∞

yn = lim supn→∞

xn.

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6. Multiplying and dividing by√n2 + n+ n gives

n2 + n− n2

√n2 + n+ n

=n√

n2 + n+ n=

n√1 + 1/n+ 1

→ 1

2.

Alternatively, using the fact that

√1 + x = 1 +

1

2x+ o(x)

as x→ 0

√n2 + n− n = n(

√1 + 1/n− 1) = n

(1 +

1

2n+ o(1/n)− 1

)=

1

2+ o(1)→ 1

2.

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Assignment 4

Due on Wednesday, February 17, 2016.

1. Investigate the behavior (convergence or divergence) of∑an when

(a) an =√n+ 1−

√n;

(b) an =√n+1−

√n

n;

(c) an = ( n√n− 1)n.

2. Find the radius of convergence of each of the following power series:

(a)∑n3zn;

(b)∑

2n

n!zn;

(c)∑

2n

n2 zn.

3. Prove that the Cauchy product of two absolutely convergent series convergesabsolutely.

4. Suppose an ≥ 0 and∑an converges. Prove that∑ √

ann

converges.

5. Let f be a continuous, real-valued function on a metric space (X, d), and let

Z(f) = x : f(x) = 0

be the zero set of f . Prove that Z(f) is closed.

6. Let f and g be continuous mappings from a metric space (X, dX) into a metricspace (Y, dY ), and let A be a dense subset of X.

(a) Prove that f(A) is dense in f(X).

(b) If f(x) = g(x) for all x ∈ A show that f(x) = g(x) for all x ∈ X.

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Solutions

1. (a) Multiplying and dividing by√n+ 1 +

√n produces.

an =(√n+ 1−

√n)(√n+ 1 +

√n)√

n+ 1 +√n

=1

√n(√

1 + 1/√n+ 1)

.

Since 1 + 1/√n+ 1 ≤ 3 for all n this gives

an ≥1

3√n.

Since∑

1/√n diverges,

∑an diverges as well.

(b) Using a similar approach an is bounded by

an =1

n√n(√

1 + 1/√n+ 1)

≤ 1

n3/2

and∑an converges.

(c) Using the root test,n√an = n

√n− 1→ 0

by Rudin’s Theorem 3.20 (b). So the series converges.

2. (a)n√n3 = ( n

√n)3 → 1, so R = 1.

(b) Using the ratio test,cn+1|z|n+1

cn|z|n=

2z

n+ 1→ 0

for all z, so the radius of convergence is =∞.

(c) n√

2n/n2 = 2/( n√n)2 → 2. So R = 1

2.

3. If∑an∑bn converge absolutely then

∞∑n=0

∣∣∣∣∣n∑k=0

akbn−k

∣∣∣∣∣ ≤∞∑n=0

n∑k=0

|ak||bn−k|.

The right hand side is the Cauchy product of∑|an| and

∑|bn| and is finite by

Rudin’s Theorem 3.50.

4. By the Cauchy-Schwartz inequality (e.g. Rudin Theorem 1.35),

∑ √ann≤(∑

an

)1/2(∑ 1

n2

)1/2

,

and both sums on the right are finite.

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5. The set 0 is closed, Z(f) = f−1(0), and f is continuous, so Z(f) is closed.

6. (a) Rudin’s Problem 4.2 asks to show that for any E ⊂ X

f(E) ⊂ f(E) :

Let x be a contact point of E. Then there is a sequence xn of points in Ethat converges to x. Since f is continuous, f(xn) converges to f(x), andf(x) is a contact point of f(E). This proves the claim.

Since A is dense A = X. So

f(X) = f(A) ⊂ f(A).

So every point in f(X) is a contact point of f(A), and f(A) is dense inf(X).

(b) If x 6∈ A then,since A is dense, there is a sequence is the limit of a sequenceof points xn in A that converges to x. Then f(xn) = g(xn) for all n, andsince f and g are continuous,

f(x) = lim f(xn) = lim g(xn) = g(x).

So if f(x) = g(x) for all x ∈ X.

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Assignment 5

Due on Wednesday, February 24, 2016.

1. Suppose f is a real-valued function defined on a closed set A ⊂ R and that f iscontinuous on A. Prove that there exists a continuous function g : R→ R suchthat f(x) = g(x) for all x ∈ A. Such a function is called a continuous extensionof f from A to R. [Hint: use the result of Rudin, Problem 2.29, and take g tobe linear on each component of the complement of A.]

2. If A ⊂ X and f is a function defined on X, then the restriction of f to a is thefunction fA defined on A such that fA(x) = f(x) for all x ∈ A. Define f and gon R2 by

f(x, y) =

0 for (x, y) = (0, 0)xy2

x2+y4otherwise

g(x, y) =

0 for (x, y) = (0, 0)xy2

x2+y6otherwise.

Prove that f is bounded on R2, that g is unbounded on every neighborhood ofthe (0, 0), and that f is not continuous at (0, 0). Prove that nevertheless therestrictions of f and g to any straight line in R2 are continuous.

3. State more precisely and prove: A uniformly continuous function of a uniformlycontinuous function is uniformly continuous.

4. If A is a nonempty subset of a metric space X define the distance from x ∈ Xto A as

ρA(x) = infa∈A

dX(x, a).

(a) Prove that ρA(x) = 0 if and only if x ∈ A.

(b) Prove that ρA(x) is uniformly continuous on X by showing that

|ρA(x)− ρA(y)| ≤ dX(x, y)

for all x, y ∈ X.

5. Suppose A and B are disjoint sets in a metric space X, A is compact, and B isclosed.

(a) Show that there exists a δ > 0 such that dX(x, b) > δ for all a ∈ A andb ∈ B. [Hint: ρB is continuous and positive on A].

(b) Show that the conclusion may fail for two disjoint closed sets if neither iscompact.

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6. A real-valued function f defined on an open interval (a, b) is said to be convexif

f(λx+ (1− λ)y) ≤ λf(x) + (1− λ)f(y)

whenever x, y ∈ (a, b) and 0 < λ < 1.

(a) Prove that every increasing convex function of a convex function is convex.(For example, if f is convex then so is ef .)

(b) If f is convex on (a, b) and a < s < t < u < b, show that

f(t)− f(s)

t− s≤ f(u)− f(s)

u− s≤ f(u)− f(t)

u− t

(c) Prove that every convex function defined on an open interval (a, b) is con-tinuous.

15


Solutions

1. The complement of A consists of a countable collection of disjoint open intervals(ak, bk), and possibly intervals of the form (−∞, a) and (b,+∞). If one or bothof these exist, for x ∈ (−∞, a) define g(x) = f(a), and for x ∈ (b,+∞) defineg(x) = f(b). On intervals of the form (ak, bk) define

g(x) = f(ak) +x− akbk − ak

f(bk).

Then g(x) = f(x) for all x ∈ A and g is by construction continuous on theopen set Ac. It remains to show that g is continuous on A. That is, we needto show that given a point x ∈ A and ε > 0, there exists a δ > 0 such that|g(y)− g(x)| < ε for any y ∈ R such that |y − x| < δ.

Given x ∈ A and ε > 0, by the continuity of f on A there exists a δ0 > 0 suchthat |f(y)− f(x)| < ε/2 for all y ∈ A with |y − x| < δ0. If there exists a z ∈ Awith with x − δ0 < z < x, then set δ1 = x − z. Then because of the linearlyinterpolated form of g,

supx−δ1<y<x

|g(y)− g(x)| = sup|f(y)− f(x)| : y ∈ A, x− δ1 < y < x

≤ ε/2 < ε.

If (x−δ0, x)∩A is empty, then g is linear on (x−δ0, x), so using continuity of thislinear function choose δ1 > 0 such that |g(y)− g(x)| < ε when x− δ1 < y < x.Combining these cases, we have found a δ1 > 0 such that |g(y)− g(x)| < ε forall y ∈ R with x− δ1 < y < x.

Similarly we can find δ2 > 0 such that |g(y) − g(x)| < ε for all y ∈ R withx < y < x+ δ2.

Taking δ = min(δ1, δ2) we obtain that |g(Y ) − g(x)| < ε for all y ∈ R with|y − x| < δ, and thus g is continuous at x..

2. Since0 ≤ (x− y2)2 = x2 + y4 − 2xy2

we have2xy2 ≤ x2 + y4

for all x, y; applying this result to −x produces

−2xy2 ≤ x2 + y4,

and thus ∣∣∣∣ xy2

x2 + y4

∣∣∣∣ ≤ 1

2.

16


But f is not continuous at (0, 0) since for all y 6= 0

f(y2, 2) =y4

y4 + y4=

1

2,

while for f(0, 0) = 0.

g is unbounded near the origin since for y 6= 0

g(y3, 3) =y5

2y6=

1

2y→∞

as y ↓ 0.

Since f and g ar3 continuous everywhere except at (0, 0) their restrictions tolines not passing through the origin are continuous.

Lines through the origin either the vertical line

V = (0, y) : y ∈ R

or the line given by the equation

Lα = (x, y) : y = αx

for some α. For the vertical line the restrictions are

fV (x, y) = gV (x, y) = 0.

For a line with slope α the restriction of f for x 6= 0 is

fLα(x, y) =α2x3

x2 + α4x4=

α2x

1 + α4x2,

and fLα(0, 0) = 0, so this is continuous at x = 0 as well. For g

gLα(x, y) =α2x3

x2 + α6x6=

α2x

1 + α6x4

for x 6= 0 and gLα(0, 0) = 0, so again the restriction is continuous.

3. A more precise statement: Suppose (X, dX), (Y, dY ), and (Z, dZ) are metricspaces and f : X → Y and g : Y → Z are uniformly continuous. Let h(x) =g(f(x)). Then h : X → Z is uniformly continuous.

Proof: Let ε > 0 be given. Since g is uniformly continuous, there is a η > 0 suchthat dZ(g(u), g(v)) < ε whenever dY (u, v) < η. Since f is uniformly continuous.there is a δ > 0 such that dY (f(x), f(y)) < η whenever dX(x, y) < δ. So ifdX(x, y) < δ, then dY (f(x), f(y)) < η and therefore

dZ(h(x), h(y)) = dZ(g(f(x)), g(f(y))) < ε.

17


4. (a) Suppose ρA(x) = 0, Since ρA(x) is defined as

ρA(x) = infa∈A

d(x, a)

there exists a sequence an ∈ A such that d(x, an) → ρA(x). If ρA(x) = 0then this means d(x, an)→ 0; thus an → x and x is a contact point of A.

Conversely, if x is a contact point of A, then there is a sequence an ∈ Asuch that an → x and thus d(x, an)→ 0. Since

0 ≤ infa∈A≤ lim inf

n→∞d(x, an) = 0,

this means ρA(x) = 0.

(b) For any a ∈ A

ρA(x) = infb∈A

d(x, b) ≤ d(x, a) ≤ d(x, y) + d(y, a).

Since ρA(x) ≤ d(x, y) + d(y, a) for all a, ρA(x) − d(x, y) is a lower boundon the set

d(y, a) : a ∈ A

and therefore less than or equal to the greatest lower bound,

ρA(x)− d(x, y) ≤ infa∈A

d(y, a) = ρA(y).

SoρA(x)− ρA(y) ≤ d(x, y).

Reversing the roles of x, y gives

ρA(y)− ρA(x) ≤ d(y, x) = d(x, y),

and threfore|ρA(x)− ρA(y)| ≤ d(x, y).

This implies uniform continuity of ρA(x).

5. (a) Since A and B are disjoint B is closed, no point of A is a limit pointof B, and therefore ρB(x) > 0 for all x ∈ A. Since ρB is continuousand A is compact, ρB attains its minimum on A and some point x∗. Letδ = ρB(x∗)/2. Then

0 < δ < ρB(s∗) = infa∈A

ρB(a) = infa∈A,b∈B

dX(a, b).

So 0 < δ < d(a, b) for all a ∈ A and b ∈ B.

18


(b) Let

A = (x, y) : x ≥ 0, y ≥ e−xB = (x, y) : x ≥ 0, y = 0

These are both closed subsets of R2, but

d((x, e−x), (x, 0)) = e−x → 0

as x→∞.

6. (a) Suppose f is convex, g is increasing, and h(x)− g(f(x)). Suppose 0 < λ <1. Since f is convex

f(λx+ (1− λ)y) ≤ λf(x) + (1− λ)f(x).

Since g is increasing, this implies that

h(λx+ (1− λ)y) = g(f(λx+ (1− λ)y)) ≤ g(λf(x) + (1− λ)f(x)).

Since g is convex,

g(λf(x) + (1−λ)f(x)) ≤ λg(f(x)) + (1−λ)g(f(y)) = λh(x) + (1−λ)h(y).

Combining these yields

h(λx+ (1− λ)y) ≤ λh(x) + (1− λ)h(y).

So h is convex.

(b) Let

λ =u− tu− s

.

Then t = λs+ (1− λ)u. The convexity inequality

f(t) ≤ λf(s) + (1− λ)f(u)

can be written as

f(t)− f(s) ≤ (1− λ)(f(u)− f(s)) =t− su− s

(f(u)− f(s))

or

f(t)− f(u) ≤ λ(f(s)− f(u)) =u− tu− s

(f(s)− f(u)).

These can in turn be rewritten as

f(t)− f(s)

t− s≤ f(u)− f(s)

u− sand

f(u)− f(t)

u− t≥ f(u)− f(s)

u− sas required.

19


(c) Using the inequality

f(t) ≤ u− tu− s

f(s) +t− su− s

f(u)

with s, u fixed and t ↓ s gives

lim supt↓s

f(t) ≤ limt↓s

(u− tu− s

f(s) +t− su− s

f(u)

)= f(s),

Similarly,lim sup

t↑uf(t) ≤ f(u).

This implies that for all x ∈ (a, b)

lim supy→x

f(y) ≤ f(x).

Writing the second inequality of the previous part as

f(s) ≥ f(u)− u− su− t

(f(u)− f(t)) =u− su− t

f(t) +s− tu− t

f(u),

fixing u, t and letting s ↑ t yields

lim infs↑t

f(s) ≥ lims↑t

(u− su− t

f(t) +s− tu− t

f(u)

)= f(t).

A similar argument fixing t, s and letting u ↓ t yields

lim infu↓t

f(u) ≥ f(t).

Together these imply that for any x ∈ (a, b)

lim infy→x

f(y) ≥ f(x).

So f is continuous at every x ∈ (a, b).

20


Assignment 6

Due on Friday, March 4, 2016.

1. Suppose f ′(x) > 0 for x ∈ (a, b). Prove that f is strictly increasing on (a, b)and let g be its inverse function. Prove that g is differentiable on f((a, b)), andthat

g′(f(x)) =1

f ′(x)

for a < x < b.

2. Suppose f is defined and differentiable for every x > 0, and that f ′(x) → 0 asx→ +∞. Let g(x) = f(x+ 1)− f(x). Prove that g(x)→ 0 as x→ +∞.

3. Suppose that f ′(x), g′(x) exist, g′(x) 6= 0, and f(x) = g(x) = 0. Prove that

limy→x

f(y)

g(y)=f ′(x)

g′(x).

4. Let f be a continuous real-valued function on R, of which it is known that f ′(x)exists for all x 6= 0 and that f ′(x) → 3 as x → 0. Does it follow that f ′(0)exists?

5. Suppose f is defined in a neighborhood of x and that f ′′(x) exists. Show that

limh→0

f(x+ h) + f(x− h)− 2f(x)

h2= f ′′(x).

Show by an example that the limit may exist even if f ′′(x) does not exist.

6. The (R, d) be the metric space of the real numbers with the usual distanced(x, y) = |x − y|, and let f : R → R be a continuous function such that thelimits

a = limx→−∞

f(x) b = limx→+∞

f(x)

both exist. Show that f is uniformly continuous.

7. Let fn be a sequence of non-decreasing real-valued functions on a closed andbounded interval [a, b]. Suppose that fn(x)→ f(x) as n→∞ for each x ∈ [a, b],and that the limit f is continuous. Show that fn converges uniformly to f . [Hint:Use the uniform continuity of f .]

21


Solutions

1. By the mean value theorem, if x < y there is a value z ∈ (x, y) such that

f(y)− f(x) = f ′(z)(y − x),

and as f ′ is strictly positive this implies

f(y)− f(x) > 0,

and thus f is strictly increasing and has an inverse.

For x ∈ (a, b) let y = f(x). For any h small enough such that y + h is in therange of f there is a value k(h) such that y + h = f(x+ k(h)). This value canbe written as

k(h) = g(y + h)− x = g(y + h)− g(y).

By Rudin’s Theorem 4.17 the inverse g is continuous and therefore k(h) → 0as h→ 0. In turn,

h = y + h− y = f(x+ k(h))− f(x).

Thereforeg(y + h)− g(y)

h=

k(h)

f(x+ k(h))− f(x)→ 1

f ′(x).

as x→ 0. So g is differentiable at y = f(x) with the specified derivative.

2. Fix ε > 0 and choose M such that |f ′(x)| < ε for x > M . By the mean valuetheorem, for each x > M there exists a y ∈ (x, x+ 1) such that

g(x) = f(x+ 1)− f(x) = f ′(y),

and therefore|g(x)| = |f ′(y)| < ε.

So g(x)→ 0 as x→ +∞.

3. Since f(x) = f(y) = 0,

f(y)

g(y)=f(y)− f(x)

g(y)− g(x)=

f(y)−f(x)y−x

g(y)−g(x)y−x

and therefore

limy→0

f(y)

g(y)=

limy→xf(y)−f(x)

y−x

limy→xg(y)−g(x)y−x

=f ′(x)

g′(x).

22


4. Consider the difference quotient

f(y)− f(0)

y − 0.

Numerator and denominator both dent to zero, and

f ′(y)

1= f ′(y)→ 3

as y → 0. So by l’Hospital’s rule the limit of the difference quotient exists andis equal to 3. So f ′(0) = 3.

5. By l’Hospital’s rule, if f ′′(x) exists, then

limh→0

f(x+ h) + f(x− h)− 2f(x)

h2= lim

h→0

ddh

(f(x+ h) + f(x− h)− 2f(x))ddhh2

= limh→0

f ′(x+ h)− f ′(x− h)

2h

= limh→0

f ′(x+ h)− f ′(x) + f ′(x)− f ′(x− h)

2h

=1

2

(limh→0

f ′(x+ h)− f ′(x)

h+ lim

h→0

f ′(x)− f ′(x− h)

h

)= f ′′(x).

Let

f(x) =

1x

for x 6= 0

0 for x = 0.

Thenf(0 + h) + f(0− h)− 2f(0)

h2= 0

for all h, so the limit exists and is zero, but f is not even continuous at x = 0.

6. Fix ε > 0. Since f(x)→ a as x→ −∞ and f(x)→ b as x→ +∞, we can finda number M > 0 such that

|f(x)− a| < ε

2

for x ≤ −M and

|f(x)− b| < ε

2.

for x ≥ M . Since f is continuous on the compact set [−M − 1,M + 1] it isuniformly continuous, so we can find δ > 0 such that δ < 1 and

|f(x)− f(y)| < ε

whenever x, y ∈ [−M − 1,M + 1] and |x− y| < δ.

23


Assignment 7


1. Suppose a ≤ c ≤ b. If f ∈ R(α) on [a, c] and f ∈ R(α) on [c, b], show thatf ∈ R(α) on [a, b], and ∫ b

a

fdα =

∫ c

a

fdα +

∫ b

c

fdα.

2. Suppose f ≥ 0, f is continuous on [a, b], and∫ baf(x)dx = 0. Prove that f(x) = 0

for all x ∈ [a, b].

3. Suppose f is a bounded real function on [a, b], and that f 2 ∈ R. Does it followthat f ∈ R? Does the answer change if we assume f 3 ∈ R?

4. Let α be a fixed increasing function on [a, b]. For u ∈ R(α) on [a, b] define

‖u‖ =

√∫ b

a

|u|2dα.

Suppose f, g, h ∈ R(α), and prove the triangle inequality

‖f − h‖ ≤ ‖f − g‖+ ‖g − h‖

as a consequence of the Schwarz inequality, as in the proof of Rudin’s Theorem1.37.

5. With the notation of the previous problem, suppose f ∈ R(α) and ε > 0. Provethat there exists a continuous function g on [a, b] such that ‖f − g‖ < ε. [Hint:For a suitable partition P = x0, . . . , xn of [a, b], define

g(t) =xi − t∆xi

f(xi−1) +t− xi−1

∆xif(xi)

if xi−1 ≤ t ≤ xi.]

25


Solutions

1. Fix ε > 0. Then, since f ∈ R(α) on [a, c] and f ∈ R(α) on [c, b], there arepartitions P1 of [a, c] and P2 of [c, b] such that

U(P1, f, α)− L(P1, f, α) <ε

2

U(P2, f, α)− L(P2, f, α) <ε

2.

Now P = P1 ∪ P2 is a partition of [a, b] and

U(P, f, α) = U(P1, f, α) + U(P2, f, α)

L(P, f, α) = L(P1, f, α) + L(P2, f, α).

So

U(P, f, α)− L(P, f, α) = U(P1, f, α)− L(P1, f, α) + U(P2, f, α)− L(P2, f, α)

< U(P, f, α)− L(P, f, α) +ε

2= ε.

So f ∈ R on [a, b].

Furthermore,

U(P, f, α) = U(P1, f, α) + U(P2, f, α)

≤∫ c

a

fdα +ε

2+

∫ b

c

fdαε

2

=

∫ c

a

fdα +

∫ b

c

fdα + ε

for all ε > 0, and therefore∫ b

a

≤∫ c

a

fdα +

∫ b

c

fdα.

Similarly using lower sums shows that∫ b

a

≥∫ c

a

fdα +

∫ b

c

fdα

and therefore ∫ b

a

=

∫ c

a

fdα +

∫ b

c

fdα.

2. Suppose there is a point c ∈ [a, b] with f(c) > 0. Chose zsuchthat0 < z < f(c).Since f is continuous, there exists a δ > 0 such that f(x) > z if |x− c| < δ. Let

u = max(a, c− δ)v = min(b, c+ δ).

26


Then a ≤ u < v ≤ b, and f(x) > z if u < y < v. Let

g(x) =

z if u < x < v

0 otherwise.

Then 0 ≤ g(x) ≤ f(x) for all x ∈ [a, b] and∫ b

a

g(x)dx = z(v − u) > 0.

This implies that ∫ b

a

f(x)dx ≥∫ b

a

g(x)dx > 0,

which is a contradiction.

3. Let

f(x) =

1 if x ∈ [a, b] is rational

−1 otherwise.

Then f 2(x) ≡ 1 ∈ R, but f 6∈ R.

If g = f 3 ∈ R then f = φ(g) with φ(x) =√

3x, so f ∈ R by Rudin’s Theorem6.11.

4. It is enough to show that for any x, y ∈ R(α)

‖f1 + f1‖ ≤ ‖f1‖+ ‖f2‖.

Now

‖f1 + f2‖2 =

∫ b

a

(f1 + f2)rdα

=

∫ b

a

f 21dα +

∫ b

a

f 22dα + 2

∫ b

q

f1f2dα

= ‖f1‖+ ‖f2‖+ 2

∫ b

a

f1f2dα.

Furthermore, by the Schwarz inequality,∣∣∣∣∫ b

a

f1f2dα

∣∣∣∣ ≤ ∫ b

a

|f1f2|α ≤(∫ b

a

f 21dα

) 12(∫ b

a

f 22dα

) 12

= ‖f1‖‖f2‖.

So ‖f1 + f2‖2 ≤ (‖f1‖+ ‖f2‖)2.

27


5. The function g for a partition P specified in the hint is a piece-wise linearinterpolation of the points (xi, f(xi)) and is therefore continuous. Since f ∈R(α) it is bounded. Let M = sup|f(x)| and choose a partition P such that

U(P, f, α)− L(P, f, α) <ε2

2M.

Let Mi and mi be the maximal and minimal values of f over [xi−1, xi]. Theng(x) ∈ [mi,Mi] for x ∈ [xi−1, xi] as well, and therefore |f(x)− g(x)| ≤Mi −mi

for x ∈ [xi−1, xi]. So

U(P, (f − g)2, α) ≤∑

(Mi −mi)2∆αi

≤ 2M∑

(Mi −mi)∆αi

= 2M [U(P, f, α)− L(P, f, α)]

< 2Mε2

2M= ε2,

and therefore ‖f − g‖2 ≤ U(P, (f − g)2, α) < ε2.

28


Assignment 8


1. Suppose x = (x1, x2, . . . , xn) ∈ Rn and for p ≥ 1 define

‖x‖p = p

√√√√ n∑i=1

|xi|p.

Show that limp→∞ ‖x‖p = max |xi|.

2. Let (X,M, µ) be a measure space and f : X → R a measurable function. Theessential supremum of f is

ess sup f = infa : µ(x : f(x) > a) = 0.

For X = [0,∞) and µ Lebesgue measure find the essential supremum of thefunctions

f1(x) =

e−x x > 0

0 x = 0f2(x) =

e−x x rational

x2 x irrational.

3. Let X be a set, f : X → R a function on X, and R be the Borel sigma algebraon R. Define

σ(f) = f−1(A) : A ∈ R.

Show that σ(f) is a sigma-algebra and that it is the smallest sigma algebra withrespect to which f is measurable.

4. Let X = −2,−1, 0, 1, 2 and let f(x) = x2.

(a) Find σ(f). [Hint: there are three points in the range of f , so the rangehas 23 = 8 possible subsets.]

(b) Suppose h is a real valued function on X and is σ(f)-measurable. Showthat there exists a function g : f(X) → R such that h(x) = g(f(x)).In other words, if h(x) is σ(f)-measurable then it can be expressed as afunction of f(x).

29


Solutions

1. Let m = max |xi|. If m = 0 then ‖x‖p = 0 for all p and the claim is true. Ifm > 0 then

‖x‖p = m p

√√√√ n∑i=1

|xi|pm

,

and

1 ≤n∑i=1

|xi|p

m≤ n,

som ≤ ‖x‖p ≤ m p

√n.

Since p√n→ 1 as p→∞ the result follows.

This is the reason the notation ‖x‖∞ = max |xi| is sometimes used.

2. For a ≥ 1 the set x : f1(x) > a is empty and has measure zero. For a < 1 theset x : f1(x) > a = (0,− log a) has positive length. So ess sup f1 = 1.

For a ≥ 1 the set x : f2(x) > a is the set of irrational numbers greater than√a, which has positive measure. So ess sup f2 =∞.

[If the rational/irrational assignments are reversed: For a ≥ 1 the set x :f2(x) > a is a subset of the rationals, which has Lebesgue measure zero. Fora < 1 set x : f2(x) > a consists of the irrational numbers in the interval(0,− log a), which has Lebesgue measure − log(a) > 0. So ess sup f2 = 1.]

3. Since f−1(∅) = ∅, we have ∅ ∈ σ(f). If A ∈ σ(f) then A = f−1(B) for someBorel set B, and

Ac = (f−1(B))c = f−1(Bc).

Since Bc is also a Borel set this means Ac ∈ σ(f); so σ(f) is closed undercomplementation. If A1, A2, · · · ∈ σ(f), then there exist Borel sets B1, B2, . . .such that Ai = f−1(Bi), and

∞⋃n=1

An =∞⋃n=1

f−1(Bn) = f−1

(∞⋃n=1

Bn

).

Since⋃∞n=1Bnis also a Borel set this means

⋃∞n=1An ∈ σ(f), and σ(f) is closed

under countable unions. So σ(f) is a sigma-algebra.

f is measurable with respect to a sigma-algebra M if f−1(B) ∈ M for everyBorel set B. So f is measurable with respect to σ(f), and if f is measurablewith respect toM, then σ(f) ⊂M. So σ(f) is the smallest sigma-algebra withrespect to which f is measurable.

4. (a) The subsets of the range and their inverse images are

30


B f−1(B)∅ ∅0 01 −1, 14 −2, 20, 1 −1, 0, 10, 4 −2, 0, 21, 4 −2,−1, 1, 20, 1, 4 −2,−1, 0, 1, 2 = X

(b) h can only have at most 5 possible values, c1, . . . , c5. Let

Ai = x : h(x) = ci.

Thenh(x) =

∑ci1Ai(x).

If h is σ(f)-measurable then Ai ∈ σ(f) for each i and therefore for each ithere is a subset Bi of the range of f such that Ai = f−1(Bi), and thus

1Ai(x) = 1Bi(f(x)).

Soh(x) =

∑ci1Bi(f(x)) = g(f(x))

withg(y) =

∑ci1Bi(y).

This argument works for any setting where h has most countably manyvalues; the result holds for any extended real-valued h and can be shownby approximating h by functions with ranges that are at most countable.

31


Assignment 9

Due on Wednesday, March 30, 2016.

1. If f ≥ 0 and∫Adµ = 0, prove that f(x) = 0 almost everywhere on A. [Hint:

Let An be the subset of A on which f(x) > 1/n.]

2. If∫Bfdµ = 0 for every measurable subset B of A, show that f(x) = 0 almost

everywhere on A.

3. If fn is a sequence of measurable real-valued functions prove that the set ofpoints x at which fn(x) converges is measurable. [Hint: write the set of pointswhere fn(x) is a Cauchy sequence in terms of countable unions/intersections ofmeasurable sets.]

4. If f ∈ L(µ) on A and g is bounded and measurable on A show that fg ∈ L(µ)on A.

5. Let f : [0,∞)→ R be a bounded measurable function and suppose f is contin-uous at the origin. Show that

limt→∞

∫ ∞0

te−txf(x)dx = f(0).

[Hint: change variables and use the dominated convergence theorem.]

32


Solutions

1. Since An ⊂ A,

0 ≤ 1

n1An ≤ f1A

for all n we have

0 ≤ 1

nµ(An) =

∫1

n1Andµ ≤

∫f1Adµ =

∫A

fdµ = 0.

and therefore µ(An) = 0 for all n. Since An ↑ A this implies µ(A) = 0.

2. Let

B1 = x ∈ A : f(x) > 0 = x : f+(x) > 0B2 = x ∈ A : f(x) < 0 = x : f−(x) > 0.

Then f+ is non-negative on B1, f− is non-negative on B2, and by the previous

problem µ(B1) = µ(B2) = 0. So

µ(x ∈ A : f(x) 6= 0 = µ(B1 ∪B2) = 0

and f = 0 almost everywhere on A.

3. The set of points s where fn(x) converges is the set of points x where fn(x) isa Cauchy sequence. Given ε > 0, the set

An,m,ε = x : |fn(x)− fm(x)| < ε

is measurable since fn− fm is measurable. The set of all points where |fn(x)−fm(x)| < ε for all n,m ≥ N

BN,ε =⋂

n,m≥N

An,m.ε

is a countable intersection of measurable sets and is therefore measurable. Theset of points where for some N we have |fn(x)− fm(x)| < ε for n,m ≥ N is thecountable union

Cε =∞⋃N=1

BN,ε

of measurable sets and is therefore measurable. The set of points where fn(x)is a Cauchy sequence is

C =⋂ε>0

Cε.

Since the Cε decrease as ε decreases,

C =∞⋂k=1

C1/k.

So C is a countable intersection of measurable sets and is therefore measurable.

33


4. Let M = supg(x) : x ∈ G. Then∫A

|fg|dµ ≤M

∫A

|f |dµ <∞

since f ∈ L(µ) on A. So fg ∈ L(µ) on A.

5. Changing variables to y = tx , which works for Lebesgue integrals as well, gives∫ ∞0

te−txf(x)dx =

∫ ∞0

e−yf(y/t)dy

Since f is bounded and f(y/t) → f(0) for all y the dominated convergencetheorem gives

limt→∞

∫ ∞0

e−yf(y/t)dy =

∫ ∞0

e−y limt→∞

f(y/t)dy =

∫ ∞0

e−yf(0)dy = f(0).

34


Assignment 10

Due on Wednesday, April 13, 2016.

1. Let (X,M, µ) be a measure space, (a, b) be an open interval on the real line,and f(x, y) : X×(a, b)→ R a real-valued function. Suppose that f(·, y) ∈ L(µ)on X for every y ∈ (a, b) and define H : (a, b)→ R by

H(y) =

∫X

f(x, y)µ(dx).

Suppose exists a non-negative measurable function g such that∫X

gdµ <∞

and for all x ∈ X and u, v ∈ (a, b) with u 6= v∣∣∣∣f(x, u)− f(x, v)

u− v

∣∣∣∣ ≤ g(x).

Suppose that for each x ∈ X the function f(x, ·) is differentiable with derivative

∂

∂yf(x, y) = h(x, y).

Show that H is differentiable on (a, b) with derivative

H ′(y) =

∫X

h(x, y)µ(dx).

2. A real-valued function f on [a, b] is absolutely continuous if it is almost every-where differentiable and for all x ∈ [a, b]

f(x)− f(a) =

∫ x

a

f ′(t)dt.

Let H be the set of all real-valued absolutely continuous functions f on [0, 1]

such that f(0) = 0 and∫ 1

0(f ′(t))2dt <∞. Let

〈f, g〉 =

∫ 1

0

f ′(t)g′(t)dt.

(a) Show H is a vector space and that 〈f, g〉 is an inner product on H.

(b) Show that H is a complete metric space.

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(c) Show that R(s, t) = min(s, t) is a reproducing kernel for H; i.e. that forany f ∈ H and any x ∈ [0, 1]

f(x) = 〈R(x, ), f〉.

3. Let H be a Hilbert space. A mapping A : H → H is linear if

A(x+ y) = Ax+ Ay

A(cx) = c(Ax)

for all x, y ∈ H and all c ∈ R. Show that a continuous linear mapping A isa projection onto some closed subspace L if and only if it is idempotent, i.e.A2x = A(Ax) = Ax for all x ∈ H, and self-adjoint, i.e. 〈Ax, y〉 = 〈x,Ay〉for all x, y ∈ H. [Hint: to show that an idempotent self-adjoint mapping is aprojection let L = N⊥ where N = x ∈ H : Ax = 0 is the null space of A.

4. Suppose (X,M, µ) and (Y,N , ν) are finite measure spaces (i.e. µ(X) <∞ andν(Y ) <∞) and λ is a measure on M×N such that

λ(A×B) = µ(A)ν(B)

for all A ∈M, B ∈ N . Show that λ(E) = (µ× ν)(E) for all E ∈M×N .

5. Suppose f is a Borel measurable non-negative function on R and define theordinate set Af and the graph Gf of f as

Af = (x, y) : 0 < y < f(x)Gf = (x, y) : y = f(x).

(a) Show that Af is a Borel subset of R2 and that the Lebesgue measure ofAf is equal to the integral of f over R.

(b) Show that Gf is measurable and that the Lebesgue measure of Gf is zero.

36


Solutions

1. Suppose y ∈ (a, b) and let un be a sequence such that un ∈ (a, b), un 6= y, andun → y. Then

H(un)−H(y)

un − y=

∫X

f(x, un)− f(x, y)

un − yµ(dx) =

∫X

hn(x)µ(dx)

Since f(x, ·) is differentiable,

hn(x) =f(x, un)− f(x, y)

un − y→ ∂

∂yf(x, y) = h(x, y).

In addition, for each x, and

|hn(x)| =∣∣∣∣f(x, un)− f(x, y)

un − y

∣∣∣∣ ≤ g(x)

and g has finite integral. So by the dominated convergence theorem

limn→∞

H(un)−H(y)

un − y= lim

n→∞

∫X

hn(x)µ(dx)

=

∫X

limn→∞

hn(x)µ(dx)

=

∫X

h(x, y)µ(dx).

Thus H is differentiable at y with derivative

H ′(y) =

∫X

h(x, y)µ(dx).

2. (a) If f, g ∈ H, c ∈ R, and h = cf + g, then

h(0) = cf(0) + g(0) = 0

and

h(x) = af + g = c

∫ x

0

f ′(t)dt+

∫ x

0

g′(t)dt

=

∫ x

0

cf ′(t) + g′(t)dt =

∫ x

0

(cf + g)′(t)dt.

So h is absolutely continuous with derivative h′ = cf ′ + g′ almost every-where. Furthermore,∫ 1

0

(h′(t))2dt ≤ c2∫ 1

0

(f ′(t))2dt+

∫ 1

0

(g′(t))2dt <∞.

37


Thus h ∈ H, and H is a vector space.

For any f, g ∈ H

〈f, g〉 =

∫ 1

9

f ′(t)g′(t)dt =

∫ 1

9

g′(t)f ′(t)dt〈g, f〉

and for any f, g, h ∈ H and c ∈ R

〈cf + g, h〉 =

∫ 1

0

(cf + g)′(t)h′(t)dt =

∫ 1

0

(cf ′(t) + g′(t))h′(t)dt

= c

∫ 1

0

f ′(t)h′(t)df +

∫ 1

0

g′(t)h′(t)dt = c〈f, h〉+ 〈g, h〉.

For any f

〈f, f〉 =

∫ 1

0

(f ′(t))2dt ≥ 0,

and 〈f, f〉 = 0 implies f ′ = 0 almost everywhere and therefore f(x) = 0for all x ∈ [0, 1]. So 〈f, g〉 is an inner product on H.

(b) Suppose fn ∈ H is a Cauchy sequence. Since L2 on [0, 1] is complete, thereis a function g ∈ L2 on [0, 1] such that∫ 1

0

(f ′n(t)− g(t))2dt→ 0.

Let

f(x) =

∫ x

0

g(t)dt.

Then f is absolutely continuous with derivative f ′ = g almost everywhereand f(0) = 0. So f ∈ H and

‖fn − f‖2 =

∫ 1

0

(f ′n(t)− f ′(t))2dt→ 0.

So H complete.

(c) For any x, y ∈ [0, 1]

R(x, y) = min(x, y) =

∫ y

0

1[0,x](t)dt.

So for any f ∈ H and x ∈ [0, 1]

〈R(x, ·), f〉 =

∫ 1

0

1[0,x](t)f′(t)dt =

∫ x

0

f ′(t)dt = f(x).

38


3. Suppose A is a projection onto a closed space L. For any x ∈ H the projection

0 = Ax is in L, and therefore Ax0 = x0, i.e. A(Ax) = Ax. So A is idempotent.Furthermore, for any y ∈ H

〈Ax, y〉 = 〈Ax, y − Ay〉+ 〈Ax,Ay〉 = 〈Ax,Ay〉

since y − Ay ∈ L⊥. Similarly,

〈x,Ay〉 = 〈x− Ax, y〉+ 〈Ax,Ay〉 = 〈Ax,Ay〉.

So 〈Ax, y〉 = 〈x,Ay〉 and A is self-adjoint.

Suppose A is linear, continuous, idempotent, and self-adjoint. Let N = x ∈H : Ax = 0 be the null space of N and L = N⊥. Since A is continuous, Nis closed, and its orthogonal complement is therefore closed as well. For anyu ∈ N and x ∈ H self-adjointness implies

〈u,Ax〉 = 〈Au, x〉 = 〈0, x〉 = 0.

so Ax ⊥ N and therefore Ax ∈ L for any x ∈ H. Since A is idempotent, forany x ∈ H

A(x− Ax) = Ax− A2x = Ax− Ax = 0.

So x− Ax ∈ N = L⊥ for all x ∈ H. Thus Ax is the projection of x onto L.

4. Let Ω be the collection of all sets E ∈ M×N such that λ(E) = (µ × ν)(E).By assumption Ω contains all measurable rectangles, which include X × Y . IfE ∈ Ω then

λ(Ec) = λ(X × Y )− λ(E) = (µ× ν)(E)− (µ× ν)(E) = (µ× ν)(Ec),

so Ec ∈ Ω. If E1, E2, · · · ∈ Ω are pairwise disjoint and E =⋃Ei, then

λ(E) =∑

λ(Ei) =∑

(µ× ν)(Ei) = (µ× ν)(E),

so E ∈ Ω. So Ω ⊂M×N is a λ-system containing the π-system of measurablerectangles; by the π−λ theorem Ω contains the sigma algebra generated by themeasurable rectangles, and thus Ω =M×N .

5. (a) Since f is non-negative and measurable there exists a sequence of non-negative measurable simple functions sn with sn ↑ f . Then Asn is measur-able since it is a finite union of measurable rectangles, and

Af =∞⋃n=1

Asn ,

So Af is measurable. Now

f(x) = m([0, f(x)]) =

∫1[0,f(x)](y)dy

39


and for all x, y1[0,f(x)](y) = 1Af (x, y).

So Fubini’s theorem gives∫fdm =

∫ ∫1[0,f(x)](y)m(dy)m(dx)

=

∫1Afd(m×m) = (m×m)(Af ).

(b) LetBf = (x, y) : 0 ≤ y ≤ f(x).

There exists a sequence of non-negative measurable simple functions sn,possibly infinite-valued for some x, such that sn ↓ f . Bsn is a finite unionof measurable rectangles and hence measurable, and

Bf =∞⋂n=1

Bsn ,

so Bf is measurable. The graph is

Gf = Bf\(Af ∪ R× 0),

so Gf is measurable. By Fubini’s theorem

(m×m)(Gf ) =

∫1Gf (x, y)(m×m)(dx, dy)

=

∫ ∫1Gf (x, y)m(dy)m(dx)

=

∫m(f(x))m(dx) = 0.

40


Assignment 11

Due on Wednesday, April 20, 2016.

1. Let µ, µ1, µ2, . . . be probabilities on the Borel subsets of the real line. Showthat µn ⇒ µ if and only if

limn→∞

∫fdµn =

∫fdµ

for all f ∈ CK , the collection of continuous functions on R with compact sup-port. [A function f has compact support if there exists a compact set K suchthat f(x) = 0 on Kc.]

2. Let Y1, Y2, . . . be independent, identically distributed real-valued random vari-ables with distribution µ and define the empirical distribution of Y1, . . . , Yn as

µn(A) =1

n

n∑i=1

1A(Yi)

for all Borel sets A. Show that µn ⇒ µ almost surely. [One possible approachis to argue that there is a countable collection of functions in CK that is densein the uniform metric and use the strong law of large numbers. Another isto consider the distribution functions at rational arguments and again use thestrong law of large numbers.]

3. Suppose fn and f are probability densites with respect to a measure m andthat limn→∞ fn(x) = f(x) for all x. Show that

limn→∞

∫|f − fn|dm = 0.

[Hint: argue that (f − fn)+ and (f − fn)− have identical integrals, and applythe dominated convergence theorem to the integral of (f − fn)+.]

41


Solutions

1. Suppose µn ⇒ µ and let f be a continuous function with compact support.Then f is bounded and therefore

limn→∞

∫fdµn =

∫dµ.

Now suppose we have convergence of integrals for all continuous functions withcompact support and let f be a bounded continuous function. Fix ε > 0 andlet M =

∑|f |. Choose A > 0 such that µ([−A,A]) > 1− ε, and let g be

g(x) =

1 −A ≤ x ≤ A

x+ A+ 1 −A− 1 ≤ x ≤ −AA+ 1− x A ≤ x ≤ A+ 1

0 otherwise.

Then g is continuous with compact support and 1[−A,A] ≤ g, so

limn→∞

gdµn =

∫gdµ ≥ µ([−A,A]) ≥ 1− ε.

Also, g ≤ 1. Now

lim supn→∞

∣∣∣∣∫ fdµn −∫fdµ

∣∣∣∣ ≤ lim supn→∞

∣∣∣∣∫ fgdµn −∫fgdµ

∣∣∣∣+ lim sup

n→∞

∣∣∣∣∫ f(1− g)dµn

∣∣∣∣+

∣∣∣∣∫ f(1− g)dµ

∣∣∣∣ .Since fg hs continuous with compact support the first term is zero. The secondterm is bounded by

lim supn→∞

∣∣∣∣∫ f(1− g)dµn


∫|f |(1− g)dµn

≤M lim supn→∞

∫(1− g)dµn

= M

(1− lim

n→∞

∫gdµn

)≤Mε.

The third term is similarly bounded by∣∣∣∣∫ f(1− g)dµ

∣∣∣∣ ≤ ∫ |f |(1− g)dµ ≤Mε.

42


Since ε > 0 was arbitrary this shows that

limn→∞

∫fdµn =

∫fdµ

for any bounded, continuous f and thus µn ⇒ µ.

2. Consider the collection F of continuous functions such that a function f ∈ F is

• equal to a polynomial with rational coefficients on [−r, r] for some positiverational number r;

• is zero on (−∞,−r−1] and [r+1,∞), and is linearly interpolates betweenf(−r − 1) and f(−r) on (−r − 1, r) and between f(r) and f(r + 1) on(r, r + 1).

Then F ⊂ CK and F is countable. Fix f ∈ CK and ε > 0. Choose a rational rsuch that f = 0 on [−r, r]c. By the Weierstrass approximation theorem we canchoose a polynomial p with rational coefficients such that |f(x)−p(x)| < ε for allx ∈ [−r, r]. Chose the g ∈ F that equals p on [−r, r]. Then f(−r) = f(r) = 0and therefore |g(−r)| < ε and |g(r)| < ε. Furthermore, g(x) ≤ g(−r) forx ≤ −r and g(x) ≤ g(r) for x ≥ r. So |g(x)− f(x)| < ε for all x ∈ R.

For a point s in the sample space S write

µsn(A) =1

n

1∑i=1

1A(Yi(s)).

For a bounded function h let Ah be the set of all s ∈ S such that

limn→∞

∫hdµsn = lim

n→∞

1

n

n∑i=1

h(Yi(s)) = E[h(Y1)] =

∫hµ.

By the strong law of large numbers P (Af ) = 1. Let

A =⋃g∈F

Ag.

Since F is countable, P (A) = 1 as well.

Fix s ∈ A, f ∈ CK , and ε > 0. Choose g ∈ F such that |f(x) − g(x)| < ε forall x ∈ R. Then

lim supn→∞

∣∣∣∣∫ fdµsn −∫fdµ


∣∣∣∣∫ gdµsn −∫gdµ

∣∣∣∣+ 2ε = 2ε

since s ∈ A ⊂ Ag. Since this holds for all ε > 0 we have

limn→∞

∫fdµsn

for all f ∈ CK and thus µsN ⇒ µ for all s ∈ A. Since P (A) = 1 this meansµsn ⇒ µ for almost all s, or µn ⇒ µ almost surely.

43


3. Since fn and f are probability densities,∫f − fndm = 0

and thus ∫(f − fn)+dm =

∫(f − fn)−dm.

This implies that∫|f − fn|dm =

∫(f − fn)+dm+

∫(f − fn)−dm = 2

∫(f − fn)+dm.

Now 0 ≤ (f − fn)+ ≤ f , the function f is integrable, and (f(x)− fn(x))+ → 0for all x, so by the dominated convergence theorem

limn→∞

∫(f − fn)+dm = 0.

44


Assignment 12

Due on Monday, May 2, 2016.

1. Let B(t) be Brownian motion on [0, 1] and set

W =

∫ 1

0

B(t)dt.

Explain why W is well defined as a Riemann integral, and why W is normallydistributed, and find the mean and variance of W . [To show that W is normal,argue that it is the limit of Riemann sums and that these are linear combinationsof jointly normal random variables. To compute the variance, write W 2 as aproduct of integrals, which in turn can be viewed as a double integral.]

2. Suppose B(t) is Brownian motion on [0,∞) and for a > 0 let

Ta = inft : B(t) ≥ a,

with Ta infinite if B(t) never crosses a.

(a) Use the reflection principle to show that

P (Ta ≤ t) = 2P (B(t) ≥ a) = 2(1− Φ(a/√t))

where Φ is the CDF of the standard normal distribution. [Argue that theevent Ta ≤ t is the same as the event sup0≤s≤tB(s) ≥ a.]

(b) Show that P (Ta <∞) = 1.

(c) Show that E[Ta] = ∞. [Show that for large enough t the probabilityP (Ta > t) can be bounded below by a constant times 1/t.]

3. Suppose f(0, 0) = 0 and for (x, y) 6= (0, 0)

f(x, y) =xy

x2 + y2.

Prove that (D1f)(x, y) and (D2f)(x, y) exist for all x, y even though f is notcontinuous at (0, 0).

4. Suppose f is a differentiable real-valued function on an open subset E of Rn,and that f has a local maximum at a point x ∈ E. Prove that f ′(x) = 0.

45


Solutions

1. W is well defined as, for each point in the sample space, it is the Riemannintegral of a continuous function. For every sample path W can be computedas

W = limn→∞

1

n

n∑k=0

B

(k

n

)= lim

n→∞Wn.

Wn is a linear combination of jointly normal random variables and is thereforenormally distributed. Limits in distribution of normal random variables arenormally distributed, so W is normally distributed. By Fubini’s theorem, themean of W is

E[W ] = E

[∫ 1

0

B(t)dt

]=

∫E[B(t)]dt = 0,

and the variance is

E[W 2] = E

[∫ 1

0

B(t)dt

∫ 1

0

B(s)ds

]=

∫ 1

0

∫ 1

0

E[B(s)B(t)]dsdt

=

∫ 1

0

∫ 1

0

min(s, t)dsdt =

∫ 1

0

(∫ t

0

sds+

∫ 1

t

tds

)dt

=

∫ 1

0

t− t2

2dt =

1

3.

2. (a) Since Ta ≤ t if and only if B(s) ≥ a for some s ∈ [0, t],

P (Ta ≤ t) = P ( sup0≤s≤t

B(t) ≥ a)

= P ( sup0≤s≤t

B(t) ≥ a,B(t) ≥ a) + P ( sup0≤s≤t

B(t) ≥ a,B(t) < a).

By the reflection principle,

P ( sup0≤s≤t

B(t) ≥ a,B(t) < a) = P ( sup0≤s≤t

B(t) ≥ a,B(t) ≥ a),

and

P ( sup0≤s≤t

B(t) ≥ a,B(t) ≥ a) = P (B(t) ≥ a) = 1− Φ(a/√t).

So P (Ta ≤ t) = 2(1− Φ(a/√t)).

(b) P (Ta <∞) = limt→∞ P (Ta ≤ t) = 1, so Ta is finite with probability one.

(c) The function f(x) = 1 − 2(1 − Φ(x)) satisfies f(0) = 0 and f ′(0) > 0 sofor some ε > 0 and b > 0 we have f(x) ≥ bx for 0 ≤ x < ε. Therefore

E[Ta] =

∫ ∞0

P (Ta > t)dt =

∫ ∞0

f(a/√t)dt ≥

∫ ∞1/ε

ba/√tdt =∞.

46


3. For y 6= 0

D1f(x, y) =y(y2 − x2)(x2 + y2)2

.

f(x, 0) = 0 for all x, so D1f(x, 0) = 0. D2f is analogous. But for y = x andx 6= 0

f(x, x) =x2

x2 + x2=

1

2.

So f is not continuous at (0, 1).

4. If f has a local maximum at x then for any v ∈ Rn

g(t) = f(x+ tv)

is defined in a neighborhood of zero and has a local maximum at zero. g isdifferentiable with derivative

g′(0) = f ′(x)v,

and g′(0) = 0 since zero is a local maxumum. So f ′(x)v = 0 for all v and thusf ′(x) = 0.

47

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