2015 Spring Comprehensive in Analysis

Problem 1.

Prove that

Proof.

We use the integral test. Define

The function is positive and decreasing for . Indeed,

for .

Therefore the convergence of the series follows from the convergence of the improper integral

Using the substitution , , we get

Letting , we obtain

Hence, by the integral test,

Problem 2.

Compute

Justify your answer.

Proof.

Let

We show that .

Set . Then . Integrating by parts with

we get

Therefore

Since , , and is integrable on , the right-hand side is bounded by for some constant independent of . Hence

Thus

Problem 3.

Assume that and

Show that for any , there exists such that

that is, is onto.

Hint: Consider and .

Proof.

Fix an arbitrary and define

Then and

We claim that

First, as , we have , so

Since , it follows that .

Second, as , we have , so

Since , it again follows that .

Thus tends to at both ends of the real line. Hence there exists such that

whenever . Since is continuous on the compact interval , it attains a minimum at some point . Because for , this minimum is attained at an interior point of .

Therefore Fermat's theorem gives

Since , we get

Because was arbitrary, is onto .

Problem 4.

(a) State Stokes' Theorem.

(b) Evaluate the following integral:

where

Proof.

(a) Stokes' Theorem says the following. If is an oriented smooth manifold with boundary , and is a smooth differential form of degree one less than , then

where has the induced boundary orientation.

(b) Let

By Stokes' Theorem,

Now

Also,

because differentiating only produces and terms, which wedge to zero with . Finally,

because .

Therefore

Hence

Now change variables

Then is obtained from the unit ball

and the Jacobian is

Thus

By symmetry,

In spherical coordinates,

Therefore

It follows that

Thus, with the standard outward boundary orientation,

Problem 5.

Prove that is uniformly continuous on .

Proof.

Let

We use the fact that is Lipschitz with constant on :

for all .

For , we have

Also,

Indeed, assuming , this becomes

which is equivalent to , true because .

Thus

Given , choose

If , then

Therefore is uniformly continuous on .

Problem 6.

Let

(a) Prove as pointwise on .

(b) Prove or disprove as uniformly on .

Proof.

(a) If , then

for every , so .

Now fix . Then

Dividing numerator and denominator by , we get

As , the numerator tends to and the denominator tends to . Therefore

Hence pointwise on .

(b) The convergence is not uniform. We show that

does not tend to .

For , compute

Thus the critical point is determined by

At this point,

Therefore

In particular,

Thus

so the supremum cannot tend to . Therefore pointwise but not uniformly on .

Problem 7.

Let be a metric space and be a compact set. Prove that is closed.

Proof.

We prove that is open.

If , then is closed. So assume .

Let . We want to find such that

For each , since , we have

Let

Then the collection

is an open cover of .

Since is compact, there exist finitely many points such that

Now define

Since this is the minimum of finitely many positive numbers, we have

We claim that

Suppose not. Then there exists . Since , there is some such that

Thus

Also, since and , we have

Therefore, by the triangle inequality,

which is impossible.

Hence

Therefore

Since every has an open ball contained in , the set is open. Hence is closed.

Problem 8.

Let be a function on the unit disc

with and existing for all . Prove or disprove each statement.

(a) If and are bounded on , then is continuous on .

(b) If and are continuous on , then is differentiable on .

Proof.

(a) The statement is true.

Assume that and are bounded on . Then there exists such that

for all .

Fix . Since is open, there exists such that the closed square

is contained in .

For in this square, write

For the first term, apply the one-variable Mean Value Theorem to the function on the segment from to . Since , we get

Similarly, applying the Mean Value Theorem to gives

Therefore

This tends to as . Thus is continuous at . Since was arbitrary, is continuous on .

(b) The statement is true.

Assume and are continuous on . We prove differentiability at an arbitrary point .

Because is open, for all sufficiently small the line segments used below remain inside . Write

By the one-variable Mean Value Theorem, for some ,

and

Therefore

Since and are continuous at , both differences in parentheses tend to as . Hence

Thus is differentiable at , with differential

Since was arbitrary, is differentiable on .

Problem 9.

Let satisfy

Prove:

(a) For any ,

(b) If and are in and , respectively, then is Lebesgue integrable and

Proof.

(a) This is the Young's inequality. We prove it directly.

Assume and satisfy

If , then the inequality is clear. So assume .

Define

We want to prove

Differentiate with respect to :

Thus the critical point satisfies

Since

this critical point gives the minimum of .

Now let

Since

we have

and

Therefore

Hence the minimum value of is , so

Therefore

which gives

(b) We prove the stronger Holder inequality

This will imply the stated inequality because

Let

If or , then almost everywhere or almost everywhere, so the result is immediate.

Assume now that and . Define

By part (a),

for almost every . Integrating, we obtain

But

Therefore

Multiplying by , we get

This proves that and gives