2022 Spring Comprehensive in Analysis

Problem 1.

Suppose is a positive number for each . Is it true that

Prove or give a counterexample.

Proof.

It is not true in general.

Define

Then for all .

Fix . If , then . Hence

for every fixed . Therefore

On the other hand, fix . If , then

The finitely many values with do not affect the limit superior as . Thus

for every fixed . Therefore

Thus the two sides are and , respectively, so they are not equal.

Problem 2.

Let be a continuous positive function, and define by

Prove that is locally, that is, restricted to a sufficiently small neighborhood of any point in , a diffeomorphism.

Proof.

Let

Since is continuous, is and

Then

Therefore is .

We compute its Jacobian matrix:

This matrix is lower triangular, so

Since is positive everywhere,

Hence

at every point .

By the inverse function theorem, for every point , there exists a neighborhood of such that is a diffeomorphism from onto its image.

Problem 3.

Let be a sequence of continuously differentiable functions on such that

for , and

for each . Prove that the sequence has a subsequence that converges uniformly on .

Proof.

We prove that is uniformly bounded and equicontinuous, and then apply the Arzela-Ascoli theorem.

First, for , we have

Since

we get

Because the function is Holder continuous on , we also have

Thus

This estimate is independent of , so is equicontinuous on .

Next we prove uniform boundedness. For any ,

Using the hypothesis , we obtain

From the derivative estimate, for any ,

Hence

for all and all .

Therefore is uniformly bounded and equicontinuous on the compact interval . By the Arzela-Ascoli theorem, there exists a subsequence that converges uniformly on .

Problem 4.

continuous on ? Explain.

Proof.

Yes, is continuous on .

Fix . We prove that the series converges uniformly on the compact interval . For ,

Consider the numerical series

By the root test,

Therefore

converges.

By the Weierstrass -test,

converges uniformly and absolutely on . Each function

is continuous, so the uniform limit on is continuous on .

Since was arbitrary, is continuous at every real number. Hence is continuous on .

Problem 5.

Let be a compact metric space, and suppose that the sequence in decreases pointwise to a continuous function ; that is,

for each and , and

for each . Prove that the convergence is actually uniform on .

Proof.

Define

Since and are continuous, each is continuous on . Also,

for all , because decreases to . Moreover,

for all , and

pointwise on .

We need to prove that

Suppose not. Then there exists such that for every there is some with

Since is continuous and is compact, attains its maximum. Thus we can choose a subsequence and points such that

By compactness of , after passing to a further subsequence, we may assume

Since , choose such that

By continuity of , for all sufficiently large ,

Also, for all sufficiently large , we have . Since decreases in ,

which contradicts

Therefore

This means

so uniformly on .

Problem 6.

Let

be the disc in centered at and radius . Let be the unit outer normal vector to at . Compute

Proof.

The vector field is

It is smooth away from the origin. For ,

Computing gives

and

Hence

away from the origin.

The disc contains the origin, because the distance from to the center is . Let be the disc of radius centered at the origin, with so small that . Apply the divergence theorem to the annular region

Since on ,

where is the outward normal for the annular region.

On the inner boundary , the outward normal for points toward the origin, so

On ,

Since on ,

Therefore

and hence

Problem 7.

Let and be two sequences of real numbers such that

Prove that

Proof.

Write

where

Then

Expanding,

Since and , their Cesaro means also converge to :

Therefore the second and third terms tend to .

It remains to show that

The sequences and are bounded, say

for all .

Let . Choose such that

whenever .

For a fixed , in the sum

all terms except possibly those with or have both indices at least . There are at most exceptional terms. Hence

Taking gives

Since is arbitrary, this term tends to .

Thus

Problem 8.

Let be a differentiable function on such that

Prove that there is an such that

Proof.

If is constant, then for every , and we are done.

Assume is not constant. Then there exists such that

First suppose

Choose such that

Since

there exists such that

whenever . Hence for ,

Thus the global maximum of is attained inside the compact interval . Since the values outside are strictly less than , this maximum occurs at an interior point . By Fermat's theorem,

Now suppose

Choose such that

For sufficiently large , whenever we have

Thus the global minimum of is attained inside , and it occurs at an interior point . Again by Fermat's theorem,

In all cases, there exists such that .

Problem 9.

Let with

(i) Prove that

(ii) Let and be bounded Riemann integrable real-valued functions on the unit ball . Prove Holder's inequality:

Proof.

(i) Fix and define

Then

The critical point occurs when

that is,

At this point,

and

Thus

Since

this critical point is the global minimum. Hence

for all , which is exactly

(ii) Let

If or , then the inequality is immediate. So assume

Apply part (i) pointwise to

Then

Integrating over , we get

By the definitions of and ,

Therefore

Finally,

Thus