2024 Spring Comprehensive in Analysis

Problem 1.

Let be a sequence of positive numbers and suppose

Show that the limit exists and has the same value .

Proof.

First suppose . Fix . Then for all sufficiently large ,

Thus, for ,

Taking th roots gives

Letting , we obtain

Since is arbitrary, .

If , then for every we have for all sufficiently large . Hence for some constant , so

Since is arbitrary, .

Problem 2.

For each , investigate convergence or divergence of the integral

Proof.

For ,

and

Hence the integral converges for .

For divergence, let

On , we have . Also, for ,

Therefore, on ,

It follows that

The last sum is comparable to

which diverges exactly when . Thus the integral converges if and only if

Problem 3.

Let be a non-empty subset of . Define by

Prove that is Lipschitz with constant , that is, for any ,

Proof.

For any and any , the triangle inequality gives

Taking the infimum over yields

Hence

Interchanging and gives

Therefore

Problem 4.

Let be a bounded set, and let be a uniformly continuous function. Prove that is bounded on .

Proof.

Suppose not. Then there is a sequence such that

Since is bounded, the sequence has a convergent subsequence in , say . In particular, is Cauchy.

By uniform continuity, for every there exists such that whenever and , we have

Since is Cauchy, it follows that is Cauchy in . Hence is bounded, contradicting . Therefore is bounded on .

Problem 5.

Let be a compact metric space and let and be continuous functions for . Assume that for all and that converges pointwise to . Show that the convergence is uniform and that this claim is not true in general if is not assumed to be continuous.

Proof.

Let

Since and are continuous, each is continuous. Also, because and pointwise,

for every .

Fix and define

Each is open. Since pointwise, we have

Also, the sets are increasing: . By compactness, finitely many cover . Since they are increasing, there is some such that

Thus for every . For , , so

for all . Therefore uniformly.

The continuity of is necessary. For example, take and

Then each is continuous and

Pointwise,

The limit is not continuous. The convergence is not uniform, since for ,

and hence

for every .

Problem 6.

Let be the open set

For which does the integral

converge, where

Proof.

First, . For , the condition

is equivalent to

Thus, near the origin,

for any sufficiently small .

As ,

Also, for small and , we have

Hence the contribution near the origin is comparable to

This converges exactly when

that is,

So convergence near the origin requires .

Next consider behavior at infinity. Since

the integral diverges for , because over this subset the integral is comparable to

which diverges when .

Conversely, if , then away from a small neighborhood of the origin,

Using , the last integral is equal to

which is finite exactly when .

Combining the origin condition and the infinity condition, the integral converges exactly for

Problem 7.

Find the flux of the vector field

in through the surface of the unit sphere, oriented by the outward-pointing normal.

Proof.

By the divergence theorem, the outward flux through the unit sphere is

Now

Therefore

by symmetry of the unit ball. Hence the flux is .

Problem 8.

Show that the mapping

is locally invertible at every point .

Proof.

The Jacobian matrix of is

Therefore

Since and , we have

for every . By the inverse function theorem, is locally invertible at every point of .

Problem 9.

Show that the set of natural numbers can be represented as a union of uncountably many distinct subsets

in such a way that for any , either , or the intersection is finite.

Proof.

We construct an uncountable almost disjoint family of subsets of whose union is .

Let denote the set of all finite binary strings. This set is countable, so choose a bijection

For each infinite binary sequence

define

That is, is the set of natural numbers corresponding to the finite initial segments of .

There are uncountably many infinite binary sequences, so the collection

is uncountable. If , then and first differ at some coordinate . Hence they have exactly the same initial segments only up to length . Therefore

is finite. In particular, the sets are distinct.

Finally, every finite binary string is an initial segment of some infinite binary sequence. Therefore every element of lies in some chain of initial segments, and hence

Thus is the union of uncountably many distinct subsets such that any two distinct subsets have finite intersection.