2019 Spring Comprehensive in Analysis

Problem 1.

Let be an open set and suppose that

Show that there exists such that

Proof.

Let

Then is compact and . Since is open, for every there is a radius such that

The balls form an open cover of . By compactness, there are points such that

Set

We claim that every point whose distance from is less than belongs to . Indeed, if satisfies , choose such that . Choose with . Then

so .

Now choose

If , then either , in which case , or , in which case

Therefore in all cases. Hence

Problem 2.

Let and be two sequences of functions defined on such that converges uniformly to , and converges uniformly to on . Does it follow that converges uniformly to ? Explain your answer.

Proof.

No, not in this generality. The issue is that the limiting functions need not be bounded, because no continuity or boundedness assumption is given.

Define

and set

for every . Then uniformly, since identically.

Now define

for every . Then uniformly on . Hence , so

However,

This sequence does not converge uniformly to . Indeed, for each fixed and for every , choosing gives

Thus the functions are unbounded on , so they cannot converge uniformly to the zero function.

Therefore uniform convergence of and alone does not imply uniform convergence of to .

Problem 3.

A metric on a space is called an ultrametric if the triangle inequality is replaced by the following stronger property: for all , we have

Let be an ultrametric space. Prove the following:

(1) If is an open ball in , then any point in is a center of . Recall that an open ball is a set of the form

is referred to as a center of the ball.

(2) Every open ball in is both open and closed.

Proof.

Let be an open ball, and let . Thus

We prove that

First let . Then . By the ultrametric inequality,

Hence . Therefore

Conversely, if , then , so again by the ultrametric inequality,

Thus , and hence

Therefore . This proves that every point of is a center of .

Now we show that every open ball is closed. Let and let . Then

We claim that

If not, then there exists . Then

and

By the ultrametric inequality,

contradicting . Hence is contained in the complement of . Therefore every point of has an open neighborhood contained in , so is open. Thus is closed.

Since is an open ball, it is open by definition. Therefore every open ball in an ultrametric space is both open and closed.

Problem 4.

Let . Show that if every continuous function attains its maximum on , i.e.,

for some , then is compact.

Proof.

We prove that is closed and bounded. Since compactness in is equivalent to being closed and bounded, this will prove the result.

First suppose that is unbounded. Define

This function is continuous on and satisfies

for every . Since is unbounded, there is a sequence such that . Therefore

Thus

but for every . Hence does not attain its maximum on , contradicting the hypothesis. Therefore must be bounded.

Next suppose that is not closed. Then there exists a point

Define

This function is continuous on . Since , there exists a sequence such that . Hence

Therefore

But , so for every , and hence

for every . Thus does not attain its maximum on , again contradicting the hypothesis. Therefore is closed.

So is closed and bounded in . By the Heine-Borel theorem, is compact.

Problem 5.

Suppose that is a function. Prove that is continuously differentiable if and only if: for every , there are open intervals such that

and such that, for each and each with and , we have

Proof.

First suppose that is continuously differentiable on . Since is continuous on the compact interval , it is uniformly continuous. Hence, for every , there exists such that

Choose finitely many open intervals covering , each with length less than .

Fix and take with and . By the mean value theorem, there exist points between and , and between and , such that

and

Since all lie in the same interval , the points and also lie in . The length of is less than , so

Therefore

This proves the required local oscillation condition for difference quotients.

Conversely, assume the stated condition. We first prove that is differentiable. Fix . Let . By the hypothesis, there is a finite open cover satisfying the stated condition with this . Choose with . Since is open, there exists such that

For with and , apply the hypothesis with

We get

Thus the difference quotients

form a Cauchy family as within . Since is complete, the limit exists. Therefore exists, with the usual one-sided interpretation at the endpoints and .

It remains to prove that is continuous. Let . Choose a finite open cover satisfying the hypothesis with this . Fix . If , then for and with , , the hypothesis gives

Passing to the limit as and , we obtain

Now fix . Choose with . Since is open, there exists such that

For every , the previous estimate gives

Thus is continuous at . Since was arbitrary, is continuous on . Hence is continuously differentiable.

Problem 6.

Let belong to , where and are open sets in . Assume the determinant of the matrix of first derivatives of is the constant function . Denote the variables in by and the variables in by . Recall that, for any differential form on , denotes the differential form on obtained by change of variables using .

(a) Show that if , then .

(b) Let . Show that

Proof.

Write

Then under the change of variables , we have

(a) Since pulls back to and pulls back to , we get

Now

and

Therefore

The coefficient is exactly . By hypothesis,

Hence

(b) Since

we have

Pullback commutes with exterior differentiation, so

By part (a),

Also,

Therefore

Problem 7.

Let be continuous. Let

and

Show that for any , there exists a nondecreasing function on such that

Proof.

Assume . Since is continuous on , the intermediate value theorem implies that for every , there exists such that

We will choose to be a nondecreasing step function with a single jump of size at .

If , define

In either case, is nondecreasing on and has exactly one jump of size at , interpreted from the side that lies inside the interval of integration.

For a continuous function , the Riemann-Stieltjes integral against a unit step at equals the value of the integrand at . Hence

Since , we obtain

Thus such a nondecreasing function exists.

Problem 8.

Let be a continuously differentiable map where is the open unit ball in centered at the origin. Suppose

for all . Show that

for all .

Proof.

Let . Since is convex, the line segment from to lies entirely in . Define

Then for all .

Now define

Since is continuously differentiable, is differentiable, and by the chain rule,

Therefore, by Cauchy's inequality and the hypothesis ,

Thus

Hence

for all .