2025 Fall Comprehensive in Analysis

Problem 1.

Let be a sequence of continuous functions that converges uniformly to . Show that

Proof.

Let

Then is Lipschitz with constant on , since

Hence, for all ,

Therefore,

Since uniformly on ,

Thus

Problem 2.

Suppose is continuously differentiable and

Define

Prove that

Proof.

By the fundamental theorem of calculus,

Therefore,

Hence

Since as , for every , there exists such that if , then

Thus if , then implies , and so

Therefore

Problem 3.

Let be compact, and suppose is a sequence of continuous functions converging uniformly to . Show that is continuous.

Proof.

Let and let . Since uniformly on , there exists such that

Since is continuous at , there exists such that if and , then

Therefore, if and , then

Hence

Thus is continuous at . Since was arbitrary, is continuous on .

Problem 4.

Define by

Determine whether is Riemann integrable on and justify the answer.

Proof.

We claim that is not Riemann integrable on .

First, is continuous at . Indeed, if , then , and if , then . Hence

Now let . Choose a sequence of rational numbers and a sequence of irrational numbers . Then

while

Since and , we have . Thus the two limits are different, so is discontinuous at every .

Therefore the set of discontinuities of is

which has positive Lebesgue measure. By the Lebesgue criterion for Riemann integrability, a bounded function on a compact interval is Riemann integrable if and only if its set of discontinuities has measure zero. Hence is not Riemann integrable on .

Problem 5.

Let be differentiable and suppose there exists a constant such that

for all . Prove that is Lipschitz continuous, namely for all ,

Proof.

Fix . Define

Thus

Define

Then is differentiable and

By the Cauchy-Schwarz inequality,

Therefore,

Hence

Thus is Lipschitz continuous with Lipschitz constant .

Problem 6.

A function is lower semi-continuous at if for all , there exists such that, for all , if , then

The function is lower semi-continuous if it is lower semi-continuous at every .

(a) Prove that is lower semi-continuous if and only if

is closed.

(b) If are lower semi-continuous functions, show that

is also lower semi-continuous.

(c) If is lower semi-continuous and bounded from below, show that attains a minimum value on any interval .

Proof.

(a) Let

First assume that is lower semi-continuous. Let

and suppose

We need to show that , that is,

Since is lower semi-continuous at , for every , there exists such that if , then

For sufficiently large, , so

Since , we have . Thus

for all sufficiently large . Letting , we get

Since was arbitrary,

Hence is closed.

Conversely, suppose is closed. We prove that is lower semi-continuous. Suppose not. Then there exist and such that for every , there exists with and

Thus we can choose a sequence such that

Let

Then

for every , so

Since and is closed, we get

Therefore

But , a contradiction. Hence is lower semi-continuous.

(b) For each , since is lower semi-continuous,

is closed. Now

Thus

if and only if

for every . Therefore

Since an intersection of closed sets is closed, is closed. By part (a), is lower semi-continuous.

(c) Let be a compact interval. Since is bounded from below, define

Choose a sequence such that

Since is compact, there exists a subsequence and a point such that

By lower semi-continuity,

Since , we get

But by definition of ,

Therefore

Hence attains its minimum value on .

Problem 7.

Let be a sequence of continuous functions such that for some ,

for all . Define by

Show that there exists a subsequence of that converges uniformly on .

Proof.

We prove that is uniformly bounded and equicontinuous. Then the Arzela-Ascoli theorem gives a uniformly convergent subsequence.

First, by the Cauchy-Schwarz inequality,

Since ,

Also,

Therefore

for all and all .

Now let . Again by Cauchy-Schwarz,

For nonnegative ,

Thus

Hence

Therefore

This bound is independent of , so is equicontinuous.

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

Problem 8.

Let be a continuous function. Prove that, for any non-decreasing continuous function , we have

Moreover, if is strictly increasing, prove that equality holds if and only if is a constant function.

Proof.

Let

We compute

Using double integrals, this equals

Symmetrizing, we obtain

Since is non-decreasing, for every ,

Therefore

Hence

Now assume that is strictly increasing. If is constant, then equality is immediate.

Conversely, suppose equality holds. Then

The integrand is continuous and nonnegative. Hence it must be identically zero:

for all . Since is strictly increasing, if , then

which is impossible. Thus

for all . Therefore is constant.

Problem 9.

Let

denote the closed ball with radius . Evaluate the following integral:

Recall that the volume of is

Proof.

Let

In polar coordinates in ,

Thus

The volume element is

because the surface area of the unit sphere in is . Therefore

Hence

Thus

Substituting

we get