2020 Spring Comprehensive in Analysis

Proof.

Let denote the set of limit points of . We prove by contradiction. Suppose that is countable.

Every point of is an isolated point of . Hence for each , there exists an open interval such that

Choose rational numbers such that

Then

The assignment

is injective from into the countable set

Indeed, if two different points were assigned the same rational interval , then would contain both and , contradicting .

Thus is countable. Since is countable by assumption, we get

is countable, because it is contained in the union of two countable sets. This contradicts the assumption that is uncountable.

Therefore must be uncountable.

Proof.

Consider the function

The function is continuous. Indeed, for , the triangle inequality gives

Thus is continuous with respect to the product metric.

Since is compact, the product is also compact. Therefore, by the extreme value theorem, attains its maximum on . Hence there exist such that

Equivalently,

Thus for every ,

as required.

Proof.

The function is a nondecreasing step function. For each , the term

has a jump of size at the point . Therefore has jumps at

and the jump size at is .

For a step-function integrator, the Riemann-Stieltjes integral is the sum of the values of the integrand at the jump points times the corresponding jump sizes. Hence

Thus

Therefore

Proof.

Define

Since and are continuous on and differentiable on , the function is continuous on and differentiable on .

Compute

Hence

By Rolle's theorem, there exists such that

But

Therefore

which is exactly

Proof.

Since the sequence is bounded on , there exists such that

for all and all .

We first show that the family is equicontinuous on . Let be any convex function satisfying on . If , then convexity implies that slopes of secant lines are monotone. In particular,

Since and , and since , we obtain

Therefore

for all . Thus the family is uniformly bounded and equicontinuous on .

By the Arzela-Ascoli theorem, a uniformly bounded and equicontinuous family of real-valued functions on the compact metric space has a uniformly convergent subsequence. Hence there exists a subsequence and a continuous function on such that

uniformly on .

Proof.

(a) Let be a Cauchy sequence in . Then is Cauchy with respect to the supremum norm. Hence for each , the sequence is Cauchy in . Since is complete, the pointwise limit

exists for each .

Moreover, because is Cauchy in the supremum norm, the convergence is uniform. Indeed, for every , there exists such that if , then

Letting , we get

for all . Thus uniformly.

Since each takes values in , the limit also takes values in . Also, if , then

for every . Passing to the limit gives

Thus is nondecreasing, so . Therefore every Cauchy sequence in converges to an element of , and is complete.

(b) We construct a sequence in with no convergent subsequence. For , define

Each is nondecreasing and takes values in , so .

If , then the jump points

are different. Assume for example that

Choose strictly between these two numbers. Then one of equals and the other equals . Hence

Therefore no subsequence of is Cauchy, and hence no subsequence converges.

In a compact metric space, every sequence has a convergent subsequence. Since the sequence has no convergent subsequence, is not compact.

Proof.

Let denote the volume of the -dimensional ball of radius . By scaling, we have

Thus it remains to compute .

The standard formula for the volume of the unit ball in is

Therefore

For completeness, we recall why this formula holds. The Gaussian integral gives

Using polar coordinates in , the same integral is

where is the surface area of the unit sphere in . Since

we get

Thus

The volume of the unit ball is

Using

this becomes

Hence

Proof.

(a) This is the Cauchy-Schwarz inequality for the Riemann integral. If , then

so except possibly on a set of measure zero, and hence

The inequality is then trivial. Similarly, the case is trivial.

Assume now that . For every real number , we have

Expanding gives

This quadratic polynomial in is nonnegative for all . Therefore its discriminant is nonpositive:

Hence

Taking square roots gives

(b) Let

Then

Using part (a), we get

Taking square roots, we obtain

Therefore