2021 Spring Comprehensive in Analysis

Proof.

Let denote the set of limit points of . We prove that is uncountable.

First we show that the set of isolated points of is countable. A point is isolated in if there is an open interval such that

For each isolated point , choose rational numbers such that

Then

The pair determines uniquely, because if two different isolated points had the same rational interval , then would contain both and , which is impossible.

Thus the isolated points of inject into the countable set

so the isolated points of form a countable set.

Now every point of is either isolated in or is a limit point of . Hence

where is the countable set of isolated points of . If were countable, then would be countable, and therefore would be countable, contradicting the assumption that is uncountable.

Therefore is uncountable.

Proof.

Let . Then

Rewrite this as

This is the right-endpoint Riemann sum for the continuous function

on . Therefore

Since , we get

Hence the sequence converges and

Proof.

Define

Since strictly decreases distances, it is continuous. Indeed, if , then for ,

so . Therefore is continuous on the compact space .

By compactness, attains its minimum at some . We claim that .

Suppose instead that

Then . Applying the strict contraction property to the distinct points and gives

That is,

which contradicts the fact that minimizes . Hence

and therefore

It remains to prove uniqueness. If and are two fixed points and , then

which is impossible. Thus the fixed point is unique.

Proof.

For , make the change of variables . Then , and

Therefore

Equivalently,

For each fixed , as we have , so by continuity of ,

Also, since is continuous on , it is bounded. Let

Then

and

By the dominated convergence theorem,

Finally,

Hence

Proof.

Since the family is equicontinuous, each is continuous. Therefore each

is continuous on .

The sequence is pointwise increasing, because

Since the family is uniformly bounded, there exists such that

for all and all . Hence

Thus for each , the increasing bounded sequence converges. Define

We next prove that is continuous. Let . By equicontinuity, there exists such that whenever ,

for every . For such , we have

for every . Taking the supremum over gives

By symmetry,

Therefore

so is continuous.

Now is an increasing sequence of continuous functions on compact , and it converges pointwise to the continuous function . By Dini's theorem, the convergence is uniform.

For completeness, we recall the proof of the needed form of Dini's theorem. For , let

Each is closed, and the sets are decreasing:

Since for every , we have

By compactness, a decreasing sequence of nonempty compact sets cannot have empty intersection. Hence for some . Thus for all and all ,

This proves uniform convergence.

Proof.

Away from , the function is a quotient of smooth functions with nonzero denominator, so it is continuous and has partial derivatives there. The only point requiring special attention is .

(a) We prove continuity at . For ,

Using

we get

Therefore

As , the right-hand side tends to . Hence

so is continuous at . Therefore is continuous on .

(b) Away from , both partial derivatives exist because the function is smooth there. At , we compute directly:

and

Thus both partial derivatives exist at every point of .

(c) The function is not differentiable at . If were differentiable at , then since both partial derivatives at are , the derivative would be the zero linear map. Thus we would need

as .

But along the curve with , we have

Also,

Therefore

This is not . Hence is not differentiable at , and therefore is not differentiable as a function on all of .

Proof.

Let

Since , it is enough to prove that is uniformly continuous on .

Observe that

The Euclidean norm is -Lipschitz, so for any ,

For nonnegative numbers , we have

Thus

Therefore

Let . Choose such that

If

then and , so

Thus is uniformly continuous.

Finally,

Hence is uniformly continuous on .

Proof.

Since

exists and is finite, we must have

Moreover,

Because is twice differentiable at , Taylor's formula with Peano remainder gives

as . Hence