2023 Spring Comprehensive in Analysis

Proof.

Let be the collection of all open balls in whose centers have rational coordinates and whose radii are positive rational numbers. Since is countable and is countable, the collection is countable.

For each ball such that , choose one point . Define

Then is countable, because it is indexed by a subcollection of the countable set . Also , and since is closed, we have

It remains to show that .

Let and let . Choose such that

Then choose a rational number such that

The ball belongs to , and it intersects because . Hence the construction gives a point . Therefore

Thus every neighborhood of contains a point of , so . Hence

Combining the two inclusions, we obtain

Therefore there exists a countable set whose closure is .

Proof.

We first show that for all . This is true for , since . If , then

so . Thus by induction,

for all .

Next,

Hence is increasing. Since it is increasing and bounded above by , it converges. Let

Passing to the limit in the recurrence gives

Therefore

so

Thus

Therefore converges, and its limit is .

Proof.

For , we have

Since

converges, the Weierstrass -test implies that

converges uniformly on . Each function

is continuous on . Therefore the uniform limit

is continuous on .

Now compute the derivative of each term:

For ,

Since

converges, the Weierstrass -test implies that the derivative series

converges uniformly on .

Also, the original series converges at , since

By the theorem on term-by-term differentiation of a series of continuously differentiable functions, the sum is differentiable on in the sense of one-sided derivatives at the endpoints and ordinary derivatives in the interior, and for ,

Thus

for . The uniform convergence of the derivative series is exactly what justifies differentiating the series term by term.

Proof.

Let

Since and are continuous on the compact interval , both functions attain their maximum values. Thus there exist such that

and

Assume, for contradiction, that there is no such that . Then the continuous function

has no zero on . Since is connected and is continuous, must have one sign everywhere. Hence either

or

If for all , then at the point where , we get

which contradicts .

If for all , then at the point where , we get

which contradicts .

Both cases are impossible. Therefore there must exist such that

Proof.

Since converges, the tails

are well-defined and satisfy

Also,

Therefore

Using summation by parts, we get

Hence

We estimate the three terms. Since ,

Also,

It remains to handle the middle term.

Let . Since , choose such that

for all . Then for ,

The first term on the right tends to because it has a fixed numerator and . For the second term,

so it is at most . Therefore the middle term tends to .

Thus all three terms tend to , and we conclude that

Proof.

For each , write

Then

For , the mean value theorem gives

Therefore

The integral on the right is

Summing over , we obtain

This proves the desired estimate.

Proof.

Define, for ,

By assumption,

as .

For , use a dyadic telescoping sum:

This identity holds because the finite partial sums telescope to

and by continuity at .

Now set

Then

Thus

Subtracting gives

Let . Since as , choose such that

whenever . If , then for every ,

Therefore

Hence

Therefore exists and

Proof.

Define

Then

Also,

so

By the implicit function theorem, there exists and a unique smooth function defined for such that

and

Equivalently,

Now differentiate

with respect to . We get

Setting and using , we obtain