2024 Fall Comprehensive in Analysis

Proof.

For each and each , we have

so

Therefore

Summing from to , we obtain

Hence

Proof.

Define

Then is continuous on . Since is even,

Therefore

for every polynomial .

Now define

Then is continuous on . By the Weierstrass approximation theorem, there are polynomials such that

uniformly on . Hence

uniformly on .

Since

for all , taking the limit gives

Thus for all . In particular,

Therefore

Proof.

The product is compact. Define

by

The map is continuous because addition is continuous in the Hilbert space .

By definition,

Since the continuous image of a compact set is compact, is compact.

Proof.

Since , the derivative is strictly decreasing.

Suppose for contradiction that there exists such that

Since is decreasing, for every we have

Thus, for ,

The right-hand side tends to as , contradicting the assumption that for all .

Therefore

for all . If for some , then since is strictly decreasing, we would have for all , again a contradiction. Hence

for all . Therefore is increasing.

Since is increasing and positive, the limit

exists and is finite. Define on by

Then is continuous on the compact interval , so it is uniformly continuous on .

On , since is decreasing, we have

for all . Hence is Lipschitz on :

for all .

Combining the uniform continuity on with the Lipschitz estimate on , we conclude that is uniformly continuous on .

Proof.

(a) False

Take

with the usual metric and

Then for ,

But has no fixed point in , since implies , and .

(b) False

Let

with the metric

This metric space is complete because is an isometry from onto with the usual metric.

Define

If , then

Thus strictly decreases distances between distinct points. But has no fixed point, since

has no solution.

(c) False

The same example as in part (a) works. The space is bounded, and strictly decreases distances, but it has no fixed point in .

(d) True

Let be compact and suppose

whenever .

First, is continuous. Indeed, if , then

Define

Since is continuous, is continuous. Since is compact, attains a minimum at some .

If , then , so

contradicting the minimality of . Hence

Therefore

so a fixed point exists.

The fixed point is unique. If and are two distinct fixed points, then

which is impossible. Hence has exactly one fixed point.

Proof.

Fix and . For small , define

By assumption,

for all sufficiently small , and

Thus is a minimum of , so

Using Taylor expansion at , we have

and

Adding these and subtracting gives

Since , dividing by and letting gives

Therefore

Proof.

The vector field

is singular at , and the square contains .

Let be the circle of radius centered at the origin, oriented counterclockwise. On , write

where . Then

Hence

Therefore

On the region between and , the vector field is smooth and has curl . Therefore the integral over the outer boundary equals the integral over the small circle:

Proof.

We write

Let

Then

Using the standard Gaussian integral formula

for positive definite , we get

Proof.

Let be an uncountable set. Define

Think of as a hedgehog with one spine for each , all joined at the point .

Define a metric on by

and

The space is bounded because

for all .

It is path connected: any point can be joined to by the path

with . Therefore any two points can be joined by going from the first point to and then from to the second point.

The space is complete. Indeed, let be a Cauchy sequence. If the sequence visits infinitely many different spines while staying away from , then distances between some terms would be bounded below by a positive number, contradicting the Cauchy property. Hence either , or eventually all lie in one spine. In the latter case, completeness follows from the completeness of .

Finally, is not separable. For each , the point belongs to . If , then

Thus contains an uncountable set of points separated from each other by distance . Therefore no countable subset can be dense in .

Hence is path connected, bounded, complete, and not separable.