2013 Spring Comprehensive in Analysis

Proof.

Let

The fixed points of satisfy

so

and hence

Since , all terms are greater than . Indeed, if , then

Also,

Thus, for , we have .

Since , then is increasing and bounded below by . In every case, converges.

Let . Passing to the limit in

gives

Since every term is greater than , the limit must be the fixed point , not . Therefore

Proof.

For each fixed , we have

Since

converges, the given series converges absolutely, hence pointwise, for every fixed .

It remains to show that the limit function is continuous. Let . On the compact interval ,

Because

the Weierstrass -test implies that the series converges uniformly on . Each function

is continuous, so the uniform limit on is continuous on .

Since every point of lies in some interval , the pointwise limit is continuous on all of .

Proof.

Since is decreasing, for every integer and every , we have

Therefore

Summing from to gives

First assume that converges. Then the partial sums on the right are bounded, so

is bounded.

Conversely, assume that is bounded. Then there exists such that

for all . From the inequality above,

Thus the partial sums of are bounded. Since all terms are positive, the partial sums are increasing, so the series converges. Adding the single term does not affect convergence. Hence

converges.

Proof.

The statement is true.

Let . Since

there exists such that

whenever .

Because is continuous on the compact interval , it is uniformly continuous there. Hence there exists such that if and

then

Let

Suppose .

If at least one of and is at most , then both and lie in , so

If both and , then

Therefore is uniformly continuous on .

Proof.

We prove the result for points in ; the two endpoints can add at most two more points.

For , since is increasing, the one-sided limits

exist. Also

Thus is discontinuous at only if

For each interior discontinuity point , choose a rational number such that

This is possible because is dense in .

We claim that different discontinuity points give different rationals. Indeed, if , then by monotonicity,

Therefore the intervals

are disjoint. Hence .

Thus the map

from the set of interior discontinuities into is injective. Since is countable, the set of interior discontinuities is countable. Adding the endpoints, is at most countable.

Proof.

Use spherical coordinates on the unit sphere:

where and . The area element is

Therefore

Thus

The first integral is

For the second integral, put . Then and

Hence

Therefore

Thus

Proof.

(a) The statement is false.

For example, on , define

Then

for all , so is Riemann integrable. However, every interval contains both rational and irrational numbers, so on every subinterval the supremum of is and the infimum of is . Hence the upper integral of is and the lower integral of is . Therefore is not Riemann integrable.

(b) The statement is true.

Let

By assumption, is Riemann integrable. Since is bounded, is bounded. The function

is continuous on every compact interval. By the standard theorem that a continuous function composed with a Riemann integrable function is Riemann integrable, is Riemann integrable. But

Therefore is Riemann integrable on .

Proof.

Apply the Mean Value Theorem to on the interval . There exists such that

Since , this gives

Now apply the Mean Value Theorem to on . There exists such that

Therefore

Since is twice differentiable, is continuous on and differentiable on . By Rolle's Theorem applied to on , there exists such that

Proof.

First, if , then is continuous, so . Thus maps into itself.

For and , the Mean Value Theorem applied to gives

Taking the maximum over , we get

Thus is a contraction with contraction constant .

The metric space with the sup metric is complete. Therefore, by the Banach Fixed Point Theorem, has a unique fixed point in .