2006 Fall Comprehensive in Analysis

Problem 1.

Analyze the convergence behavior of each of the following infinite series. This means: determine whether the series converges or diverges; if it converges, determine whether it converges absolutely or conditionally; if it diverges, determine whether or not the sequence of terms converges to .

(a)

(b)

Justify your answers.

Proof.

(a) The series can be written as

Its absolute series is bounded by

Both

are convergent geometric series. Hence the given series is absolutely convergent.

In fact,

but the main conclusion is that the series converges absolutely.

(b) The series is positive, so ordinary convergence and absolute convergence are the same. Consider

Use the integral test with

The function is positive and decreasing for . Moreover,

Therefore the series converges. Since all terms are positive, it converges absolutely.

Problem 2.

Throughout this problem let be a function with continuous first partial derivatives at each point of .

(a) Prove that is continuous at each point of .

(b) Suppose that has exactly one critical point, namely at a certain point . Suppose further that

Prove or disprove: the function takes on its global minimum at the point .

Proof.

(a) Fix . Since the first partial derivatives of are continuous at each point, they are bounded on some small closed box

around . Thus there is a constant such that

for all and all .

Let be small enough so that . Move from to one coordinate at a time. By the one-variable mean value theorem applied in each coordinate direction,

As , the right-hand side tends to . Hence

so is continuous at . Since was arbitrary, is continuous on .

(b) The statement is true.

Because

we may choose so large that and

whenever .

The closed ball is compact, and by part (a), is continuous. Therefore attains a minimum on at some point .

The point cannot lie on the boundary , because on that boundary we have

while lies inside the ball. Hence is an interior point of . Since has a local minimum at and is differentiable, we must have

Thus is a critical point. By assumption, the only critical point is . Therefore

So attains its global minimum at .

Problem 3.

Prove or disprove: if is a continuous real-valued function such that

then for all .

Proof.

The statement is true.

Suppose, toward a contradiction, that there exists such that

Then

Since is continuous, is continuous. Therefore there exists an interval containing and having positive length such that

for all .

Hence

which contradicts the assumption that

Therefore no such exists, and so

for all .

Problem 4.

Let be a real-valued function with domain , and let and be elements of . Prove that if

then there is a sequence in such that

and

Proof.

We use the standard definition

For , define

Then decreases as , and

For each , choose such that

and

By the definition of supremum, there exists satisfying

and

Since , we get

Also, because lies in the punctured -neighborhood of ,

Thus

Therefore

Also,

This proves the claim.

Problem 5.

Suppose that is an infinite sequence of positive real numbers, and for each positive integer let denote the finite product

The infinite product

is said to be convergent if converges to a nonzero finite value .

(a) Show that the infinite product

is convergent.

(b) Estimate the value of the infinite product in part (a) correct to one decimal place. Justify your estimate.

Proof.

(a) Let

Since every factor is greater than , the sequence is increasing. We prove it is bounded above.

Taking logarithms,

For ,

Therefore

Thus is increasing and bounded above, so it converges to some finite number . Hence

Since , the limit is positive and finite. Therefore

converges.

(b) A convenient exact evaluation comes from Euler's product formula

Thus

Therefore the value correct to one decimal place is

One can also justify the decimal estimate directly from partial products. For example, if

then

With , direct computation gives

and hence

This also gives the one-decimal estimate .

Remark.

Here is the proof of the Euler's product formula for :

Let

The product converges locally uniformly, since

converges uniformly for in any bounded interval. Hence is well-defined, and we may differentiate its logarithm term by term:

We now compute this sum by a real Fourier series argument. Consider

Since is even, its cosine series has the form

The constant coefficient is

For , we have

Evaluating the Fourier series at , and using , gives

Dividing by , we obtain

Therefore

Thus

On the other hand,

Hence

It follows that

is constant. Since

the constant is . Therefore

Problem 6.

Prove, from the definition of Riemann integrability, that if and are Riemann integrable on , then is also Riemann integrable on .

Proof.

Since and are Riemann integrable on , they are bounded. Choose constants such that

for all .

For a subinterval , let

denote the oscillation of on . If , then

Therefore

Taking the supremum over , we get

Let . Since is Riemann integrable, there exists a partition such that

Since is Riemann integrable, there exists a partition such that

Let be a common refinement of and . Refining a partition does not increase the corresponding upper-minus-lower sum, so

and

Using the oscillation estimate,

Thus, for every , there is a partition such that

By the Riemann criterion for integrability, is Riemann integrable on .

Problem 7.

Consider the sequence

defined recursively by

and

Determine the convergence properties of this sequence. That is, determine whether it is convergent; if it is convergent, determine its limit; if it is not convergent, determine how it diverges. Justify your answer.

Recall that

Proof.

We first note that

For , we have

Indeed, for , and the inequality follows from the fact that

satisfies

and

for .

Therefore, if , then

By induction, the sequence is positive and strictly decreasing. Hence it is bounded below by , so it converges. Let

Then . Passing to the limit in

and using continuity of , we obtain

For , we have , so the equality is impossible. Thus

Therefore the sequence converges monotonically decreasing to .

Problem 8.

Consider the sequence of functions , , given by

(a) Show that this sequence converges pointwise on to some limit function , and determine the formula for .

(b) Determine whether the sequence converges uniformly to the function obtained in part (a). Justify your answer.

Proof.

(a) Fix . Then

As ,

Therefore

Thus the pointwise limit function is

(b) We check whether

tends to . Since

we have

The supremum is not attained at a finite , but it is approached as .

Since the supremum is for every , it does not tend to . Therefore does not converge uniformly to on .

Problem 9.

Let be a continuous function, with values in the positive real numbers, defined on the standard closed unit ball in .

Determine the convergence properties of the infinite series

That is, determine whether the series converges pointwise on ; and if it does, determine whether the convergence is uniform, and compute the formula for the function to which the series converges.

Justify your answer.

Proof.

Because is compact and is continuous, attains its minimum on . Since takes values in , this minimum is positive. Thus there exists such that

for all .

For each fixed , the series is geometric:

Since , we have

Therefore the pointwise sum is

Now we prove uniform convergence. Since ,

for all and all . The numerical series

is a convergent geometric series. Hence, by the Weierstrass -test,

converges uniformly on .

Thus the series converges pointwise and uniformly on , and its sum is