2009 Spring Comprehensive in Analysis

Problem 1.

Let be a continuous function which satisfies

for all . Prove the following statements:

(a) If is positive at one point of , then is positive at every point of .

(b) If is differentiable at one point of , then is differentiable at every point of .

Proof.

(a) First note that

so or .

Suppose that is positive at one point, say . Then is not identically zero, so , and hence

For any ,

It remains to rule out the possibility that for some . If , then for every ,

so would be identically zero. This contradicts . Therefore

for every .

(b) If is identically zero, then it is differentiable everywhere. Thus assume is not identically zero. Then, as above, , and for every .

Assume that is differentiable at some point . Since , the limit

exists. But

Since , it follows that

exists.

Now let be arbitrary. Then

Letting gives

Thus is differentiable at every point of .

Problem 2.

Let , , and be Riemann integrable real-valued functions defined on . For each of the following statements, determine whether the statement is true or not; prove your claims:

(a) If

then

(b) If

then

Proof.

(a) The statement is false.

Let on , and define

Each is Riemann integrable. Then

However,

for every . Therefore the second limit is not .

(b) The statement is true.

Let

By the Cauchy-Schwarz inequality,

Since

we have

The right-hand side tends to by assumption. Hence

Problem 3.

Suppose that is the vector-valued function defined by

for all .

(a) Compute the determinant of the Jacobian matrix of at the point .

(b) Is there an open neighborhood of such that is one-to-one in this neighborhood? If your answer is yes, give the reason and find an explicit example of an open neighborhood of on which is one-to-one.

Proof.

(a) The Jacobian matrix is

At this becomes

Therefore

(b) Yes. Since

the inverse function theorem implies that is one-to-one on some open neighborhood of .

We now give an explicit neighborhood. Let

Suppose and

Comparing the three components gives

and, since the exponential function is one-to-one,

Thus

Subtracting the second equation from the first equation gives

Using , we obtain

Since

we get

But and , so

Hence , and therefore

Then , and the first equation gives . Thus

Therefore is one-to-one on the explicit open neighborhood of .

Problem 4.

Let be a real-valued function defined on a closed bounded interval . We define the graph of to be the set of points such that and .

Prove or disprove: The function is continuous on if and only if the set is a closed subset of .

Proof.

The statement is false as written. One implication is true, but the converse is false.

First suppose is continuous on . Let be a sequence in converging to . Since and is closed, we have . By continuity,

But also . Hence , so . Thus is closed.

The converse is false. Take and define

This function is not continuous at .

We show that its graph is closed. Suppose and

If , then eventually , and continuity of on gives

If , then it is impossible to have a subsequence with and , because then , contradicting the convergence of to the finite number . Therefore, for all sufficiently large , , and hence .

Thus every convergent sequence in the graph has its limit in the graph, so is closed. Hence a closed graph does not imply continuity in this setting.

Problem 5.

Prove that the set of all positive integers has the same cardinality as the Cartesian product .

In other words, construct an explicit example of a one-to-one map of onto .

Proof.

Every positive integer can be written uniquely in the form

where . Indeed, is the largest power of dividing , and is the remaining odd factor.

Define

This is well-defined because the factorization above is unique.

The map is one-to-one: if

then

The map is onto: given any , the number

is a positive integer and satisfies

Therefore and have the same cardinality.

Problem 6.

Suppose that is continuously differentiable on the closed interval , and suppose that is a monotonic function such that

Prove or disprove: There exists a constant such that

for all .

Proof.

The statement is true, assuming .

Since , the derivative is continuous on , so

For any , the mean value theorem gives

Because is monotonic and , , the value lies between and . Hence

Again by the mean value theorem,

Therefore

Now

Thus the desired estimate holds with

Problem 7.

Define a sequence recursively by setting

Prove that the sequence converges, and compute its limit.

Proof.

We solve the recurrence. The characteristic equation is

or equivalently

Thus

so the two roots are

Therefore the sequence has the form

Using gives

Subtracting the first equation from the second gives

Therefore

Hence

Since

the sequence converges and

Problem 8.

Let be a real-valued function defined on the real line.

(a) Prove or disprove: If is uniformly continuous on , then is uniformly continuous on .

(b) Prove or disprove: If is uniformly continuous on , then

is uniformly continuous on .

Proof.

(a) The statement is false.

Take

Then is uniformly continuous on . However,

is not uniformly continuous on . Indeed, let

Then

but

This contradicts uniform continuity. Hence need not be uniformly continuous.

(b) The statement is true.

Define

Then

The derivative is bounded on ; for example, there exists such that

for all . By the mean value theorem,

for all . Thus is Lipschitz, hence uniformly continuous in a strong form.

Now suppose is uniformly continuous on . Given , choose such that

if . If , the conclusion is immediate. Then

implies

Therefore is uniformly continuous on .

Problem 9.

Suppose is a Riemann integrable function on . Prove that

Proof.

Since is Riemann integrable on , it is bounded. Thus there exists such that

for all .

Therefore

Since

we obtain

Hence

Remark.

For improper integrable functions, the statement is also true if is absolutely integrable, meaning

The proof follows from the fact that for any , we have