2004 Spring Comprehensive in Analysis

Problem 1.

Let be a function of , defined by the system of equations

Compute at the point where . Justify the existence of this derivative.

Proof.

Define

and

Then the system is

At , the Jacobian matrix with respect to is

Its determinant is

Therefore, by the implicit function theorem, and are differentiable functions of near this point.

Differentiate the two equations with respect to . From the first equation,

At , this becomes

Thus

Therefore

Problem 2.

Suppose

and assume that this series converges whenever .

(a) Prove that the series

converges for .

(b) Prove that there exists a constant such that

for any .

(c) Prove that there exists a constant such that

for any .

Proof.

(a) Since the series converges at , its terms tend to . Hence

In particular, there exists such that

for all . Therefore

Since

converges, the series

converges absolutely. Hence it converges.

(b) Since the series converges at , we have

Thus the sequence

is bounded. Hence there exists such that

for all . Therefore

Increasing if necessary, we may write

(c) From part (b), there exists such that

Since lies inside the radius of convergence, termwise differentiation is valid at . Thus

Therefore

Write . Then

Since

we get

Using

with , we obtain

Hence

Taking

we get

Problem 3.

(a) Define a contractive mapping. State the Contractive Mapping Principle.

(b) For , define by setting

for . Prove that is a contractive mapping on .

(c) Use the Contractive Mapping Principle to prove that the system of equations

has a unique solution on .

Proof.

(a) A mapping on a metric space is called contractive if there exists a constant such that

for all .

The Contractive Mapping Principle says that if is a complete metric space and is contractive, then has a unique fixed point. That is, there exists a unique such that

(b) We use the sup norm

on . For ,

Thus

Taking the supremum over gives

Since

is a contractive mapping.

(c) Since is complete under the sup norm and is contractive, the Contractive Mapping Principle implies that has a unique fixed point .

The fixed point condition means

Putting , we get

Differentiating the equation gives

Therefore

Hence the fixed point is a solution of the system.

Conversely, any solution of the system satisfies

Integrating from to gives

Since , this becomes

Thus any solution is a fixed point of . Since the fixed point is unique, the solution is unique.

Problem 4.

State and prove the Intermediate Value Theorem in one variable.

Proof.

Intermediate Value Theorem. Let be continuous. If is between and , then there exists such that

Assume, without loss of generality, that

Define

Then is nonempty because , and is bounded above by . Let

We claim that

First suppose . By continuity of at , there exists such that if , then

In particular, for some with , we have . Thus , contradicting the fact that is an upper bound of .

Now suppose . By continuity of at , there exists such that if , then

Then no point of belongs to , contradicting the fact that .

Therefore neither nor is possible. Hence

This proves the theorem.

Problem 5.

Suppose is a subset of a complete metric space , and is a uniformly continuous -valued function defined on . Prove that there exists a uniformly continuous function

such that

Proof.

Let . Choose a sequence in such that

Since is Cauchy and is uniformly continuous, the sequence is Cauchy in . Because is complete, converges.

Define

We must show that this definition is independent of the sequence chosen. Suppose and are two sequences in such that

Then

By uniform continuity of ,

Hence the two limits are equal. Therefore is well-defined.

If , we may take the constant sequence , and then

Thus

It remains to show that is uniformly continuous. Let . Since is uniformly continuous, there exists such that if and

then

Let

If and

choose sequences and in with

For sufficiently large ,

Then

Therefore

Letting , we get

Thus is uniformly continuous on .

Problem 6.

Suppose is Riemann integrable on , and assume that there exists a positive constant such that

for all . Prove that is Riemann integrable.

Proof.

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

for all .

The function

is continuous on the compact interval . Therefore is uniformly continuous on .

A standard theorem says that if is Riemann integrable and is continuous on an interval containing the range of , then is Riemann integrable. Applying this theorem to , we get that

is Riemann integrable on .

Remark.

This can be generalized to the improper integral case.

Problem 7.

For , define

Suppose maps onto in such a way that

for any . Suppose is an open subset of . Prove that is open.

Proof.

First, is injective. Indeed, if

then

Since is also onto, is bijective. Let

For , write

Then

Therefore is Lipschitz continuous, hence continuous.

Now let be open. Since

and is continuous, is open. Hence

is open.

Problem 8.

Suppose is an open subset of containing a point . Let and be real-valued functions defined on such that is continuous, is differentiable, and

Prove that the product is differentiable at .

Proof.

Since , we have

For ,

Since , this becomes

As ,

because is differentiable at , and

because is continuous at . Therefore

Thus is differentiable at , and

Problem 9.

(a) State the Heine-Borel Theorem for subsets of .

(b) Construct a sequence of real numbers as follows: set

and let

for . Does the sequence converge? If it does, compute

Proof.

(a) The Heine-Borel Theorem says that a subset is compact if and only if is closed and bounded.

(b) We first show that for all . Define

Then

and

for . Hence

for every . Since

and , it follows by induction that

for all .

Next, since , we have

Therefore

Thus is decreasing. Since , it is bounded below by . Therefore converges. Let

Then . Passing to the limit in

we get

Thus

Therefore