2010 Spring Comprehensive in Analysis

Proof.

The statement is false.

The open unit ball

is connected. Indeed, it is convex, since if and , then

Every convex subset of is connected.

If is continuous, then must be connected because the continuous image of a connected set is connected. However, is not connected as a subset of . For example,

is a separation of into two nonempty relatively open subsets.

Therefore no such continuous function exists.

Proof.

(a) Let be bounded. Choose such that

Since , the second derivatives of are bounded on the compact ball . By Taylor's theorem at , using and , there is a constant such that

for all .

For and , we have , so

Since

converges, the Weierstrass -test implies that

converges uniformly on . Hence the series converges uniformly on every bounded subset of .

(b) In general, the series need not converge uniformly on all of .

Take

Then , , and . The series becomes

For a tail of the series,

For every fixed , the factor

is positive. Therefore

Thus the tails do not converge uniformly to , so the series is not uniformly convergent on in general.

Proof.

Yes, is differentiable.

Let . Then

Expanding, we get

Therefore

The terms that are linear in are

Define

This map is linear in .

It remains to check that the remaining terms are . Using any matrix norm compatible with multiplication, we have

Thus

Dividing by gives

as .

Hence is differentiable at every , and its derivative is

Proof.

(a) The Contraction Mapping Theorem states:

Let be a nonempty complete metric space. Suppose is a contraction, meaning that there exists a constant such that

for all . Then has a unique fixed point , meaning

Moreover, for any starting point , the sequence

converges to .

(b) Choose any and define

for . Since is a contraction,

By induction,

If , then by the triangle inequality,

Since , the right-hand side tends to as . Therefore is a Cauchy sequence. Because is complete, there exists such that

We now prove that is a fixed point. Since is Lipschitz, it is continuous. Hence

Thus is a fixed point.

Finally, suppose is another fixed point. Then

Since , this implies

Therefore , so the fixed point is unique.

Proof.

This is Cauchy's condensation test.

Since is decreasing and positive, for each and each integer satisfying

we have

Therefore

It follows that

Thus if

converges, then

also converges.

Conversely, for and

we have

There are terms in this block, so

Equivalently,

Hence, if converges, then

Therefore the two series converge or diverge together.

Proof.

(a) Define

At every point , this function is differentiable, so both partial derivatives exist.

At , we compute the partial derivatives directly:

and

Thus both first partial derivatives exist everywhere.

However, is not continuous at . Along the line ,

for , while . Hence

So is not continuous on .

(b) Let . Since is open, there is such that the closed rectangle

is contained in .

Assume

on . Let be close enough to so that the horizontal and vertical line segments used below remain inside the rectangle. Then

By the one-variable Mean Value Theorem applied in the -direction,

By the one-variable Mean Value Theorem applied in the -direction,

Therefore

The right-hand side tends to as . Hence is continuous at . Since was arbitrary, is continuous at every point of .

Proof.

(a) One form of the Implicit Function Theorem is as follows.

Let be continuously differentiable near , where and . Suppose

and the matrix

of partial derivatives with respect to the second group of variables is invertible. Then there are neighborhoods of and of and a unique continuously differentiable function

such that

and

for all . Moreover,

(b) Define by

At the point ,

Now compute the Jacobian with respect to the unknowns :

At this becomes

Its determinant is

Therefore the matrix is invertible. By the Implicit Function Theorem, there exist neighborhoods of and and unique functions

with

such that the given system is satisfied.

(c) We differentiate the two identities

and

First differentiate the first equation with respect to :

At , with and , this gives