2022 Fall Comprehensive in Analysis

Proof.

(a) The Axiom of Completeness says that every nonempty subset that is bounded above has a least upper bound in . That is, there exists such that

and if is any upper bound for , then

(b) Let be monotone increasing.

First suppose that . Then every convergent sequence is bounded. In particular, there exists such that

for all . Hence for ,

The finitely many terms are also bounded above. Thus the whole sequence is bounded above.

Conversely, suppose is bounded above. Let

Then is nonempty and bounded above, so by completeness it has a supremum. Let

We claim that .

Let . Since is the least upper bound, is not an upper bound for . Therefore there exists such that

Since the sequence is increasing, for all ,

Also, since is an upper bound for ,

for all . Hence for all ,

so

Thus .

Proof.

Assume, for contradiction, that

converges uniformly on .

Then the partial sums are uniformly Cauchy on . Thus for every , there exists such that for all ,

Therefore, for all ,

For fixed and , the function

is continuous on , because it is a finite sum of continuous functions. Taking the limit as gives

Thus the numerical series

is Cauchy, hence converges. This contradicts the hypothesis that it diverges.

Therefore

does not converge uniformly on .

Proof.

Let

By hypothesis,

Since is bounded, there exists such that

for all .

From

we have

Iterating this identity gives, for any integer ,

Hence

Let . Choose so large that

Since , there exists such that

for all . Then for all , every , so

Since

we get

Therefore .

Proof.

First we prove local uniform boundedness and local equicontinuity.

Fix and suppose . Then

Thus the family is equicontinuous on .

Also, since , taking gives

Therefore, for ,

So is uniformly bounded on .

By the Arzela-Ascoli theorem, from we can extract a subsequence that converges uniformly on . From that subsequence, extract a further subsequence that converges uniformly on . Continuing inductively, for each we obtain a subsequence that converges uniformly on .

Now take the diagonal subsequence. More explicitly, if the -th subsequence is denoted by

with each subsequence taken from the previous one, define

Then for each fixed , all but finitely many terms of lie in the -th subsequence, so converges uniformly on .

Since every real number belongs to some interval , it follows that converges pointwise on . Thus has a pointwise convergent subsequence on .

Proof.

Since and are strictly decreasing and converge to zero, define

Then

Thus

Because and , we can write the tails as

and

Therefore

Let . Since , there exists such that for all ,

Then for ,

Since all , we get

Thus

Proof.

Yes. We compute the Jacobian determinant.

Write

where

By the fundamental theorem of calculus,

and

Therefore

Its determinant is

If , then . Hence and

Thus

So

In particular, the Jacobian determinant is nonzero at every point of .

By the inverse function theorem, is a local diffeomorphism near every point .

Proof.

Assume that is oriented by the outward normal. The given integral is the flux integral of the vector field

Indeed,

Let be the solid ellipsoid

By the divergence theorem,

Now

Hence

Use the change of variables

Then maps to the unit ball

and the Jacobian is

Therefore

By symmetry,

Also,

Thus

Similarly,

Therefore

Hence

If the opposite orientation is used, the answer has the opposite sign.

Proof.

The statement is false. We disprove uniform continuity by constructing two sequences of points whose distances go to zero while the corresponding function values stay separated.

Let

where

Then

Now compute the difference of the function values. Since

and

we have

But