2021 Fall Comprehensive in Analysis

Proof.

First observe that is injective. Indeed, if , then

so .

A continuous injective function on an interval is strictly monotone. Hence is either strictly increasing or strictly decreasing on .

Suppose first that is strictly increasing. For , using the given inequality with , we have

Since is increasing, , so

Therefore

as .

Similarly, for , since , we have

and hence

as .

Thus is continuous and its values tend to in one direction and in the other direction. By the intermediate value theorem, every real number is attained by .

If is strictly decreasing, the same argument gives

and

Again, the intermediate value theorem implies that is surjective.

Therefore is surjective.

Proof.

As written, the recurrence defines a real sequence only when the expression under the square root is nonnegative at every step. We use this fact below.

The fixed points of the recurrence are the solutions of

Squaring gives

so

Thus the only possible limits are

We first note that, if the sequence is defined as a real sequence for all , then necessarily

Indeed, suppose . Since the sequence is defined for all , we must have for all . For ,

because this inequality is equivalent to

or

which is true for . Hence would be decreasing and bounded below by , so it would converge to some . Passing to the limit in the recurrence would give

so or , a contradiction. Therefore .

If , then

and so for all . Hence

Now suppose .

If , then for ,

because

is equivalent to

Also, if , then

Thus if , then

So is increasing and bounded above by , hence convergent.

If , then for all .

If , then for ,

and

Thus is decreasing and bounded below by , hence convergent.

In all cases with , the limit satisfies

and since , the limit cannot be . Therefore

Thus, for a real sequence defined for all , the answer is

Proof.

Since is continuous on the compact interval , it attains its maximum and minimum. Let

It is enough to prove

because then, for any ,

Choose such that

Assume without loss of generality that . By the mean value theorem and the bound ,

On the other hand, since ,

Therefore, again using the mean value theorem,

Hence

and

Adding these two inequalities gives

Thus

Therefore

for all .

Proof.

(a) The statement is false.

For example, define

where is fixed. This function is bounded and is discontinuous at . However, it is Riemann integrable, and

Indeed, for every partition, the lower sum is , and by choosing a partition with a very small interval containing , the upper sums can be made arbitrarily small. Hence is Riemann integrable but not continuous.

Therefore (a) is false.

(b) The statement is also false.

Let be the standard Cantor set, and define

The Cantor set is uncountable, closed, has empty interior, and has Lebesgue measure zero.

The function is discontinuous at every point of . Indeed, if , then arbitrarily close to there are points outside , so the function values near include both and .

If , then since is closed, there is a small open interval around disjoint from . On that interval, , so the function is continuous at .

Thus the set of discontinuities of is exactly , which is uncountable. Nevertheless, since has measure zero, is Riemann integrable and

For a general interval , one may use an affine copy of the Cantor set inside .

Therefore (b) is false.

Proof.

The hypothesis implies that the sets

are bounded for every .

Thus, for every , there exists such that

In other words, as .

We now show that, in fact, at infinity or at infinity.

Take . There exists such that

The exterior region

is connected. Its image under the continuous function is therefore connected. But it lies in

which has two connected components. Hence the image must lie entirely in one of these components.

Therefore either

or

Suppose first that

Let . Choose so large that

Since also implies , we have there. Therefore

Thus no global minimum can occur outside the closed disk

Since is continuous and is compact, attains a minimum on . This minimum is also the global minimum on .

The other case is similar. If

then choosing and large enough gives

So no global maximum can occur outside . By compactness, attains a maximum on , and hence on all of .

Therefore attains either a maximum or a minimum value.

Proof.

Since is continuous on the compact set , it is uniformly continuous.

Let . By uniform continuity, there exists such that whenever

and , we have

Fix with .

For every ,

Taking the maximum over gives

Similarly, for every ,

Taking the maximum over gives

Therefore

Thus is continuous on .

Proof.

Yes. Under this hypothesis, is pathwise connected.

Fix a point , and let be the path component of . Thus

We show that is both open and closed in .

First, let . Since is locally pathwise connected in the stated sense, there exists an open pathwise connected neighborhood of . For any , there is a path inside from to . Since , there is also a path from to . Concatenating these paths gives a path from to . Hence . Therefore

so is open.

Next, let . Again, choose an open pathwise connected neighborhood of . If met , say , then there would be a path from to , and also a path inside from to . Concatenating them would give a path from to , contradicting . Thus

so is open. Therefore is closed.

Thus is a nonempty subset of that is both open and closed. Since is connected, we must have

Therefore every point of can be joined to by a path, and is pathwise connected.

Proof.

By symmetry of the unit sphere,

and

Therefore

By symmetry again,

Since on we have

it follows that

The area of the unit sphere is

Hence

Therefore

Proof.

Define

Then

We claim that is Lipschitz on . For ,