2016 Spring Comprehensive in Analysis

Proof.

We separate the sequence into its odd and even subsequences.

For , applying the hypothesis first with and then with gives

and

Therefore

Hence the odd subsequence is decreasing. Since every is nonnegative, this odd subsequence is bounded below by . Thus it converges; say

for some .

Now we show that the even subsequence has the same limit. From

we get

On the other hand, from

we get

Since , it follows that

Therefore

Both the odd and even subsequences converge to the same limit , so the full sequence converges.

Proof.

We prove that has the intermediate value property. Let and let be a real number strictly between and . We must show that there exists such that

Define

Then is continuous on , differentiable on , and

If , then

Since , for sufficiently small we have

so cannot be a point where attains its minimum on . Since , for sufficiently small we have

so cannot be a point where attains its minimum. By compactness, attains a minimum on , and the preceding argument shows that this minimum is attained at some interior point . By Fermat's theorem,

Thus

so

If , the same argument is applied to a maximum of on : now , so neither endpoint can be the maximum, and Fermat's theorem again gives an interior point with .

Therefore every value between and is also a value of . Hence the range of is an interval, possibly a point or an unbounded interval, and is therefore connected.

Proof.

Fix any . Let

The maximum exists because is continuous on the compact interval .

We claim that . Suppose instead that . Choose such that

Since and , we must have . By the mean value theorem applied to on , there exists such that

Therefore

Using the hypothesis,

Since , we have and . Hence

But and , so , a contradiction.

Thus . Hence for all . Since was arbitrary, for all . Finally, continuity at gives

Therefore on .

Proof.

For each fixed ,

Since

as , with , we have

Thus the pointwise limit is

We justify passing the limit through the integral. Since for all real ,

The function is integrable on . Therefore, by the dominated convergence theorem,

Finally,

Therefore the value of the limit is .

Proof.

We use the Arzela-Ascoli theorem.

First, the functions are uniformly bounded. Since

and

for , we have

Thus

for all and all .

Next, the functions are equicontinuous. If and, without loss of generality, , then

This bound is independent of , so the family is equicontinuous.

The interval is compact. Therefore, by the Arzela-Ascoli theorem, every uniformly bounded and equicontinuous sequence of real-valued functions on has a uniformly convergent subsequence. Hence has a subsequence that converges uniformly on .

Proof.

(a) Let

Suppose, for contradiction, that is disconnected. Then there exist disjoint nonempty sets and , separated in the relative topology of , such that

For each , the set is connected and is contained in . Therefore must be contained entirely in or entirely in .

Choose one index . Without loss of generality, suppose

Now take any . By hypothesis,

Since , the set meets . But is connected and must lie entirely in either or . Since it meets , it cannot lie in . Hence

Because was arbitrary, every is contained in . Hence

which contradicts the fact that is nonempty. Therefore is connected.

(b) Yes, the union is path connected.

Let

Take any two points . Then there exist indices and such that

By hypothesis,

Choose a point

Since is path connected, there is a path in from to . Since is path connected, there is a path in from to . Concatenating these two paths gives a path from to lying entirely in

Thus any two points of can be joined by a path in , so is path connected.

Proof.

Because is compact, the open cover

has a finite subcover. Thus there exist points such that

Set

Now take any set of distinct points in . Since these points are covered by only balls, the pigeonhole principle implies that two of the points, say and , lie in the same ball .

Therefore

Hence every set of distinct points in contains at least two points whose distance is less than .

Proof.

Write

The curve starts and ends at the origin:

For , we have

and

Thus the curve is one-to-one on , so it is a simple closed curve except for the common endpoint. The area is the absolute value of the signed area

We compute

Therefore

Expanding,

Hence the signed area is

Thus

The negative sign means the curve is oriented clockwise. Therefore the area is .

Proof.

Yes, is differentiable everywhere.

Let

The set is the closed disk centered at the origin with radius . In polar coordinates,

Define

Since is continuous, is continuous. Therefore the one-variable function

is differentiable and

For , the chain rule gives

and similarly

Thus, for ,

where

Equivalently,