2026 Spring Comprehensive in Analysis

Problem 1.

Let be any sequence of positive real numbers. Denote

Prove that the series

converges and that its sum is less than or equal to .

Proof.

Since

we have

Therefore

Hence the partial sums telescope:

Since , we get

The partial sums are increasing because all terms are positive, and they are bounded above by . Therefore the series converges and

Problem 2.

Suppose is uniformly continuous on and satisfies

for any .

(a) Show that

(b) Does the above claim still hold if uniformly continuous is replaced by continuous? Explain your answer.

Proof.

(a) Suppose not. Then there exist and a sequence such that

for every .

Write

where and . Since is compact, after passing to a subsequence we may assume

for some .

Because is uniformly continuous on , we have

By assumption, since ,

Therefore

contradicting . Hence

(b) No. The claim can fail if is only continuous.

For each , let

Define a triangular spike function

Now define

The supports of the are very small intervals centered at . They are locally finite, so is continuous on .

Also,

for every , and . Hence

However, for each fixed , we have

Indeed, the only possible spikes near have centers of the form . For fixed , the fractional parts tend to , while the widths of the spikes are of order . Thus for all sufficiently large , the point lies outside every spike support, and so

Therefore continuity alone is not enough.

Problem 3.

Does the integral

converge? Justify your answer.

Proof.

The only possible difficulty occurs near the zeros of . These zeros are

Away from these points, is bounded below by a positive constant on subintervals, and the integrand is bounded by a constant multiple of , which is integrable at infinity.

It remains to estimate the contribution near each . For near , write

Then

For , we have

when is small, and in any case this gives the required local estimate. Also, for large , is comparable to on the interval

Thus

The last integral is bounded by

Using the change of variables , we get

Therefore

Since

the integral converges. Hence

converges.

Problem 4.

(a) Show that if is a compact connected metric space, and is continuous, then the image is a finite closed interval.

(b) Suppose is a metric space such that for any continuous function , the image is a finite closed interval. Show that must be compact and connected.

Proof.

(a) Since is compact and is continuous, is compact in . Since is connected and is continuous, is connected in .

The compact connected subsets of are exactly the finite closed intervals. Therefore

for some real numbers .

(b) First we prove that is connected. Suppose is not connected. Then there exist nonempty disjoint open sets such that

Since and are also closed in , the function

is continuous. But

which is not an interval. This contradicts the hypothesis. Hence is connected.

Now we prove that is compact. Let be any fixed point of . Then the function must be bounded by assumption. Let be a sequence of , if for any such sequence there is a subsequence which is convergent to a point , then must be compact. If not, then is an unbounded continuous function on , whcih is a contradiction.

Problem 5.

Show that for any continuously differentiable function and any , there exists a polynomial such that

and

Proof.

Since , we have . By the Weierstrass approximation theorem, there exists a polynomial such that

Define

Since is a polynomial, is also a polynomial.

Now

Therefore

and

Also, for every ,

Thus

Hence

Problem 6.

Consider a linear operator defined by

where

Is bounded? If yes, find .

Proof.

Define a sequence by

Then

Since

we have

and

By the Cauchy-Schwarz inequality,

Thus is bounded and

To show equality, take

Then

and

Therefore

Problem 7.

Show that there exist and such that for any matrix that is -close to identity, there exists a unique matrix in the -neighborhood of the identity matrix such that

Proof.

Let denote the vector space of real matrices. Define

Then is continuously differentiable. Also,

We compute the derivative of at . For ,

and

Therefore

Thus

is the linear map

which is invertible.

By the inverse function theorem, there exist neighborhoods and of such that

is a bijection with a continuously differentiable inverse. Therefore, there exist and such that if

then there exists a unique matrix with

such that

That is,

Problem 8.

Evaluate the integral

Proof.

The region of integration is

Equivalently,

Therefore

Now

Thus

Hence

Therefore

Problem 9.

A fisherman's net has a rim, which is a circle of radius . He fixes it in the sea in such a way that the rim is in the -plane with center at the origin. The velocity of water is given by the vector field given by

Find the flux of the water across the net.

Proof.

The rim is the circle

in the plane . Since

we get

Therefore the flux through any surface spanning the rim is the same as the flux through the flat disk

Choose the normal vector

On the disk , we have , so

Thus

Therefore the flux is

Hence the flux across the net is

depending on the orientation chosen.