2017 Spring Comprehensive in Analysis

Proof.

Fix a point , and let

Clearly , so is nonempty.

We first show that is open. Let . Since is locally pathconnected, there is a pathconnected neighborhood of . If , then there is a path in from to . Since , there is also a path from to . Concatenating these two paths gives a path from to . Hence , so . Thus is open.

We next show that is open. Let . Again, choose a pathconnected neighborhood of . If met , then for some there would be a path from to . Since is pathconnected, there would also be a path from to . Concatenating these paths would give a path from to , contradicting . Hence , so . Therefore is open.

Thus is both open and closed in . Since is connected and is nonempty, we must have . Therefore every point of can be joined to by a path. If , join to and then to ; after reversing the first path if necessary, this gives a path from to . Hence is pathconnected.

Proof.

Yes, is continuous on .

For each fixed , consider the series

By the root test,

Hence the series converges absolutely for every , so is well-defined on all of .

To prove continuity, it is enough to prove uniform convergence on every compact interval. Let and assume . Since , there exists such that for all . Then for all and all ,

The series converges, so by the Weierstrass -test the given series converges uniformly on .

Each function

is continuous. Therefore is the uniform limit of continuous functions on every compact interval . Hence is continuous on every compact interval, and consequently is continuous on .

Proof.

We use the convention that and .

Because is open, there is such that the rectangle

is contained in . For nonzero with , define the second difference quotient

We will show that as .

Fix such . For fixed , apply the one-variable mean value theorem to the function . There exists , depending on and , such that

Now apply the one-variable mean value theorem in the second variable to . There exists such that

Therefore

Since as and is continuous at , we get

Now fix sufficiently small. By the definition of ,

Since is close to uniformly for all sufficiently small nonzero , passing to the limit gives

Thus the derivative of with respect to the first variable exists at , and

Proof.

Since is bounded, there exists such that

for all .

Fix . We prove that is continuous at . Choose and restrict attention to satisfying . Then lies in the compact interval between and , which is contained in

On this interval, is bounded by , and . Therefore

For such , we have

Hence

Thus is continuous at . Since was arbitrary, is continuous on .

Solution.

As stated, this assertion is false. The assumptions imply pointwise convergence, but they do not imply uniform convergence.

Indeed, let

with the usual metric, and define

Each is continuous on . Also, for every ,

Thus for all and . Moreover,

so the sequence is uniformly bounded.

For each fixed ,

Thus the pointwise limit is

This limit function is discontinuous at .

If converged uniformly to , then would be continuous as the uniform limit of continuous functions. But is not continuous at . Therefore the convergence is not uniform.

Equivalently, one can see the failure directly: at ,

which tends to , not to . Hence

does not tend to .

Therefore the proposed statement is false without an additional hypothesis, such as continuity of the pointwise limit.

Proof.

Define

Since is continuous, the fundamental theorem of calculus gives

Let

Then

Differentiating gives

Also,

Now define

Again by the fundamental theorem of calculus,

and also

Thus for every and . Therefore is constant and equals . Hence

for all .

Proof.

(a) The unit sphere is a closed oriented surface; it has no boundary. Thus

One way to see this explicitly is to write as the union of the upper and lower hemispheres. Each hemisphere has boundary equal to the equator, but the induced orientations on the equator are opposite. Therefore the two boundary curves cancel, and the total boundary is zero:

(b) Since is exact, there exists a -form on such that

By Stokes' theorem,

Using part (a), , so

Hence

Proof.

Since is increasing on , it is bounded:

for all .

We prove Riemann integrability using upper and lower sums. If , then is constant and hence integrable. Assume .

Let . Choose a partition

with equal subinterval lengths

Because is increasing, on each interval we have

Therefore

Since all subintervals have the same length,

Thus

Choose large enough so that

Then . Hence is Riemann integrable on .