2006 Spring Comprehensive in Analysis

Proof.

The statement is false. Suppose that

where each is a straight line. Each line has one direction, and there are only countably many directions among the . Choose a line whose direction is not equal to the direction of any .

Then intersects each in at most one point. Hence

is countable. But itself is uncountable. This is a contradiction. Therefore the union of the cannot be all of .

Proof.

(a) The series is absolutely convergent. Indeed,

because and . Since converges, the given series converges absolutely.

(b) Let . Since , Taylor's formula with integral remainder gives

Thus

for all . Therefore, for each fixed ,

Hence

converges absolutely for every .

Moreover, if , then

By the Weierstrass -test, the series converges uniformly and absolutely on every bounded interval.

On the whole real line, the convergence is not uniform unless . Indeed, if is not identically zero, then

for every , so the terms do not converge uniformly to zero. Hence the series cannot converge uniformly on .

Proof.

(a) Bolzano-Weierstrass theorem: every bounded sequence of real numbers has a convergent subsequence.

Heine-Borel theorem: A subset of is compact if and only if it is closed and bounded.

(b) Since , both sequences remain positive. By the arithmetic-geometric mean inequality,

Thus for all .

Also,

so is decreasing, and

so is increasing. Since , both sequences are bounded. Therefore both sequences converge. Let

Passing to the limit in the recurrence gives

so . Therefore both sequences converge to the same limit. This common limit is the arithmetic-geometric mean of and .

Proof.

Let

The desired series is . Since

we get

Therefore

Here the Taylor expansion for comes from

Thus

Proof.

Let

Since on , is strictly increasing.

We use the identity

Hence

Therefore

Integrating by parts gives

Since is monotonically decreasing and , we have

Thus

Now, using , we get

Therefore

The last integral equals

Hence

Since , it follows that

In particular,

Proof.

At , the system becomes

Thus and .

Define

The Jacobian with respect to is

At , this is

whose determinant is . Hence the implicit function theorem gives local functions and .

To compute , differentiate the system implicitly. For the -derivatives,

Now

Thus at the base point,

Therefore .

Similarly, for the -derivatives,

and both vanish at the base point. Hence . Therefore

Proof.

(a) A family of functions is uniformly equicontinuous on if for every there is a such that for all and all ,

The same must work for every function in the family.

(b) The family is not uniformly equicontinuous. We show that the definition fails with, for example, .

Let be arbitrary. Choose

Then , , and . Also

Therefore

Since this happens for every , the family is not uniformly equicontinuous on .

Proof.

(a) We regard the space of real matrices as . The set

is closed because the map is continuous and is closed.

Also, if , then each column of has Euclidean norm . Hence every entry of has absolute value at most , so is bounded in . By the Heine-Borel theorem, is compact.

(b) Let be a tangent vector to at . Then there exists a differentiable curve in such that

Since ,

Differentiating with respect to gives

At , this becomes

Thus , so is skew-symmetric.

Proof.

(a) Let be functions on an open set . Suppose is a local maximum or minimum of subject to the constraints

If the constraint gradients

are linearly independent, then there exist real numbers such that

(b) Let

Then at the maximum point, which exists by compactness, we have

for some . Therefore we have for all . Since

we have . Thus

Apprent the maximum value is obtained when , which is .