2007 Spring Comprehensive in Analysis

Proof.

We split into two cases.

First, suppose that for infinitely many . For each such ,

Hence the terms of the series do not even tend to along this subsequence, so the series diverges.

Second, suppose that for all sufficiently large . Then for all sufficiently large ,

and therefore

Since diverges, its tail also diverges. By comparison,

diverges.

Thus in all cases diverges.

Proof.

We prove both implications.

(a) (b). Assume is continuous at in the - sense. Let and suppose . Given , choose such that

Since , there exists such that for all ,

Therefore, for all ,

Thus .

(b) (a). Assume the sequential condition holds. We prove the - condition by contradiction. Suppose is not continuous at in the - sense. Then there exists such that for every , there exists with

and

For each , choose such that

and

Then , but does not converge to , contradicting the sequential condition.

Therefore the two definitions are equivalent.

Proof.

(a) The sequence converges uniformly on to a function if for every , there exists such that for all and all ,

Equivalently,

(b) The statement is true.

Assume uniformly on . Since

the mean value theorem implies that for all real numbers ,

Therefore, for every ,

Taking suprema over , we get

The right-hand side tends to , so uniformly on .

Proof.

(a) The statement is false.

Let with the usual metric. The set of rational numbers is dense in , and the set of irrational numbers is also dense in . However,

which is not dense in . Thus the intersection of finitely many dense subsets need not be dense.

(b) The statement is true.

Let be open dense subsets of . Their intersection

is open because finite intersections of open sets are open.

It remains to prove that is dense. Let be any nonempty open subset of . Since is dense, is nonempty. Also is open. Since is dense,

is nonempty and open. In fact, let

Since is dense in and is a nonempty open subset of , we have

Thus is nonempty. Also, since both and are open, is open.

Pick any point . Since is open, there exists such that

Since is dense in the metric space , every nonempty open ball in intersects . Therefore

But , so

Hence

or equivalently,

Continuing inductively, we obtain

Thus every nonempty open set intersects , so is dense in .

Proof.

As the statement gives , the expression must be interpreted as a scalar. We interpret it as the squared Euclidean, or Frobenius, norm

Thus

Let . We use the Frobenius inner product

Then

Therefore

Since as , and since

we get

Hence

Therefore is differentiable at , and its derivative is the linear map

Equivalently, the gradient is

Proof.

(a) Since is continuous on the compact interval , it is Riemann integrable and uniformly continuous.

The number is the right-endpoint Riemann sum for on the uniform partition

We prove directly that

Let be the modulus of continuity of :

Since is uniformly continuous, as .

For , we have

so

Therefore

The right-hand side tends to . Hence converges, and

(b) Suppose for all , and for some . By continuity, there exists a closed subinterval of positive length and a constant such that

for all .

Hence

From part (a),

Proof.

Let . By the mean value theorem applied to on , there exists such that

Since , this gives

By the mean value theorem applied to on , there exists such that

Because and is monotonically increasing,

Therefore

Multiplying by , we get

Thus

and hence

Since were arbitrary, is monotonically increasing on .

Proof.

Although the problem says to use the Inverse Function Theorem, the standard equivalent tool here is the Implicit Function Theorem, which follows from the Inverse Function Theorem.

Define

At , the equation becomes

so the roots are , , and .

We compute

At the three base roots, we have

and

All three values are nonzero.

By the Implicit Function Theorem, there exist neighborhoods of and continuously differentiable functions

such that

and

and

Shrinking the neighborhood of if necessary, the three values , , and remain in disjoint neighborhoods of , , and , respectively. Hence they are three distinct real roots of the cubic.