2019 Spring Real Analysis

Problem 1.

Let be a sequence of functions in that converge to a function in . Prove, or disprove by example, that:

(a) converge to in .

(b) converge to in .

Proof.

(a) is true. It follows from the Cauchy inequality

(b) is not true. For example, let for . Then is convergent to in but not is convergent in .

Problem 2.

(a) Show that any sequence of non-negative integrable functions on with

must converge to zero almost everywhere.

(b) Is there a sequence of non-negative integrable functions on satisfying

which does not converge to zero almost everywhere? Explain.

Proof.

(a). Since

must be almost everywhere finite. Hence for almost all , as .

(b). We consider intervals . Let be the characteristic function of . Write any uniquely as , where . Then as , we have . Let . Then . However, doesn't tend to zero almost everywhere.

Problem 3.

Assume that and are two finite positive measures on a measure space . Prove that is absolutely continuous with respect to if and only if .

Proof.

If , then for any with , we must have

Hence .

On the other hand, if , then by Radon–Nikodym theorem, there is an integrable function such that

for any measurable set . Thus

Since is integrable, we have

by the Dominated convergence theorem.

Problem 4.

We say a function is convex if

for all and .

(a) Let be a convex function on . Show that exists Lebesgue almost everywhere.

(b) Does there exist a convex function on such that equals zero almost everywhere, but is not linear? Either construct or prove it is impossible.

Proof.

We assume that . From (1), we have

Then we have

Thus

Similarly, we have

and hence

As a result, we have whenever , i.e., is monotonically increasing. As a monotonic function, its derivative exists almost everywhere. This proves (a).

Let be a singular function, that is, is strictly monotonically increasing function whose derivative is zero almost everywhere. Let . Then is convex and equals to zero almost everywhere.

Problem 5.

Let and let be the 1-periodic function on with for . Show that

Proof.

We observe that

Then the anti-derivative function

is a -periodic function. For any , let such that

Using integration by parts, we obtain

Since is periodic, and is a function with compact support, we have

Thus for , we must have

Since , by combining the above, we get

for , completing the proof.

Problem 6.

Let be a twice continuously differentiable function on with compact support. Show that

Proof.

Since is of compact support, we have

from which we conclude that

Therefore, we have