# Test 2 with solutions

This test is closed book. You are not permitted to bring any books, notes or such material with you. You may use theorems, lemmas and propositions from the book or from class.

Note that most of the solutions are relatively short - this is likely to be the case in the final as well!

1. If is a convergent sequence, with respect to the norm show that there is a subsequence which converges pointwise almost everywhere.

Solution:- Since has finite measure we know that and in fact

Now, from results in class we know that any sequence which converges in has a subsequence which converges almost everywhere. It follows form the estimate above that if converges in it converges in and hence has a subsequence which converges almost everywhere in

2. Suppose that has Fourier coefficients satisfying

 (44)

Show that is continuous on

Solution:- Since we know that its Fourier series converges to in

Now, consider the Fourier series itself, under the assumption (1). If then

where we have used Schwarz' inequality. From the assumed convergence it follows that the right hand side is arbitrarily small if is large enough. That is, the Fourier series itself is Cauchy in the uniform norm, hence uniformly convergent. It follows that the limit of this series is continuous and that it is a representative of (this is what continuity of means, it has a continuous representative).

3. Let be the sequence obtained by orthonormalization (the Gramm-Schmid process) of Show that for each the Fourier transform is linearly dependent on

Solution:- The Gram-Schmid process replaces the by where each is a linear combination of the for Thus it is enough to show that itself is linearly dependent on the Fourier transforms Now, we also know that for some non-zero constant so the statement is true for We can proceed by induction, assuming that we have already shown that the statement is true for and then just prove it for In fact the Fourier transform satisfies

Again by induction we therefore know (in fact we showed in class) that

where is a polynomial of degree at most and Thus by induction we see that the Fourier transform of is a linear combination of the for

4. Show that if is a bounded measurable function which satisfies for all non-negative integers then there is an odd function such that for almost all

Solved:- Since is bounded and measurable, it is in Consider the function This is even and in Moreover it satisfies

For even powers this follows from the assumption on since

On the other hand for odd powers

Now, we know that polynomials are dense in so we can choose a sequence in Thus it follows that in and hence that

almost everywhere.

That almost everywhere implies that

where is an odd function.

Richard B. Melrose 2004-05-24