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!
- 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
- 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).
- 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
- 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