We will use the completeness of proved last week. Let be a Cauchy sequence in Consider first the set where at least one of the 's is non-zero. This is a countable union of measurable sets, hence measurable. It is in fact a countable union of sets of finite measure, for instance
Now that we have our putative limit we can use Fatou's Lemma. Applied to the sequence of measurable, non-negative, functions with fixed and it states that
See if you can show that the limit is independent of the order chosen for
Thus we can proceed by induction to define an orthonormal set, which will either be finite or countable (depending on In either case it is complete. To see this, suppose there is some element orthogonal to all the By the density of for any there exists such that However, is in the (finite) span of the so This however implies, by Pythagorus' theorem, that Thus so in fact and hence proving the completeness.
So I need to go back and check some version of their equality. The following will do for present purposes.
This argument only needs slight modification to show that every Riemann integrable function on is Lebesgue integrable and that the integrals are equal; it is done in Adams and Guillemin.
Thus we know that the Fourier functions do indeed form a countable orthonormal set for We still need to know that it is complete. This involves some more work.
However this will take some work.
Before worrying about the proof of this, let us apply it to and for some On the left in (26) the inner integral is our definition of in (25) except for the missing constant. On the right in (26) we compute, using equality of Riemann and Lebesgue integrals for continuous functions, finding
if satisfies (20) | (26) |
We have to prove (26). Again this is a return to basics. Notice first that both sides of (26) do make sense if since then the integrated functions
Now, (26) is separately linear in and So as usual we can assume these functions are positive, by first replacing the functions by their real and imaginary parts, and then their positive and negative parts. On the left the we can take a sequence of non-negative simple functions approaching from below and we get convergence in Similarly on the right the integrated functions approach uniformly so the integrals converge. The same argument works for so it suffices to prove (26) for simple functions and hence, again using linearity, for the characteristic functions of two measurable sets, measurable. Recalling what it means to be measurable, we can approximate say in measure, by a sequence of elementary sets, each a finite union of disjoint intervals. Using Lemma 1 (several times) the resulting integrals converge. Applying the same arguement to it suffices to prove (26) when and are the characteristic functions of intervals. In this case our identity has become rather trivial, except that we have to worry about the various cases. So, suppose that is the characteristic function of a subinterval of (open, half-open or closed does not matter of course). We are trying to prove
The reason for looking at is that, from (36),
I should have commented a little on the case
Richard B. Melrose 2004-05-24