## For Lectures 11-14

I
Our discussion of Hilbert space is relevant because of:-

Proof. We have shown that is a linear space and that the inner product on is well-defined and satisfies the needed properties. Thus is a pre-Hilbert space and we only need to show that it is complete for the norm

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

 (2)

Let us set and so obtain the union as where each has finite measure and Then implies, by the Cauchy-Schwarz inequality

 (3)

Applying this to we see that is Cauchy in for each fixed So, by completeness of it converges to Recall that we may change the values of the on a set of measure zero in so that on a subsequence pointwise on Passing to the diagonal subsequence in the uniqueness of the limit (modulo values on sets of measure zero) means that there is one function defined on the whole of such that almost everywhere. Taking on and again changing values (to on a set of measure zero if necessary we can arrange that pointwise on

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

 (4)

Given the fact that is Cauchy in allows us to choose such that for This implies that, for the right side of (3) is less than or equal to On the other hand the integrand on the left side is pointwise convergent so

 (5)

This shows that in (and in particular that

II
Now, suppose we are in a general Hilbert space Suppose that is a countable - either finite or countably infinite - orthonormal set. That is, each element has norm one and they are pairwise orthogonal so if Then for each element we can consider the constants

 (6)

Proof. Take an ordering of so that we can replace it by and for any finite consider the finite sum

 (7)

Expanding out the inner product using the sesquilinearity gives

 (8)

using the orthonormality of the 's. Doing the same thing for where we find

 (9)

On the other hand if we write we see that for Since is a sum of these 's, This in turn means that

since the cross-terms vanish in the linear expansion. Thus

 (10)

Going back to (8) we conclude from (11) that the series has an upper bound independent of When is inifinte this means it converges (in From (9) it follows that the sequence is Cauchy in Since is complete, being a Hilbert space, it must converge and the limit must satisfy (6).

See if you can show that the limit is independent of the order chosen for

III
A (countable) orthonormal set is said to be complete if for all Notice that is the limit of as and whenever Taking the limit we see that

 (11)

for all where we have used the fact that if in then for each - this follows from Schwarz inequality since

 (12)

Thus another way of stating the completeness of the orthonormal set is that

IV
Existence of complete orthonormal bases.

Proof. Recall that a metric space is separable if it has a countable dense subset. So we can suppose there is a countable set with Let be an enumeration of We extract a complete orthonormal basis from by applying the Gramm-Schmid procedure. First consider If it is zero, pass to If it is non-zero, set which gives an orthonormal set with one element, now pass to Proceeding by induction, suppose at stage we have an orthonormal set with elements such that each of the is linearly dependent on these elements. Now consider If it is dependent on the for pass on to If not then

 (13)

is well-defined, such that is orthonormal and such that the for are dependent on this new orthonormal set.

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.

V
The basic result we will prove on Fourier series is that for the special case of computed with respect to Lebesgue measure, the exponentials

 form a complete ornomormal set (14)

First we want to check that it is indeed an orthonormal set. Since the norm is easy enough to compute

 (15)

since we do know how to integrate constants. The orthogonality would seem almost as easy,

 (16)

However, this is proof by abuse of notation since here we are using the Riemann integral (and Fundamental Theorem of Calculus) and we are supposed to be computing the Lebesgue integral.

So I need to go back and check some version of their equality. The following will do for present purposes.

Proof. We can split into real and imaginary parts if it is complex valued and the result then follows from the real case; so assume that is real. Since we do know how to integrate constants (being simple functions) we can add to the constant (or something larger) and we can then assume that Now, the Riemann integral is defined as the common value of the upper and lower integrals (look this up in Rudin [2], I am not going to remind you of all of it.) One result for continuous functions (in fact general Riemann integrable functions) is that given there exist a partition of such that the difference of lower and upper partial sums satifies

 (17)

Notice here that the lower sum is actually for a simple function which is smaller than So, directly from the definition of the integral we know that

 (18)

where the integral on the left is Riemann's and on the right is Lebesgue's. On the other hand if we simply divide into equal parts and take the simple function with value on each interval we get a sequence of simple functions increasing to (in fact uniformly on The Riemann lower partial sum for this partition is bounded above by the Riemann integral, and by the monotone convergence theorem this sequence converges to the Lebesgue integral. This gives the opposite inequality to (19) so the two integrals are equal.

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.

VI
To prove the completeness of the Fourier functions we need to show that any function which satisfies

 (19)

itself vanishes almost everywhere, so is zero in In fact we will show something a little stronger,

However this will take some work.

VII
First we make the following observation directly from our construction of the integral.

Proof. This is really just a reminder of what we have done earlier. We might as well assume that is non-negative, since we can work with the real and imaginary parts, and then with their positive and negative parts, separately. It suffices to show that every subsequence of the real sequence in (21) has a convergent subsequence with limit the integral over Since we are not assuming anything much about the 's it is enough to show that there is a subsequence converging to the integral over and then apply the argument to any subsequence. Since the measure of the symmetric difference we can pass to a subsequence (which we then renumber) so that

 (20)

Since it suffices to show that these sequences tend to zero. So, set from (22) it follows that has finite measure and has measure tending to zero, with Thus it suffices to show that since this sequence dominates by monotonicity. Finally then we write which is a decompostion into a countable collection of disjoint open sets. By the countable additivity of the integral

 (21)

Thus the series of non-negative terms on the right converges which implies that the series `of remainders'

 as (22)

This shows that the as the other half of (22) can be handled in the same way, so we have a subsequence with the correct convergence and hence have proved the Lemma.

VIII
So, let's apply Lemma 1 directly as follows.

Proof. By Lemma 1, as

IX
Now, go back and consider which satisfies (20). We will show that there is a continuous function (not obviously zero) which also satisfies (20). Namely consider

 chosen so (23)

The constant term here is added so that We have to work harder to show that the other To do so we use the integration by parts identity

 (24)

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

 (25)

This is an linear combination of our Fourier exponentials, so

 if  satisfies (20) (26)

Since we already have arranged that and the constant does not change the for once we prove Proposition 4 we will know that

 given by (25) satisfies (20) if  does so. (27)

X
Proof of Proposition 4.

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

 and (28)

are both continuous. This means that the products and are both integrable (why exactly).

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

 (29)

Using the constraints on the domains (these are now Riemann integrals anyway) this is equivalent to

 (30)

We can replace and both by from the support properties and similarly we can replace and by Calling the new end points and again we are down to

 (31)

which is just a very special case of Riemannian integration by parts. Thus (26), and hence Proposition 4, is proved in general.

XI
Okay, so now we know that if has all the then in (25) is continuous and satisfies the same thing. Of course we can continue this, and integrate again, replacing by and get a new function which satisfies

 (32)

Thus is continuously differentiable and still has We will show that this implies by proving a convergence result for Fourier series, this is [1] Theorem 2 on p.140, more or less.

XII

Proof. Pick a point and consider the partial sum of the series we are interested in

 (33)

Here we have just inserted the definition of the So consider the function

 (34)

To see this, multiply the sum definining by and observe that it becomes where Also integrating term by term we find that

 (35)

The reason for looking at is that, from (36),

 (36)

From (38) we can write the expected limit in a similar way

 (37)

where we use the second part of (38) to reorganize the integral. Combining these two we get

 (38)

Now, the function in the denominator of in (37) vanishes precisely at in and does so simply. That is, is continuous if we define it to take the value of the derivative at This means that the quotient

 (39)

is bounded and continuous on if we define it correctly at In particular this means it defines an element Then (41) can be written

 (40)

We already know, by Bessel's inequality, that the series converges so as Thus we have proved Theorem 3.

I should have commented a little on the case

XII
So, now we know that for each if is differentiable. In fact the proof shows a little more than this, so let us record it:

Proof. Just check that this is all we really used.

XIII
Returning to our efforts to prove Proposition 3 we now know that if has for all then given by (25) must be constant, since its indefinite integral vanishes. Thus the integrals are independent of the end point so must all vanish (since we know that the limit as is zero). Thus we have a function with integral zero. The proof of Proposition 3 is therefore finished by

Proof. First, we can take the difference

 (41)

to conclude that

 (42)

the ring of subsets of consisting of finite unions of disjoint intervals. But then, by Lemma 1, we see that the integral vanishes for all Lebesgue subsets of since they can be approximated by such finite unions. Now, take to be the measurable set on which and then the set on which and we conclude that almost everywhere in and hence in

XIV
Finally then we have proved Proposition 3 which can be restated as

 No two distinct elements of  have the same Fourier coefficients, (43)

or, if you prefer, that the Fourier coefficients determine a function in For a function in we have the stronger statement that

Richard B. Melrose 2004-05-24