Abstract Fourier series and Parseval’s identity

Jordan Bell
April 3, 2014

1 Orthonormal basis

Let H be a separable complex Hilbert space.11 1 One talk do everything we are doing and obtain the same results for nonseparable Hilbert spaces, but one has to define what uncountable sums mean. This is done in John B. Conway, A Course in Functional Analysis, second ed., chapter I. If eiH, i1, and ei,ej=δi,j, we say that the set {ei} is orthonormal. If span{ei:i1} is a dense subspace of H, we say that {ei:i1} is an orthonormal basis for H. We can write this in another way. If Sα,αI are subsets of H, let αISα be the closure of the span of αISα. To say that {ei} is an orthonormal basis for H is to say that {ei} is orthonormal and that H=i1{ei}.

2 Abstract Fourier series

If ak and the sequence k=1nakek converges in H, we denote its limit by


This is a definition of an infinite sum in H. Since H is complete, one usually shows that a sequence converges by showing that the sequence is Cauchy, and hence to show that k=1nakek converges it is equivalent to show that


as m,n. And showing this is equivalent to showing that


as m,n. This is equivalent to


as m,n, and this is equivalent to the series


converging. Thus, the series k=1akek converges if and only if the series k=1|ak|2 converges.22 2 Furthermore, using the triangle inequality rather than the orthonormality of the ek, one can check that if the series k=1|ak| converges then the series k=1akek converges.

Let {ei:i1} be an orthonormal basis for H; it is a fact that one exists. Let vH and define


If 1in then




It follows that

k=1n|v,ek|2 = sn,sn
= v,v,

where we used v-sn,sn=0 in the third line. Therefore the series k=1|v,ek|2 converges, and so the sequence sn converges to some v=k=1v,ekekH. Since sn converges to v, in particular it converges weakly to v, i.e., for any wH,


Therefore for any j,


this is because for nj we have v-sn,ej=0 and hence v,ej=sn,ej. As v-v,ej=0 for all j, it follows that v-v=0, i.e. v=v. Hence,


We call this an abstract Fourier series for v.33 3 If H=L2(𝕋), one checks that ek=eik,k, is an orthonormal basis for H. Then, f,ek=12π02πf(t)e-ik𝑑t and f is the limit in H of k=0nf,ekeik. Thus in H, f=kf,ekeik. It can be written as


and thus can be written without v as


ekekB(H) is a projection with rank 1, and the above series conveges in the strong operator topology on B(H). Writing the identity map in this way is called a resolution of the identity.

3 Parseval’s identity

On the one hand


On the other hand,




which is Parseval’s identity.