# What is a Form Domain of an Operator?

I tried to look this up on Wikipedia, but I couldn't find anything.

I am reading Barry Simon's book "Schrödinger Operators", where he brings up the concept of a form domain $Q(A)$ of a self-adjoint operator $A$. I looked in his mathematical physics series, but couldn't pinpoint a definition.

In parallel to stating some perturbation theorems about form domains, he also states theorems about operator domains, which are quite self-explanatory. I've never heard of form domains however.

What are they? How and where do they come up in operator theory? Why are they important, generally speaking?

-
I believe it might have something to do with quadratic forms... – r.g. Jul 11 '11 at 7:37
Well, this is all discussed in detail in Reed and Simon vol. I. Section VIII.6 (also, Simon gives a reference...) – t.b. Jul 11 '11 at 10:33
Thanks Theo. I was looking in the second volume, because in the book I was reading the chapter on Self-Adjointness, and the second volume of Reed-Simon series is about Fourier Analysis and Self-Adjointness. – r.g. Jul 11 '11 at 11:26

Essentially the form domain is the "largest domain" on which a self-adjoint operator can "make sense" in a weak way. Formally speaking given an unbounded self-adjoint operator $A$ on some Hilbert space $\mathcal{H}$, you can consider the quadratic form $\psi \mapsto (\psi,A\psi)$. The largest dense subspace of $\mathcal{H}$ which we denote by $Q(A)$ on which this quadratic form makes sense is called the form domain of $A$.
As an example, let $\mathcal{H}$ be $L^2(\mathbb{R})$. We known that the Laplacian does not makes sense on all of $L^2$, it's domain $D(A)$ is roughly speaking the Sobolev space $H^2(\mathbb{R})$ of functions in $L^2$ whose second derivative is also in $L^2$. But if you consider the quadratic form $(\psi,\triangle\psi) = (\nabla\psi,\nabla\psi)$ (a formal integration by parts; this can be made more precise using the spectral representation of $A$), we see that the form domain $Q(A)$ is more like $H^1$; we just require the first derivative now to be in $L^2$.
Note that $Q(A)$ necessarily contains $D(A)$. But it is often possible to extend the quadratic form to a larger domain.