8
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ I believe it might have something to do with quadratic forms... $\endgroup$ – r.g. Jul 11 '11 at 7:37
  • $\begingroup$ Well, this is all discussed in detail in Reed and Simon vol. I. Section VIII.6 (also, Simon gives a reference...) $\endgroup$ – t.b. Jul 11 '11 at 10:33
  • $\begingroup$ 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. $\endgroup$ – r.g. Jul 11 '11 at 11:26
14
$\begingroup$

The definition is given in Reed & Simon, Methods of Mathematical Physics, volume 1, page 276; see also example 2 on page 277.

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.

(For why and how this is used, you should see that section of Reed and Simon, as well as vol 2, Chapter X, the section on quadratic forms.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.