5
$\begingroup$

I suspect I am asking a very stupid question.

Suppose you have self-adjoint negative-definite operator $L$ densely defined on a space $L^2(\pi)$, with $Lf = \nabla \cdot ( A(x)\nabla f)$, for some symm. pos. def matrix A. Here assume that differentiation $\nabla = (D_i)_{i=1,..,n}$ is a skew-adjoint operator densely defined on $L^2(\pi)$, that is, $\mathcal{D}(D_i) = \mathcal{D}(D_i^*)$ and $D_i^* = -D_i$, for $i=1,\ldots n$.

Now, I'm trying to make sense of the statement that $f \in \mathcal{D}((-L)^\frac{1}{2})$. This would imply that $((-L)f, f) < \infty$, but I'm not sure if we can ALWAYS write this as:

$$((-L)f, f) = \int_\Omega \nabla f\cdot A(x) \nabla f \space \pi(dx)$$

I'm sorry if this is a stupid question but for some reason I can't convince myself of this fact.

$\endgroup$
6
  • 2
    $\begingroup$ Am I right that you want to study this problem in the abstract (i.e. $\pi$ is a measure on some unspecified measurable space etc.)? Then your assumption that the operators $D_i$ are only skew-Hermitian seems too weak to me. A more natural assumption would be that the $D_i$ are skew-adjoint, that is, the domain of the adjoint $D_i^*$ equals that of $D_i$ (in addition to the skew-symmetry). $\endgroup$
    – Florian
    Jun 30, 2011 at 15:11
  • $\begingroup$ Yes, actually I should have mentioned that assumption also. Thanks for pointing this out. $\endgroup$ Jun 30, 2011 at 15:14
  • $\begingroup$ @Florian. Yes I would to know of results where $\pi$ is as general as possible (for example the underlying $\sigma$-algebra$ not necessarily countably generated, etc) $\endgroup$ Jun 30, 2011 at 15:19
  • $\begingroup$ What is the question? I just read it three times, and am not sure what the question is. My best guess is: "What does $f\in D((-L)^(1/2))$ mean?" $\endgroup$
    – Helge
    Jul 7, 2011 at 3:46
  • $\begingroup$ The question is: Given f with the properties specified can I write $(-Lf,f)$ as $\int \nabla f \cdot A(x) \nabla f \pi(dx)$ $\endgroup$ Jul 7, 2011 at 14:13

3 Answers 3

4
$\begingroup$

It is possible to make sense of $T^{1/2}$ without some of the particulars mentioned, when $T$ is a positive self-adjoint (densely-defined) operator on a Hilbert space. Namely, Friedrichs' argument (as in Riesz-Nagy, for example) shows that the resolvent $(T-\lambda)^{-1}$ exists and is a bounded operator for $\lambda$ not positive real. In particular, $T^{-1}$ is a bounded operator. It is also positive, so by standard (bounded-operator) spectral calculus admits a positive square root, whose inverse is the desired $T^{1/2}$.

Edit: after seeing some reactions, it is worth clarifying, as follows. Again, the square root of a positive unbounded (densely defined) _truly_self-adjoint_ operator exists without necessarily expressing the square root in terms of differential operators. The domain $D(\sqrt{T})$ is the same as the domains $D(\sqrt{T-\lambda})$ for $\lambda$ not in $[0,\infty)$.

There is the subordinate issue of whether one knows that the operator is genuinely self-adjoint, or only "symmetric". In the latter case, the question of self-adjoint extensions is non-trivial, depending on things abstracting "boundary conditions", tho' there is always the Friedrichs "minimal" extension.

In a similar vein, if we truly know a-priori that the operators $D_i$ are well-behaved in the sense that their domains and their adjoints' domains agree (the skew-adjoint version of self-adjoint), and assuming the domain(s) of the various expressions genuinely agree with that of the original operator, the seemingly formal computations are (by fiat) correct.

In practice, yes, there would be a non-trivial issue of specifying a common domain for the $D_i$ so that they are "genuinely" skew-adjoint, and so that the implied domain of the symmetric second-order operator is equal to that of its adjoint (so it is truly self-adjoint).

G. Grubb's book "Distributions and Operators" discusses many concrete examples of such things.

$\endgroup$
1
$\begingroup$

In short, "yes, probably, but you should be careful about boundary conditions."

The long version:

First a cautionary result. Let the Hilbert space be $L^2(0,1)$ and let $Lf =f''$ on the domain of functions $f\in L^2(0,1)$ with two derivatives in $L^2$ and such that $f(0)=-f'(1)$ and $f(1)=f'(0)$. This is a self-adjoint operator and by integration by parts

$$(f,(-L)f)= 2 \mathrm{Re}(\overline{f(1)}f(0))+ \int_0^1 |f'(x)|^2 dx.$$

However, this counter example is crazy, since it is not possible to interpret $L$ as $- D^\dagger D$ for an operator $D$ which takes one derivative. Also the boundary conditions I used would be very unlikely to arise in practice. Nevertheless it shows that some caution is needed.

So, let me interpret your question as follows:

"Let $L$ be the (unique!) operator defined on the domain $\mathcal{D}(L)\subset > \mathcal{D}(\nabla)$ of functions $f$ such that $A(x)\nabla f \in > \mathcal{D}(\nabla^\dagger)$, with $Lf=-\nabla^\dagger A(x) \nabla f$. Does the identity $$((-L)f,f)=\int \nabla f \cdot A(x)\nabla f d\pi$$ hold for $f\in \mathcal{D}(L)$?"

To begin with it may not be obvious that such an operator exists, or that it is self-adjoint, however this is the case and further more your integration by parts identity always holds. In fact, under the standard construction -- originally due to Friedrichs I think -- the answer is trivially yes since integration by parts is essentially the definition of $L$!

The Friedrichs construction is based on a theorem from functional analysis that says that any closed, positive quadratic form $q$ on a Hilbert space is the quadratic form of a positive self-adjoint operator. (See, for example, Thm. VIII.15 in Reed and Simon Vol. I.) In the present case we would define the quadratic form

$$q(g,f)= \int \overline{\nabla g(x)} \cdot A(x) \nabla f(x) d\pi(x) $$

which is easily shown to be closed and positive so long as $\nabla$ is a closed operator and $A(x)$ is symmetric positive definite as you assume. The proof of the theorem goes by showing that the domain $\mathcal{D}$ of functions $f$ such that $|q(g,f)| \le C \|g\| $ is dense so that it makes sense (by the Riesz thm. on linear functionals) to define an operator $L$ on this domain by the identity

$$q(g,f)= (g,(-L)f).$$

Note that the identity you want is a special case of this defintion!

It is easy to see now that the domain of $L$ consists of all functions $f$ such that $A(x)\nabla f \in \mathcal{D}(\nabla^\dagger)$ and that the quadratic form domain agrees with the domain of $\sqrt{-L}$ so that we have

$$\|\sqrt{-L}f\|^2 =q(f,f).$$

Note that, none of what was done above relied on the derivatives being implemented as anti-self-adjoint operators as you asked for. Returning to the one-dimensional case with Hilbert space $L^2(0,1)$ and $A=1$, first let $\nabla = d/dx$ on the domain of functions in $L^2(0,1)$ with one derivative in $L^2$. The resulting operator $L_N$ is the Neumann second derivative defined on the domain $\mathcal{D}(L_N)$ of twice differentiable functions with derivatives that vanish at $0$ and $1$ and the identity holds, however $\nabla$ is not anti-self-adjoint.

$\endgroup$
1
  • $\begingroup$ The BC described at the beginning is not crazy. It represents an operator interpolating between periodic and anti-periodic boundary conditions. $\endgroup$
    – Helge
    Jul 6, 2011 at 18:09
0
$\begingroup$

Some comments:

  • Self-adjointness of $L$ imposes a non-trivial condition on the measure $\pi$. I believe it has to be essentially Lebesgue. For example try $A = 1$, $\pi = 1 + x$, and $M = [0,1]$. Then $L$ is not self-adjoint.

  • You should ask if for $f \in \mathcal{D}( (-L)^{\frac{1}{2}})$, one has $$ ( (-L)^{\frac{1}{2}} f, (-L)^{\frac{1}{2}} f) = \int \nabla f \cdot A \nabla f \pi(dx), $$ since $-L f$ is only defined for $f \in \mathcal{D}(L)$.

    Is this an accurate interpretation of your question?

  • The formulation above is non-trivial, since one doesn't have that $$ (-\Delta)^{\frac{1}{2}} = - i \nabla. $$ It is given by multiplication by $|k|$ in the fourier basis. But my best guess is that for the usual cases the formula you stated is still correct... But I am also not completely sure how to check if for non-constant $A$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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