Integration by parts for a general negative-definite self-adjoint operator.

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

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.

-

2
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). – Florian Jun 30 2011 at 15:11
Yes, actually I should have mentioned that assumption also. Thanks for pointing this out. – RadonNikodym Jun 30 2011 at 15:14
@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) – RadonNikodym Jun 30 2011 at 15:19
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?" – Helge Jul 7 2011 at 3:46
The question is: Given f with the properties specified can I write $(-Lf,f)$ as $\int \nabla f \cdot A(x) \nabla f \pi(dx)$ – RadonNikodym Jul 7 2011 at 14:13
show 1 more comment

3 Answers

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.

-

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.

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.

-

The BC described at the beginning is not crazy. It represents an operator interpolating between periodic and anti-periodic boundary conditions. – Helge Jul 6 2011 at 18:09

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 $A$.

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 80, "mathjax_display_tex": 8, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9426164031028748, "perplexity_flag": "head"}