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I'd like to know if one can define a pertinent Hilbert space where the operator $$A_p v := -\frac{1}{2} v" + (vF + v\int_\mathbb{R} Sp + p\int_\mathbb{R} Sv )'$$ is symmetric. Here, $p$ satisfies the following differential equation with $p \in L^2_1(\mathbb{S})\cap C^\infty(\mathbb{S}) $ $$\frac{p'}{2p} = F + \int Sp $$ and $u,v$ and $F,S$ can be taken to satisfie at least the following weak regularity conditions:

  • $u,v \in L^2_0(\mathbb{R})\cap C^2(\mathbb{R})$
  • $F,S \in C^\infty(\mathbb{R})$ and odd

but the regularity will probably be restricted much more by choosing the nice Hilbert space that we wish. We note that $\int_\mathbb{R} Sp= \int_\mathbb{R} S(x) p(x) dx$.

My motivation for a possible positive answer for this question comes from the following similar problem. Given a fixed $p \in L^2_1(\mathbb{S})\cap C^\infty(\mathbb{S}) $ ($\mathbb{S} = \mathbb{R}/(2\pi\mathbb{Z})$ and $p \neq 0$), solution of $\frac{p'}{2p}= S*p$ ($S(\theta) = \sin(\theta)$) for, the operator

$$B_p v := -\frac{1}{2} v" + ( v\int_\mathbb{S} S*p + p\int_\mathbb{S} S*v )'$$

a such Hilbert space is $H_ {-1,1/p}$ provideded with $\left\langle f,g\right\rangle_{H_ {-1,1/p}} := \int \frac{\mathcal{F}\mathcal{G}}{p} $ where $\mathcal{F}$ and $\mathcal{G}$ are respectively the primitives in $ L^2(\mathbb{R})$ of $f$ and $g$ such that $\int \frac{\mathcal{F}} {p} = \int \frac{\mathcal{G}} {p} =0$ . Here $H_ {-1,1/p}:= H_{1,p}'$ (notation:$V'$ is the dual space of $V$), $H_{1,p}:= ${$ \overline{ f \in C^1(\mathbb{S}): \int f =0 } ^H$} (wich is also a Hilbert space with scalar product $\left\langle f,g\right\rangle_{H_ {1,p}} := \int (f'g'p)$ ) and $H =L^2 _0 (\mathbb{S})$.

We can prove for this case that $$\left\langle f,B_pg\right\rangle_{H_ {-1,1/p}} = \left\langle B_pf,g\right\rangle_{H_ {-1,1/p}}=-\frac{1}{2}\int\frac{fg}{p}+\int fS'*g$$ then $B_p$ is clearly symetric in $H_ {-1,1/p}$

I expecte had well motivated my question. I've tried several integral and derivative restrictions under spaces similar to $H_ {-1,1/p}$ but I've not been sucsecefull yet. I'd be glad for some advice. If you visit the topic please leave a message with your opinion.

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    $\begingroup$ Is this a homework problem? I see that you included a "hint". $\endgroup$
    – Nik Weaver
    Jun 18, 2012 at 16:44
  • $\begingroup$ No, I've just tried to make a clear question giving as a hint the technic that I've saw in a article in which a similar problem is solved. So it inspires me to triyng solve this subproblem of my real and bigger problem of reaserch. $\endgroup$
    – Paul
    Jun 18, 2012 at 23:45
  • $\begingroup$ If this question seems to be a homework question for you, you must be able to help me. If you want I can show you, Nik Weaver, the solution of the problem that motivates me (which I give as Hint). Do you think you have an idea? thank you $\endgroup$
    – Paul
    Jun 18, 2012 at 23:49
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    $\begingroup$ Paul, I suggest you don't use the word "hint" (and there is no need to feel insulted by Nik Weaver's suggestion, it was just a reaction to your unfortunate choice of words). Instead, call it a "Observation which may be helpful", and perhaps even include a reference to the paper where you saw this. $\endgroup$
    – Yemon Choi
    Jun 19, 2012 at 0:14
  • $\begingroup$ Thanks Yemon Choi. I really don't want to be misinterpreted. Just to make clear I did not feel insulted by Nik Weaver's comment, but at the same time I didn't understended that it was a sugestion, differently of yours comment explicitly sugesting me to change it. Sorry, I'm new here. And I still need help guys ... $\endgroup$
    – Paul
    Jun 19, 2012 at 0:49

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I don't completely understand the question, but I would answer "no". If you take $f = 1$ and $u = S = 0$ then you get the operator $Av = -{1\over 2}v'' + v'$, which is not even formally symmetric.

I'm especially confused by the motivating example, where the inner product seems to depend on the vector $v$.

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  • $\begingroup$ Thank's for you answer and for make me constact some errors of typing. $\endgroup$
    – Paul
    Jun 19, 2012 at 0:56
  • $\begingroup$ Actually, it's $u$ where before was $v$ in the definition of the spaces and scalar product. Please, see if now it makes more sense, because before it had any. $\endgroup$
    – Paul
    Jun 19, 2012 at 1:02
  • $\begingroup$ Okay, it makes more sense now. I still don't see how to help because your operator isn't formally symmetric --- the $v'$ part is skew symmetric --- but if the inner product can be anything I'm not sure how to answer you. $\endgroup$
    – Nik Weaver
    Jun 19, 2012 at 1:34
  • $\begingroup$ You have reason. The operator is formally not symetric but it can maybe be symetrized by chosing a good scalar product $\endgroup$
    – Paul
    Jun 19, 2012 at 11:19
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    $\begingroup$ yes. it is the same trick to represent the ornstein-uhlenbeck operator $u\mapsto \Delta u+\nabla B\cdot \nabla u$ as the gradient of the energy functional $u\mapsto \frac{1}{2} \int_\Omega |\nabla u|^2 e^{-B}dx$ -- this is done in the space $L^2(\Omega)$, taking the inner product $(u|v):=\int_\Omega u v e^{-B}dx$. $\endgroup$
    – Delio Mugnolo
    Feb 8, 2013 at 8:16

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