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Dear MOs,

I am now considering the following norm:

$$ ||f||_{H}^2 := \iint f(x) H(x,y) f(y) d x d y\:. $$

where the integral is over the whole space $R^{2d}$ and $H(x,y)$ is some non-negative definite kernel. Denote the space of functions with finite $||\cdot ||_{H}^2$ norm to be $L_H^2(R^d)$. If $H(x,y)=\delta_0(x-y)$, then the above norm reduces to the standard $L^2(R^d)$ norm and in this case $L^2(R^d)=L_{H}^2(R^d)$. For convenience, you can also assume that $H(x,y)$ is non-negative.

First question

Is there a general theory on this space? What are the weakest assumptions we need to put on the kernel $H(x,y)$?

Clearly, it can be more than functions, say measure (Dirac delta). I think that (if you assume that $H$ is non-negative)

  • $H(x,y)$ cannot go beyond measures.

  • $H(x,y)$ should be locally integrable. (Recall Dirac $\delta$ is locally integrable.)

When $H(x,y)$ has the form $H(x,y) =h(x-y)$ for some function or measure $h$, then there are theories by Gel'fand and Vilenkin (Generalized functions, Vol.4) that uses the Fourier transform to find the spectral measure $\mu$ corresponding to $h$:

$$ ||f||_{h}^2 = \int \widehat{f}(\xi) \overline{\widehat{f}(\xi)} \mu(d \xi)\;, $$

where $\widehat{f}$ is the Fourier transform of $f$ and the overline is the complex conjugate.

Second question

My second question is closely related to my motivation of this post. As we know that, if $f\in L^2(R^d)$,then $f*J_\epsilon\in L^2(R^d)$ and also

$$ ||J_\epsilon * f|| \le ||f||,\quad\text{and}\quad \lim_{\epsilon\rightarrow 0_+}||J_\epsilon*f-f||=0\;, $$

where $J$ is a molifier ($J$ is nonnegative, real-valued, belonging to $C^\infty(R^d)$ with compact support, $\int J(x) d x=1$, $J(x)=0$ if $|x|\ge 1$), $J_\epsilon(x)=\epsilon^{-d}J(x/\epsilon)$ and * denotes the convolution. The question is whether this result can be extended to the case of $L_H^2(R^d)$. More precisely, whether the following statement is true:


If $f\in L^2_H(R^d)$,then $f*J_\epsilon\in L^2_H(R^d)$ and also

$$ \lim_{\epsilon\rightarrow 0_+}||J_\epsilon*f-f||_H = 0\;. $$


Certainly, we need to put appropriate assumptions on $H$, which is related to the first question.

Thank you very much for any hints and help! :-)

Anand

share|improve this question
    
For the second question, have you tried working through the usual proof? –  Michael Albanese Jul 5 '12 at 12:11
2  
@Michael: clearly it strongly depends on the choice of $H$. Let $H(x,y)$ be a $C^\infty$ function supported in a $\epsilon$ neighborhood of $|x-y| = 10$. Then if $f\in C^\infty_0(B_1)$ we have that $\|f\|_H^2 = 0$. But for a sufficiently wide mollifier you can get $\|J_\epsilon f\|_H^2 \neq 0$. –  Willie Wong Jul 5 '12 at 12:18
1  
@Michael, I tried. It fails since I can't apply Holder or Schwartz inequality in this case. –  Anand Jul 5 '12 at 12:33
1  
Nice example. I didn't mean to criticise, I was just curious as to where the proof broke down. –  Michael Albanese Jul 5 '12 at 12:44
2  
$H$ could be a distribution that is not a measure. For example $H=h(x-y)$ with $h$ the negative Laplacian of a delta measure at 0. The Fourier transform of h is $|\xi|^2$. –  Jeff Schenker Jul 5 '12 at 14:14
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1 Answer

up vote 2 down vote accepted

Formally, we can define an operator $A$ on $L^2({\bf R}^d)$ by setting $$Af(x) = \int H(x,y)f(y) dy.$$ Then $$\langle Af, f\rangle = \int\int f(x)H(x,y)f(y) dxdy = \|f\|^2_H$$ (I am assuming real scalars, as I think you are). That is, you are just defining $\|f\|_H$ to be $\|A^{1/2}f\|$. The kernel $H(x,y) = \delta_0(x,y)$ corresponds to the case where $A$ is the identity operator.

I don't know if there's a definitive answer to the question of what the weakest possible assumptions on $H$ are, but any bounded positive operator $A$ will define a seminorm by the formula $\|A^{1/2}f\|$. It will be a norm if and only if $A$ has no kernel.

For your second question, as long as $\|J_\epsilon*f - f\| \to 0$, we will have $\|A^{1/2}(J_\epsilon*f - f)\| \to 0$ for any bounded positive operator $A$, that is, $\|J_\epsilon*f - f\|_H \to 0$.

I guess the point is that your problem can be approached by looking at whether integrating against $H$ defines a bounded operator on $L^2({\bf R}^d)$.

share|improve this answer
    
Thanks Professor Nik Weaver. Your operator point of view is very nice. In Gal'fand's book, they treat $H$ as a bilinear functional. I am used to their approach and forgot to use this operator method. Indeed, in your method, the function space $L^2(R^d)$ is fixed in the beginning. One looks for positive bounded operators over $L^2(R^d)$. In Gel'fand's approach, the function space is a completion of certain test functions, which usually are subset of $L^2(R^d)$ (for example some Sobolev spaces). Thanks a lot for your answer. :-) –  Anand Jul 6 '12 at 7:34
1  
Sure, well, for someone with my background, this is the natural way to look at it. –  Nik Weaver Jul 6 '12 at 13:27
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