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! :-)


  • $\begingroup$ For the second question, have you tried working through the usual proof? $\endgroup$ Jul 5, 2012 at 12:11
  • 2
    $\begingroup$ @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 $\|J_\epsilon f\|_H^2 \neq 0$. $\endgroup$ Jul 5, 2012 at 12:18
  • 1
    $\begingroup$ @Michael, I tried. It fails since I can't apply Holder or Schwartz inequality in this case. $\endgroup$
    – Anand
    Jul 5, 2012 at 12:33
  • 1
    $\begingroup$ Nice example. I didn't mean to criticise, I was just curious as to where the proof broke down. $\endgroup$ Jul 5, 2012 at 12:44
  • 2
    $\begingroup$ $H$ 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 $|\xi|^2$. $\endgroup$ Jul 5, 2012 at 14:14

1 Answer 1


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 $L^2({\bf R}^d)$.

  • $\begingroup$ 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. :-) $\endgroup$
    – Anand
    Jul 6, 2012 at 7:34
  • 1
    $\begingroup$ Sure, well, for someone with my background, this is the natural way to look at it. $\endgroup$
    – Nik Weaver
    Jul 6, 2012 at 13:27

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.