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Hello MathOverFlow

I have some questions about frames and reproducing kernels and here they are:

For a Hilbert space $H$ spanned by a frame $\lbrace f_n \rbrace$ there exists a reproducing kernel $K(y,y')$ such that

$f(y) = \int f(y') K(y,y') dy'$

where

$K(y,y') = \sum_n f_n(y) g_n(y')$

and $\lbrace g_n\rbrace$ is the dual frame to $\lbrace f_n \rbrace$.

I assume that in general the kernel will not be a convolution kernel (ie $K(y,y′)=K′(y−y′)$). Is this true? Under what conditions is this frame-based kernel a convolution kernel?

Thank you

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Could you be more specific about your setup? An RKHS needs to be a space of functions defined on some underlying set $E$. In order for your question to make sense, this set $E$ has to be a group (abelian?) so $y-y'$ is defined. Also, the reproducing property is $f(y) = \langle f, K(\cdot,y)\rangle_H$, but the inner product in $H$ is typically not the $L^2$ inner product. Indeed, $L^2(\mathbb{R})$ is not an RKHS because it is not a space of functions and even if it were, evaluation could not be made continuous. Perhaps you meant something more general by the integral? –  Noah Stein Jul 18 '12 at 0:04
    
Hi Noah Let me give it another try. For a frame $\lbrace f_n\rbrace$ in $L^2$ define the quantity $K(y,y')$ such that $K(y,y') = \sum_n f_n(y) g_n(y')$ where $\lbrace g_n\rbrace$ is the dual frame to $\lbrace f_n\rbrace$. I believe that it is true (correct me if I am wrong) that $K(y,y')$ has the following property for all $f(y)$ in $L^2$ $f(y) = \int f(y') K(y,y') dy'$ What I would like to know is: Under what conditions is $K(y,y′)$ a convolution kernel. For example: Is it always a convolution kernel? Is it only a convolution kernel for tight frames? That sort of thing. –  user25187 Jul 18 '12 at 17:25
    
OK I have investigated reproducing kernels a bit and I now understand that the math of RKHS is not going to be of use to me in answering the question I need to answer. So here is the question that I want to answer: Suppose $\lbrace f_n \rbrace$ is a frame on $L^2$. Then the sum in my definition of $K(y,y')$ is infinite. When the sum is truncated what can be said about the form of $K(y,y')$? Can it be said that in general $K(y,y') \ne k(y-y')$? In the case where the frame is the discrete-time fourier transform basis then $K(y,y') = k(y-y')$. But will this be true for general frames? –  user25187 Jul 19 '12 at 22:43

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