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Consider the following convolution of a product of two functions $f(x)$ and $g(x)$:

$\int f(x')g(x')K_n(x-x') dx'$

where the kernel $K_n$ is a sequence of functions that approach a Dirac delta function as $n$ goes to $\infty$. In the limit as $n \rightarrow \infty$ we can write:

$\int f(x')g(x')K_{\infty}(x-x') dx' = \int f(x')K_{\infty}(x-x') dx' \int g(x')K_{\infty}(x-x') dx'$

that is this convolution of a product equals the product of two convolutions.

But for finite $n$ the convolution of a product is not in general equal to the product of convolutions. However for large $n$ we expect this to be approximately true. My question: How can one go about estimating the error in the following equation:

$\int f(x')g(x')K_n(x-x') dx' = \int f(x')K_n(x-x') dx' \int g(x')K_n(x-x') dx' + \epsilon_n(x)$.

I have faced this question in three different imaging scenarios this past year and I have found no treatment of the problem in the literature. It could be that I just don't know where or how to look.

Any help would be much appreciated,

DJS

D_J_S
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  • Or perhaps the approximation is better when written as this: $∫f(x′)g(x′)K_n(x−x′)dx′=∫f(x′)K^{1/2}_n(x−x′)dx′∫g(x′)K^{1/2}_n(x−x′)dx′+ϵ_n(x)$. I don't know. – D_J_S Feb 11 '14 at 20:49
  • Talking to myself here but ... The approximation with the 1/2 power kernel doesn't appear to make much sense if $f(x)=1$. So maybe my original question is best. – D_J_S Feb 11 '14 at 21:04

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