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Afer some experiments, it seems to me that:

the integral (in the entire domain) of a convolution $(f*s)(x)$, of a filter function $f(x)$ with a a signal function $s(x)$, multiplied by a weight function $w(x)$,

$$ \int (f*s)(x) \cdot w(x) \, dx $$

, when all functions are integrable,

is equal to the integral of the signal function $s(x)$ weighted by the cross-correlation $(f\star w)(x)$ of the filter function $f(x)$ with the weight function $w(x)$,

$$ \int (f\star w)(x) \cdot s(x) \, dx $$

Is that right?

If it is, what is the proof?

If it is not, is there something similar that let one merge the filter and the weight functions to create a new weight function?

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If I am not mistaken, this is just changing the order of integration (which has to be justified by integrability conditions): \begin{align} \int (f*s)(x)w(x) dx & = \int\int f(y-x)s(y)dy\, w(x)dx\\ & = \int\int f(y-x)w(x)dx\, s(y)dy\\ & = \int (f\star w)(y)s(y)dy. \end{align}

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  • $\begingroup$ Thanks! I was almost reaching that proof, but your answer has a much cleaner formulation. $\endgroup$ – Cristóvão D. Sousa Oct 29 at 15:21

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