the fourier transform of a “double convolution”

Suppose I have a function

$$m(x) = f(x)\int_{-\infty}^{\infty} h(w)g(w-x)dw = f(x)h*g(x)$$

I want to find the Fourier transform of m(x) in terms of the Fourier transforms of $f,h,g$ but for the life of me I can't figure out a way to do it. If anyone has some insight, please help?

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2 Answers

In general, we have the product-to-convolution formulas: \begin{align*} \mathcal{F}[f\cdot g]=\hat{f}*\hat{g}\\ \mathcal{F}[f* g]=\hat{f}\cdot\hat{g} \end{align*} In your case, we have

$$\mathcal{F}[f\cdot(h*g)]=\hat{f}*\mathcal{F}[h *g]=\hat{f}*(\hat{h}\cdot\hat{g})$$ So essentially it swaps the convolution and the product.

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alright thanks, this is precisely what i needed - cheers –  user61038 Feb 5 '13 at 5:37

Let $q(x) = h(x) * g(x)$. You want the FT of $f(x)q(x)$. This is $\hat{m}(k) = \hat{h}(k) * \hat{q}(k)$. But you know that $\hat{q}(k) = \hat{g}(k) \hat{h}(k)$, so that $\hat{m}(k) = \hat{f}(k) * [\hat{g}(k) \hat{h}(k)]$.

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