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I am wondering how to compute the functional derivative of an ordinary scalar function with itself, say

$$[f\star f](x)=\int dy~f(y)f(x-y).$$

My attempt would give

$$\delta_{f}[f\star f](x)=2f(x-y).$$

If this is correct, then the functional derivative of the $n$-fold convolution should give $$\delta_{f}[f\star...\star f](x)=nf^{\star(n-1)}(x-y).$$ But I am not sure, as I had guessed the argument of the last expression to involve more variables than just $2$. Has anybody come across such variations?

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    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$ – Qmechanic May 22 '17 at 18:17
  • $\begingroup$ Well, imagine we deal with an action functional with some interaction potential involving a convolution, then this is certainly relevant to physics. $\endgroup$ – Hamurabi May 22 '17 at 18:36
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    $\begingroup$ Yes, it is correct. Computing the wanted functional derivative at $z$ is equivalent to computing the standard $a$-derivative for $a=0$ of your functional where each $f$ is replaced with $f(x) + a \delta(x-z)$. Since $\delta$ acts as the identity with respect to the convolution, you easily get the results you wrote. $\endgroup$ – Valter Moretti May 22 '17 at 18:48

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