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What methods do there exist to solve equations of the type:

$$\int_{-\infty}^{\infty} f(f(\tau)) f(t-\tau)d\tau = g(t)$$

That is: to try and find a function that convolved with itself applied twice becomes a given function $g(t)$

Jean Marie
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mathreadler
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  • Based on a rather long experience in different forms of convolution, it is almost certain that there are no solution, even for very simple $g$. Could you say us why you are interested in this problem ? – Jean Marie Jun 13 '17 at 12:01
  • @JeanMarie: Just curious. I have not found any place to apply it yet. – mathreadler Jun 13 '17 at 15:13

1 Answers1

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Here are some simple examples of non-uniqueness:

If $g\equiv 0$, there are too many solutions. Let $f:\mathbb R\to (-\infty,0]$ satisfy $f(x)=0$ for $x\le 0$, with $f(x)\le 0$ arbitrary for $x>0$. Then $f\circ f\equiv 0$. (Formally, $f$ could even be non-measurable!)

If $g\equiv const.$ is nonzero, then there are still many solutions. If $g>0$, let $f(x)=ab\chi_{[0,a]}(x)$ for some $a>0$ and $b\in(0,1)$. Then $f\circ f\equiv ab$, and $(f\circ f)\ast f=ab\int f=a^2b$, so put $a=\sqrt{g/b}$.

user254433
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  • I don't agree: $f○f$ is $ab\chi_{[0,a]}$, and the convolution gives a triangle function, not a constant. – Jean Marie Jun 13 '17 at 18:46
  • Sorry, here's my answer. I thought that $f\circ f=ab$ since $f(\mathbb R)\subset [0,a]$. Could you explain where your claim $f\circ f=f$ comes from? – user254433 Jun 13 '17 at 19:54
  • By the way, $\chi$ indicates the closed interval $[0,a]$, not the open interval $(0,a)$. I agree that if $g=ab\chi_{(0,a)}$, then $g\circ g=g$. – user254433 Jun 13 '17 at 20:05
  • Now I understand, You have found the precise reason of my misunderstanding ! Thank you very much. – Jean Marie Jun 13 '17 at 20:07