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If you have an equation of the form

$$ a(x) \star f(x) + b(x)f(x)=0 $$

Where $\star$ is convolution over the real line and $a,b$ are given functions where you want to solve for the set of possible $f$.

Is there some way to express $f$ as an operator-equation in terms of $a,b$? The standard intuition of taking a fourier transform doesn't work here since it reduces one convolution to multiplication and transforms the other multiplication into a convolution.

This feels like a fundamental operation that should exist and among other things a discrete variant of it would help resolve this question.

Superficially it also appears to be related to the pauli problem

jd27
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Sidharth Ghoshal
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  • Thanks for your question! – mathoverflowUser Nov 13 '23 at 17:57
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    $f = 0$ is always a solution. If you are looking for non-trivial solutions, $f$ is never unique since multiplying $f$ by a constant gives another solution. – Tzimmo Nov 13 '23 at 18:02
  • Ah yes, given this is a homogenous linear equation that makes sense. I am interested in the set of non trivial solutions and/or generating just one non trivial solution – Sidharth Ghoshal Nov 13 '23 at 18:04
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    "The standard intuition of taking a fourier transform doesn't work here since it reduces one convolution to multiplication and transforms the other multiplication into a convolution." Just throwing out odd ideas, but would a (1/2) Fractional Fourier Transform get both A-F and B-F terms into the same kind of half-convolved-half-multiplied situation? – DotCounter Nov 13 '23 at 18:17
  • Hmm that is a very interesting idea! – Sidharth Ghoshal Nov 13 '23 at 18:18
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    You can compare with https://math.stackexchange.com/questions/4768288/is-there-a-function-whose-autoconvolution-is-its-square-g2x-gg-x – dsh Nov 13 '23 at 21:43

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