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Given a function $f(x)$ and its Fourier transform $\tilde f(k)$, I know from the convolution theorem that the Fourier transform of $g(x):=\sqrt{f(x)}$ fulfils $$\tilde g(k)\ast\tilde g(k) = \tilde f(k)$$ up to some normalization factors. But now I need to find the inverse of auto-convolution. Does this help somehow or is it rather the other way around that the "convolution-squareroot" is obtained via the Fourier transform?

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  • $\begingroup$ Although your statement is hard to read, I believe your conclusion is correct. Take Fourier transform, square root of transform, and take inverse transform. $\endgroup$ – herb steinberg Mar 31 '18 at 21:27

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