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I try to compute the convolution root of a symmetric, positive definite, nonnegative, one dimensional function $f: \mathbb R\to \mathbb R^+_0$. Furthermore I assume $f$ is bounded and $\int_{\mathbb R} f(x) d x<\infty$.

There is a function $g:\mathbb R\to \mathbb R_0^+$ satisfying

$$g*g=f,$$

where $*$ is the convolution by taking the Fourier transform ($\int_{\mathbb R} e^{-i\xi x} (\cdot) d x$):

$$\mathcal F [g] ^2 = \mathcal F [f].$$

My problem is the following: If i take the square root, i do not know which branch of the square root to take in order to guarantee that $g$ is positive again.

Here are two examples: if $f=\frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}$, then

$$\mathcal F[f](\xi)= e^{- \frac{\xi^2}{2}}.$$

Taking the positive square root gives again a positive definite function and therefore the inverse Fourier transform is positive again. However if $f=tri(x)$ is the triangular function (https://en.wikipedia.org/wiki/Triangular_function) then

$$\mathcal F[f](\xi)= \bigg( \frac{2sin(x/2)}{x}\bigg)^2.$$

Here i have a problem, since I can't take only the positive square root, because $\big| \frac{2sin(x/2)}{x}\big|$ is not positive definite and therefore $g$ is not positive again.

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  • $\begingroup$ the convolution root of a triangular function $f$ is a box function $g$, which is positive. Of course, the Fourier transform $G$ of a box function is not positive, but that is OK, only $g$ needs to be positive. $\endgroup$ – Carlo Beenakker Apr 25 '18 at 10:29
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    $\begingroup$ @CarloBeenakker My understanding is that the OP knows that perfectly well. What he is really asking is how to recover $g$ from $f$ using the Fourier transform (or, perhaps, something else) if it is not known but known to exist. $\endgroup$ – fedja Apr 25 '18 at 11:37

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