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Given a well-behaved convex function $f(t):\mathbb{R}\to \mathbb{R}$, its Fourier transform (FT) $\hat{f}(\omega)=\mathcal{F}[f(t)](\omega)$ is positive (and decreasing) proof here.

It follows that the fourth square root of the FT $\sqrt[4]{\hat{f}(\omega)}$ is invertible.

I was wondering if it is possible to determine the function $g(t)=\mathcal{F}^{-1}\Big[\sqrt[4]{\hat{f}(\omega)}\Big](t)$ directly from $f(t)$, without making use of the Fourier transform.

In other words, knowing $f(t)$, I would like to determine the function $g(t)$ such that $$\mathcal{F}[g(t)](\omega)=\sqrt[4]{\hat{f}(\omega)}.$$

I had a look at the convolution theorem and at the fractional Fourier transform but I do not see a straightforward application of these tools to solve this problem.

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    $\begingroup$ You should specify that you are considering the cosine transform $\int_0^\infty f(t)\cos(\omega t)\, dt$, as convex functions cannot be integrable on the full line $(-\infty, \infty)$, which immediately rings an alarm when one reads your first sentence. This said, I am afraid that there is not a satisfactory answer to your question. The square root of the Fourier/cosine transform is a nonlinear operation, even a non-analytic one, and these are known to pose substantial difficulties. $\endgroup$ – Giuseppe Negro Oct 22 '18 at 13:55

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