Fourier transform and fourth root

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.

• 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. Oct 22, 2018 at 13:55