Fourier transform of $\sqrt{f(t)}$

Question

If the Fourier transform of $f(t)$ is $F(f)$, can you conclude that the Fourier transform of $\sqrt{f(t)}$ is $\sqrt{F(f)}$?

Probably this is not always the case, but what are the cases in which this is true?

Answer 1

Counterexample: Let $f=\chi_{[-1,1]}.$ Then $F(f)(x)=2(\sin x)/x$. But here $\sqrt f =f.$ If the result held, you'd have

$$\sqrt {2(\sin x)/x} = 2(\sin x)/x$$

for all $x,$ contradiction.

Answer 2

Interesting question. The first thing we can remark is that if it is true then $f$ and $\hat f$ are nonnegative.

Here is already a partial answer.

Assuming that $f$ is integrable (I.e. $\sqrt f$ is in $L^2$), we also get by Plancherel, $f(0) = \int \hat f = \int f = \hat f(0)$, and $\hat f(x) \leq \int f = \hat f(0)$, so the maximum of $\hat f$ is $\hat f(0)$ and the same for $f$. Without loss of generality, we can assume $f(0) =1$.

Let $g = \sqrt f$. Remark that the maximum of $g$ is also $g(0) = \sqrt f(0) =1$ and the same for $\hat g$, so that $0\leq \hat g\leq 1$ and $\int g =1$. Therefore $g-g^2$ is a nonnegative function verifying $$ \int g^2 - g = 1- 1 = 0 $$ So that $g^2=g$. Therefore, $g = \mathbb{1}_A$ is the indicator function of some set $A$. And the same result holds for $\hat g$. But since $g$ is integrable, $\hat g$ should be continuous and go to $0$ at infinity. This is impossible.

Therefore, $f$ cannot be an integrable function, and cannot be continuous near $0$.

• If $f$ is not locally integrable, then it is a more general tempered distribution for which $\sqrt f$ does not have a clear meaning.

• It remains to check the cases when $f\in L^1_{\mathrm{loc}}$ and $\hat f$ also, but none of them is in $L^1$.