Is $F^{-1}((F\psi)^\lambda)$ a P.D.F for all $\lambda$?

Question

Suppose $\psi$ has the properties $\psi \geq 0 $ and $\int_S\psi \ dS = 1$

How can I show (or disprove) that

$$F^{-1}((F\psi)^\lambda)$$ has the same properties for all $\lambda$?

$F$ denotes fourier transform. If $\lambda$ is a positive integer it's easy to show as the expression becomes repeated convolution (by the convolution theorem). Otherwise I'm stuck.

Answer 1

$$||F^{-1}(F\psi)^\lambda||_1 = \int_\mathbb{R}\int_\mathbb{R}F\psi^\lambda(y)\ e^{2i\pi uy}dydu $$ $$= \int_\mathbb{R}F\psi^\lambda(y)\int_\mathbb{R}\ e^{2i\pi uy}dudy $$ $$= \int_\mathbb{R}F\psi^\lambda(y)\delta(y)dy $$ $$= F\psi^\lambda(0) $$ $$= (\int_\mathbb{R}\psi(x) e^{0}dx)^\lambda = 1^\lambda = 1$$

Still missing the positiveness property though