# Positivity of the Fourier transform: prove or disprove that $\operatorname{Re}(\overline{\widehat{u}}(\xi) \widehat{F\circ u}(\xi))\geq0$

Let $$F:[0,\infty) \to[0,\infty)$$ be increasing, $$C^1$$ and $$L-$$Lipschitz with $$F(0)=0$$. Let $$u\in L^1 (\Bbb R^d)$$, $$u\geq0$$ so that $$F\circ u\in L^1 (\Bbb R^d)$$

I would like to prove (or disprove) that \begin{align*} \operatorname{Re}\big(\overline{\widehat{u}}(\xi) \widehat{F\circ u}(\xi)\big)\geq0\qquad\text{for all \xi\in \Bbb R^d} \end{align*} Note that for $$\tilde{u}(x)=u(-x)$$, \begin{align*} \operatorname{Re}\big(\overline{\widehat{u}}(\xi) \widehat{F\circ u}(\xi)\big)= \operatorname{Re}\big( \widehat{\tilde{u}* F\circ u}(\xi)\big)= \widehat{\tilde{u}* F\circ u}(\xi)+ \widehat{F\circ \tilde{u}* u}(\xi). \end{align*}

PS. This problem is suggested by the decay of generalized porous medium equation $$\partial_t u= \Delta F\circ u.$$ In the Fourier variables we expect the time decay $$|\widehat{u(t)}|^2\leq c |\widehat{u(0)}|^2$$. The latter is true if we have $$\partial_t |\widehat{u}(\xi)|^2= -2|\xi|^2\operatorname{Re}\big(\overline{\widehat{u}}(\xi) \widehat{F\circ u}(\xi)\big)\leq 0.$$

My main problem can be resolved if one shows that $$|\widehat{u(t)}|^2\leq c |\widehat{u(0)}|^2,\, c>0$$.

Any idea, partial answer or reference it warmly appreciated.

• for f(x)= x it works. This problem is suggested by the decay of generalized porous medium equation $\partial_t u= \Delta F\circ u$ is the Fourier variables we expect the time decay $|\widehat{u(t)}|^2\leq c |\widehat{u(0)}|^2$. The latter is true if we have $\partial_t |\widehat{u}(\xi)|^2= -|\xi|^2\operatorname{Re}\big(\overline{\widehat{u}}(\xi) \widehat{F\circ u}(\xi)\big)\leq 0$ Sep 21 at 9:28
• In the main part you are certainly asking too much. Take $d=1$ and $u(x)=e^{-|x|}$, so $\hat u>0$. Then $F\circ u$ is essentially any even decreasing on $[0,\infty)$ $L^1$-function (in the sense that you can approximate any such function in $L^1$ by a function of your kind with arbitrarily small error). However, among those functions there are plenty with sign-changing Fourier transform. Sep 27 at 21:31
• I don't see any reason why you would have the decay. Note that here, $|\widehat{u(t)}|^2\le c|\widehat{u(0}|^2$ doesn't imply decay, as $c$ may be greater than $1$. It just implies a kind of stability for the Fourier transform Sep 29 at 9:53