Let $\phi:\Bbb R^d\setminus\{0\}\to [0,\infty)$ be a continuous and symmetric, i.e., $\phi(-\xi)=\phi(\xi)$. Let $F:\Bbb R\to[0,\infty)$ be increasing and $L-$Lipschitz with $F(0)=0$. Consider the space $ H^\phi(\Bbb R^d)=\{u\in L^2(\Bbb R^d)\,:\, \langle u,u\rangle_\phi<\infty\}$ where we define \begin{align*} \langle u,v\rangle_\phi= \int_{\Bbb R^d}\widehat{u}(\xi), \overline{\widehat{v}(\xi)}\phi(\xi)d \xi. \end{align*} I would like to show (or disprove) that for $u\in H^\phi(\Bbb R^d)$ we have \begin{align*} 0\leq \langle F\circ u, u\rangle_\phi\leq L\langle u, u\rangle_\phi. \end{align*}
The intuition behind this inequality is that it is true in the particular case $\phi(\xi)=|\xi|^{2s}, s\in (0,1)$. Indeed, $H^\phi(\Bbb R^d)= H^{s}(\Bbb R^d)$ and it can be shown that \begin{align*} \langle u,v\rangle_\phi= \int_{\Bbb R^d}\widehat{u}(\xi), \overline{\widehat{v}(\xi)}\phi(\xi)d \xi= c_s\iint_{\Bbb R^d\Bbb R^d} (u(x)-u(y))(v(x)-v(y))|x-y|^{-d-2s}d yd x. \end{align*} Since $F$ is increasing and Lipschitz we have \begin{align*} 0\leq (u(x)-u(y))(F\circ u(x)-F\circ u(y))\leq L (u(x)-u(y))^2. \end{align*} From this, it follows that \begin{align*} \langle F\circ u, u\rangle_\phi\leq L\langle u, u\rangle_\phi. \end{align*} The inequality can also be easily shown when $\phi=1$ and $\phi(\xi)=|\xi|^2$.
Any comment or reference regarding this type of inequality is welcome.
