Let us consider the linear Schrödinger equation in $\mathbb{R}^N$
$$ (i\partial _t+\Delta)\,u=0\mbox{ ,}\quad u(0,x)=f$$
with $f\in L^2(\mathbb{R}^N)$, and let $u(t,x)=e^{it\Delta}f$ be its solution.
We have that
$$\frac{1}{T}\int_0^Te^{it\Delta}fdt\rightarrow 0\quad\mbox{in }L^2(\mathbb{R}^N)$$
The proof I have is "geometric", based on the following idea:
The operator (once one proved it exists)
$$P:\,f\mapsto\lim_{t\rightarrow +\infty}\frac{1}{T}\int_0^Te^{it\Delta}fdt$$
satisfies $P\circ e^{it\Delta}=P$ and $e^{it\Delta}\circ P=P\quad \forall t\in\mathbb{R}.$
This implies $\mathrm{Im}\, P = \mathrm{Fix}\{e^{it\Delta}\}_{t\geq 0} = \mathrm{Ker}\, i\Delta=\{0\}$, that is $P=0$.
I'm looking for a more analytical proof, based on the decay estimates for ${e^{it\Delta}}$. Infact I would try to understand under which hypotesis I can extend the result to the nonlinear case, where a geometric approach seems to be unavailable. Does anyone has some ideas?
