By definition of $e^{it\partial_x^2}, \partial_x^2 u \in L^2({\mathbb{T}}) $..

Question

My proffesor in class said that "By definition of $e^{it\partial_x^2}, \partial_x^2 u \in L^2({\mathbb{T}}) $ if we initially have that $u \in L^2({\mathbb{T}}) $" where $\mathbb{T}$ is the 1-dim torus and $e^{it\partial_x^2}$ is the infinitesimal generator of $i\partial_x^2$. Could someone please help me showing me how come is that true.

Answer 1

\begin{align*} |\left<\partial_{x}^{2}u,\varphi\right>|&=|\left<u,\partial_{x}^{2}\varphi\right>|\\ &\leq\int|u||\partial_{x}^{2}\varphi|dx\\ &\leq\|u\|_{L^{2}}\|\partial_{x}^{2}\varphi\|_{L^{2}}\\ &<\infty \end{align*} for all $\varphi\in C^{\infty}({\bf{T}})$, and all such functions are $L^{2}$ dense, so $\partial_{x}^{2}u$ induces a bounded linear functional on $L^{2}$, by Riesz Representation Theorem, there is a corresponding $L^{2}$ function realizing the bounded linear functional, which we again denote by $\partial_{x}^{2}u$.

So the semigroup is defined as \begin{align*} t\rightarrow e^{it\partial_{x}^{2}} \end{align*} in such a way that \begin{align*} e^{it\partial_{x}^{2}}:u\rightarrow e^{it\partial_{x}^{2}u}, \end{align*} I think here we need $u$ to be twice differentiable. As a result, \begin{align*} \|e^{it\partial_{x}^{2}u}\|_{L^{2}({\bf{T}})}=1<\infty \end{align*} for real-valued $u$ because of that $|e^{it\partial_{x}^{2}u}|=1$.

For real-valued $u\in L^{2}({\bf{T}})$, one can correspond an $L^{2}({\bf{T}})$ function $\partial_{x}^{2}u$ to it, and then we still have $|e^{it\partial_{x}^{2}u}|=1$. Note that this correspondent $\partial_{x}^{2}u$ is real-valued if we are dealing with real Hilbert space $L^{2}({\bf{T}})$.