Let be $u(x,t) = e^{it\partial_x^2}u_0$ be the linear solution of \begin{align*} i\partial_t u + \partial_x^2 u &= 0\\ u(x,0) &= u_0 \end{align*} so i would like to know if we have $u_0 \in L^2(\mathbb{T})$, $\varepsilon >0$ and if we define $u_\varepsilon = \dfrac{e^{i\varepsilon \partial_x^2}u_0 - u_0}{\varepsilon}$. It is $u_\varepsilon $ bounded uniformly on $H^{-2}(\mathbb{T})$ in $\varepsilon \to 0$ .
$\mathbb{T}$ is the one dimension torus. Hope u can help me.
Thank you in advance
As the problem is in the torus one way is to use a Fourier decomposition.
With: $$u_0=\sum_{k \in \Bbb Z} a_k e^{ikx}$$
$$\sum_k |a_k|^2=\|u_0\|_{L^2}^2 < \infty$$
you have the explicit solution: $$ e^{it\partial_x^2}u_0=\sum_{k \in \Bbb Z} a_k e^{-ik^2 t} e^{ikx}$$ so your question can be rewritten as:
And using the fact that $\left|\frac{e^{ix}-1}{x} \right| \leq 1$ for all $x \in \Bbb R$: $$\sum_{k \in \Bbb Z}\frac{1}{k^4} \left|a_k \frac{e^{-ik^2 \epsilon}-1}{\epsilon} \right|^2=\sum_{k \in \Bbb Z}\left|a_k \right|^2 \left| \frac{e^{-ik^2 \epsilon}-1}{\epsilon k^2} \right|^2 \leq \sum_{k \in \Bbb Z}\left|a_k \right|^2< + \infty$$ so it is indeed bounded in $H^{-2}$ uniformly with respect to $\epsilon$.