Consider the Sobolev space defined by Fourier transform $H^s(\mathbb{R}^n):=\{u \in \mathcal{S}': \int_{\mathbb{R}^n} (1+|\xi|^2)^{s} |\hat{u}(\xi)|^2 d\xi <+\infty \}.$ Here $\mathcal{S}'$ means tempered distribution. Now consider a sequence of distribution $u_n \in H^{s+2}$ satisfying that $u_n \rightarrow u$ in $H^s(\mathbb{R}^n)$ and $\nabla u_n \rightarrow \nabla u$ in $H^s(\mathbb{R}^n),$ and $\Delta u_n \rightarrow f$ in $H^s(\mathbb{R}^n),$ can we deduce that $u \in H^{s+2}(\mathbb{R}^n)$ and $\Delta u=f$?
If $s$ is an integer, the Sobolev space coincides with the traditional Sobolev space defined by the weak derivative, then it is obvious that the above statement is true via testing smooth compactly supported functions.
My attempt is, since $\widehat{\Delta u_n}=-|\xi|^2\hat{u}_n(\xi),$ then we have $$ \int_{\mathbb{R}^n} (1+|\xi|^2)^{s} |-|\xi|^2\hat{u_n}(\xi)-\hat{f}( \xi)|^2 d\xi \rightarrow 0. $$ We also know that $$ \int_{\mathbb{R}^n} (1+|\xi|^2)^{s} |\hat{u_n}(\xi)-\hat{u}( \xi)|^2 d\xi \rightarrow 0, \qquad \int_{\mathbb{R}^n} (1+|\xi|^2)^{s} |\xi|^2||\hat{u_n}(\xi)-\hat{u}( \xi)|^2 d\xi \rightarrow 0. $$ Then how can I deduce that $\hat{f}=-|\xi|^2\hat{u}(\xi)$?