I have a confusion. I have read that $\Delta$ operator on $\mathcal{S}$ is not closed but if $\Delta:D(\Delta)\to L^2$ with $D(\Delta)=\left\{\varphi\in L^2:\Delta\varphi\in L^2\right\}$ is closed?
My attempt: Let $\left\{\varphi_k\right\}\subset D(\Delta)$ with $\varphi_k\to \varphi$ in $L^2$ and $\Delta\varphi_k\to \psi$ in $L^2$. I want proves that $\varphi\in D(\Delta)$ and $\Delta \varphi=\psi$.
$\int |\varphi|^2\leq \int |\varphi-\varphi_k|^2+\int |\varphi_k|^2$ then $\int |\varphi|^2\leq \lim_k \int |\varphi_k|^2$ because $\varphi_k\to \varphi$ in $L^2$ but I don't know how to ensure that $\lim_k \int |\varphi_k|^2<\infty.$
Also I do not know if it is fulfilled (formally) that $$\Delta \varphi=\Delta(\lim \varphi_k)=\lim \Delta \varphi_k=\psi$$ with $L^2$ convergence.
Integration by parts.
Verify that your definition in terms of Fourier transforms is equivalent to saying that $\Delta \phi = \psi$ if and only if $\int \psi f = \int \phi \Delta f$ for every $f \in C^\infty_c$. So $$\int \psi f = \lim \int (\Delta \varphi_k) f = \lim \int \varphi_k \Delta f = \int \varphi \Delta f$$ for every $f \in C^\infty_c$. Thus $\psi = \Delta \varphi$ and in particular $\varphi \in D(\Delta)$.