# Does convergence of tempered distributions implies convergence in $\mathcal{S}(\mathbb{R}^4,\mathbb{R})/\mathcal{S}_{0}$?

We can define the following symmetric semi-definite positive bi-linear form on $$\mathcal{S}(\mathbb{R}^{4},\mathbb{R})$$ with values in $$\mathbb{C}$$, $$$$\label{prodintespaciales} (h_{1},h_{2})_{\mathcal{S}}=2i\int_{\mathbb{R}^{4}\times\mathbb{R}^{4}} h_{1}(x)\Delta^{(+)}(x-y)h_{2}(y)dx dy,$$$$ where $$\Delta^{(+)}(x)=\frac{-i}{(2\pi)^{3}}\int e^{-ipx} \delta(p^2-m^2)\theta(p^0) d^4p$$ and $$p\cdot x=p^0x^0-\vec{p}\cdot\vec{x}$$. This can be alternatively written as \begin{align}\label{prodintmomento} (h_{1},h_{2})_{\mathcal{S}} & =2\int_{\mathbb{R}^{4}} \mathscr{F}(h_{1})^{\ast}(p)\mathscr{F}(h_{2})(p) \delta(p^2-m^2)\theta(p^0) d^4p \\ & =\int_{\mathbb{R}^{3}} \mathscr{F}(h_{1})^{\ast}(\omega(\vec{p}),\vec{p})\mathscr{F}(h_{2})(\omega(\vec{p}),\vec{p}) \frac{d^3p}{\omega(\vec{p})}, \end{align} where $$\omega(\vec{p})=\sqrt{m^2+\vec{p}^2}=:\omega_p$$. There is a set of functions such that $$(h,h)_\mathcal{S}=0$$, namely those with Fourier transform that vanishes on the hyperboloid $$H_m:=\left\{p^2=m^2 / p^{0}>0\right\}$$. We denote this set by $$\mathcal{S}_{0}$$. In order to obtain a Hilbert space we quotient by $$\mathcal{S}_0$$, $$\mathcal{S}(\mathbb{R}^4,\mathbb{R})/\mathcal{S}_{0}$$, and complete using $$(,)_{\mathcal{S}}$$.

Now suppose I have a sequence of Schwarts functions $$\{g_{n}\}_{n\in \mathbb{N}}$$ such that $$g_{n}\to g$$ in the sense of tempered distributions. Does it imply convergence of $$\{\bar{g}_{n}\}_{n\in \mathbb{N}}$$ in $$\mathcal{S}(\mathbb{R}^4,\mathbb{R})/\mathcal{S}_{0}$$ with te topology induced by the inner product? if the answer is yes, what is the limit?, and if the answer is not, is there any additional condition I can put over the sequence $$\{g_{n}\}_{n\in \mathbb{N}}$$ to conclude convergence for $$\{\bar{g}_{n}\}_{n\in \mathbb{N}}$$?