For clarity I am using the following definition of Schwartz space $\mathscr{S}$
Let $\mathscr{S}$ be the set of functions f(x) s.t. the family of norms $$||f(x)||_{r,s}=\sup\left|x^r\frac{d^s}{dx^s}f(x)\right|<\infty \quad \forall r,s\in \mathbb{Z}_+$$
So we let $f(x)\in \mathscr{S}$ and set out to show that $f_\epsilon (x)$ is a member of this space, namely
$$f_\epsilon(x):=e^{-\epsilon x^2}f(x)\in\mathscr{S}, \quad \epsilon \geq0.$$
In addition (moreover a subsequent question) I would like to show that $f_\epsilon(x)$ converges in $\mathscr{S}$. Thus I need to show that the limit of the norms tends to zero as $\epsilon \rightarrow 0^+$, i.e. $$\lim_{\epsilon \to 0^+}||f-f_\epsilon||_{r,s}=0$$
for all of the indices.
So far I have tried using the Leibniz rule to show that $f_\epsilon$ belongs to Schwartz space but was not able to bound the norm. Is this the usual way to show a product of functions belongs to $\mathscr{S}$?
As for the second problem of showing convergence, this is an area where I am quite weak in analysis and am struggling to get off the ground. Could someone provide the guiding path to get started with this question?
Also, I'd be interested to hear how the topology induced by the norm here shows S is complete. Will this notion be helpful for proving the convergence?
It is not hard to see that, for any $r,s\in\mathbb Z^+$, there exist polynomial $p_1,\ldots,p_s$, such that $$ x^r \frac{d^s}{dx^s}\left(\mathrm{e}^{-\varepsilon x^2}f(x)\right)=x^r\left(\,f^{(s)}(x)+\varepsilon\sum_{k=1}^{s} p_k(x,\varepsilon)\,f^{(s-k)}(x)\right)\,\mathrm{e}^{-\varepsilon x^2}. $$ Thus $$ x^r \frac{d^s}{dx^s}\left(\mathrm{e}^{-\varepsilon x^2}f(x)\right)\in{\mathscr S}(\mathbb R), $$ for any fixed $\varepsilon>0$.
Next $$ \|\mathrm{e}^{-\varepsilon x^2}f(x)-f(x)\|_{r,s}=\sup_{x\in\mathbb R} \left|\,\varepsilon x^r\sum_{k=1}^{s} p_k(x,\varepsilon)\,f^{(s-k)}(x)\,\mathrm{e}^{-\varepsilon x^2}\,\right|\le \varepsilon \sup_{x\in\mathbb R} \left|\,x^r\sum_{k=1}^{s} p_k(x,\varepsilon)\,f^{(s-k)}(x)\right|, $$ and clearly the right hand side of the above tends to zero, as $\varepsilon\to 0$.