Space of rapidly decreasing functions Let H_n(x) be the Hermite Polynomials defined as in
http://en.wikipedia.org/wiki/Hermite_polynomials
The Hermite Polynomials form an orthonormal basis of the space of the rapidly decreasing
functions.
Define the function
$f(t,x):=\sum _{n=0}^{\infty }c_{{n}}{{\rm e}^{-nt}}H_{n}(x)$
with $c_{{n}} = (g,H_{n})$ which denotes the scalar product. 
I'm able to show that f converges uniformly in $L^2$. 
I'm trying to show that f is in the space of the arbitrarily often differentiable functions and that for a fixed t, say $t_0$, $f(t_0,x)$ is in the space of the rapidly decreasing functions for all $t_0 > 0$. 
Any ideas? I'm not too familiar with functional-analysis. 
 A: I have the feeling this question follows me whereever I go ;-)
Just this morning I found a proof, which is essentially the same that was hinted by Scott Carnahan. I you understand german (or want to guess the proof just by the formulas) you can find it here at www.matheplanet.com. Maybe that question there is even your own or from someone who works on the same PDE-problem?
A: It's (still) not completely clear what space you are working in since you haven't clarified the confusions from Yemon Choi and fedja's comments.  But it may help to note an analogous problem that is well-defined and where there is a clear strategy.  That strategy may adapt to your situation (whatever that is).
This analogous situation is of periodic functions using the usual basis of periodic exponentials, $e^{2\pi i n x}$.  The key here is:

$g$ is $C^\infty$ if and only if the Fourier coefficients $c_n(g) = \int_0^1 g(x) e^{-2\pi i n x} d x$ are rapidly decreasing.

Then for $g$ in $L^2$, $c_n(g)$ is square summable and so $c_n(g) e^{-n t}$ for $t > 0$ is rapidly decreasing (that is, $\lim_{n \to \infty} n^k c_n(g) e^{-n t} = 0$ for each $k$) since exponentials beat polynomials.
So I would try to find out a characterisation of the Schwartz space of rapidly decreasing functions in terms of the coefficients of their expansions with respect to the Hermite polynomials (suitably weighted).  I would expect such a characterisation to be well-known, but I don't know it off the top of my head.
A: My initial interpretation is that your inner product comes with a Gaussian weight.  If $g=H\_0$, then $c_n = \delta\_{n,0}$, and for any fixed $t\_0$, the sum is a nonzero constant function, which is smooth, but does not decrease rapidly.
Alternatively, I might assume you meant to multiply each Hermite polynomial by a suitable Gaussian factor, i.e., we replace each $H\_n(x)$ in your question with the Hermite function $\psi\_n(x)$.  Assuming this interpretation holds, and that $g$ is Schwartz, you get the result you wanted, because the weighted coefficients $c\_n e^{-nt\_0}$ decay exponentially in $n$, while the derivatives of $\psi\_n(x)$ are dominated by polynomials in $n$.
A: $$
f(t,x)=e^{t/2}\sum_{n\ge 0} e^{-(n+\frac12)t}\langle{g},{H_{n}}\rangle H_{n}(x)
$$
The function $F(t,x)=e^{-t/2}f(t,x)$ is the solution 
of
$$
\frac{\partial F}{\partial t}+\bigl(-\frac{d^2}{dx^2}+\frac{x^2}4\bigr) F=0,\quad
F(0,x)=g(x).
$$
since the projection on the eigenvector $H_{n}$ of the harmonic oscillator
$-\frac{d^2}{dx^2}+\frac{x^2}4$
gives
$$
\frac{dF_{n}}{d t}+(n+\frac12)F_{n}=0,\quad
F_{n}(0)=\langle{g},{H_{n}}\rangle.
$$
As a result, with $\mathcal H=-\frac{d^2}{dx^2}+\frac{x^2}4$
$$
e^{-t/2}f(t,x)=F(t,x)=(e^{-t\mathcal H} g)(x).
$$
This implies in particular that for all $N\ge 0$ and $t>0$
$$
\mathcal H^N F=t^{-N}e^{-t\mathcal H} (t\mathcal H)^Ng\Longrightarrow
\Vert\mathcal H^N F(t,\cdot)\Vert_{L^2(\mathbf R_{x})}\le t^{-N}N!
\Vert g\Vert_{L^2(\mathbf R_{x})}
$$
and $f(t,\cdot)$ in the Schwartz class for any $t>0$, provided the initial datum $g$ is in $L^2$.
