Define $\psi_n(x) = c_n H_n(x) e^{-x^2/2}$ as in http://en.wikipedia.org/wiki/Hermite_polynomials . Also define the differential operator $H u = - u'' + x^2 u$. Then the $\psi_n$ form an othonormal basis of $L^2$ and $H \psi_n = (2n + 1) \psi_n$.
Warning: As coudy points out below: one needs $\|H f\| < \infty$ and not just $\langle f, Hf\rangle < \infty$. So the computations below need to be changed.
Rest of original post
Given now $f$ such that $$ A = \langle f,Hf \rangle =\int \overline{f(x)} (Hf)(x) dx $$ is finite. Then by writing $f(X) = \sum_{n \geq 0} f_n \psi_n(X)$, we obtain $$ A = \langle f,Hf \rangle =\langle f, \sum_{n \geq 0} f_n H\psi_n(X) \rangle = \langle \sum_{n \geq 0} f_n \psi_n(X) , \sum_{n \geq 0} f_n (2 n + 1)\psi_n(X) \rangle $$ Now using orthonormality of the $psi_n$, we conclude that $$ A = \sum_{n \geq 0} |f_n|^2 (2n + 1). $$ Now using that the $\psi_n(x)$ are all bounded by $2$ it follows that the sequence converges uniformly!
Now, what does $\langle f,Hf \rangle < \infty$ mean for $f(x) = e^{-x^2/2} g(x)$. This can be computed to mean $$ \int |g'(x) + \frac{x}{2} g(x)|^2 e^{-x^2} dx. $$
On a philosophical level, this is not about the $H_n$ being orthogonal polynomials, but about them being eigenfunctions of a self-adjoint operator. (well the $\psi_n$ are).
