A dense subset of a Hilbert space I am curious about the following problem:
Consider the Hilbert space (a weighted $L^2(\mathbb{R})$ space):
$$\mathscr{H}=\bigg\{f: \mathbb{R}\to\mathbb{R}\text{ Lebesgue measurable}\,\bigg|\,\int_\mathbb{R} \big|\,f(x)\,\big|^2\exp\Big(-\frac{x^2}{2}\Big)\,\mathrm{d} x<\infty\bigg\}$$
on which the inner product is defined as: for every $f,g\in\mathscr{H}$,
$$\langle f, g\rangle=\int_\mathbb{R} f(x)g(x)\,\exp\Big(-\frac{x^2}{2}\Big)\,\mathrm{d}x.$$
I want to find a dense subset of $\mathscr{H}$. Obviously, $\mathscr{H}$ contains $L^2(\mathbb{R})$. My question is: is any of the following three a dense subset of $\mathscr{H}$? Can anyone give me some hints about the proof? Or if none of them is, what could be a good candidate? Thanks a lot !!
(1) The set of simple functions in $\mathscr{H}$;
(2) The set of continuous functions in $\mathscr{H}$;
(3) The set of polynomial functions in $\mathscr{H}$.
 A: Since the measure for $\mathscr{H}$ is the Lebesgue measure with a (nice) weight function, you get an isometric isomorphism by multiplication with the square root of the weight:
$$T \colon \mathscr{H} \to L^2(\mathbb{R}); \quad T(f) \colon x \mapsto f(x)\exp \left(-\frac{x^2}{4}\right).$$
That isometry gives you a bijection between the dense subsets of $\mathscr{H}$ and $L^2(\mathbb{R})$. Since the weight is continuous, $T$ and $T^{-1}$ map continuous functions to continuous functions. Since the set of continuous functions is dense in $L^2(\mathbb{R})$, the set of continuous functions in $\mathscr{H}$ is also dense. The space of simple functions is dense in every $L^p(X,\mu)$ for $p < \infty$ (at least for $\sigma$-finite $\mu$), the space of simple functions is dense in $\mathscr{H}$. I think - but I'm not sure about it - that the set of polynomials is also dense in $\mathscr{H}$.
A: Notice that if you take a function $f$ in $\mathscr{H}$ and take $f_k = f\cdot 1_{[-k,k]}$ you have $f_k \to f$. Hence the subspace $L^2$ is dense in $\mathscr{H}$ and (1) and (2) are also dense, being dense in $L^2$. Also (3) is dense in $\mathscr{H}$ because you can truncate your $f$ as before, approximate with a continous function and hence with a polinomial (by Stone-Weierstrass).
