Weakly dense subset spans the space?

Question

If a set $\mathcal{S}$ is weakly dense in a Hilbert space $\mathcal{H}$, is it true that $\mathcal{H}=Span(\mathcal{S})$? I am trying to use this to show that separatability and weak separatability are equivalent, but I can't see why this claim is true.

Answer 1

Any dense set is weakly dense and a dense set need not span the whole space. For example the space of finitely non-zero sequences is weakly dense in $\ell^{2}$. However it is true that the close subspace spanned by a weakly dense set is equal to the whole space.

This last fact follows from the following: if $x_n \to x$ weakly in a Banach space then there exist a sequence of convex combinations of $x_n$'s which converges to $x$ in the norm. Equivalently, the weak closure of a convex set equals its norm closure.

Answer 2

Kavi`s answer solves my main concern here, and the last fact noticed in the answer also gives a proof of the equivalence of two types of separability.

Given $x_{n}\rightarrow x$ weakly, one construction of the converging sequence is $y_{k}=(x_{n_{1}}+...+x_{n_{k}})/k$ for some well chosen $x_{n_{i}}\in(x_{n})$. Thus, if a countable set $\mathcal S$ is weakly dense, the collection of $(x_{1}+...+x_{k})/k$ of any k and any choices of $x_{i}\in\mathcal S$ is a countable dense set. So weak separable implies separable. Converse is trivial.