Q:

Weakly dense subset spans the space? 

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. 

A:

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. 