Weak$^*$ convergence on dense subspace of Hilbert space Let $H$ be a separable infinite dimensional Hilbert space over $\mathbb{C}$
Let $V$ be a dense subspace of $H$
Let $\{T_n\}_{n \in \mathbb{N}} \subset H^*$ be a sequence of continuos linear functionals such that 
$$
\forall v \in V: \lim_{n \to \infty} T_n(v) = 0
$$
I would like to know if is it true that $T_n \stackrel{\ast}{\rightharpoonup} 0$ i.e. 
$$
\forall w \in W: \lim_{n \to \infty} T_n(w) = 0
$$

No. Let $H=\ell^2(\mathbb N)$, and $V$ the subspace of sequences with finitely many nonzero elements. Denote by $\{e_n\}$ the canonical basis, and let 
$$
T_n(x)=n\,x_n. 
$$
Then, for any $x\in V$, eventually $x_n=0$, so $T_n(x)\to0$. On the other hand, if $x=(1/n)_n$, then $T_n(x)=1$ for all $n$. 
The assertion becomes true if the norms of the $T_n$ are bounded. If $\|T_n\|\leq c$ for all $n$, then given $\varepsilon>0$ and $w\in H$ choose $v\in V$ with $\|v-w\|<\varepsilon$. Then 
$$
|T_n(w)|\leq|T_n(v)|+|T_n(v-w)|\leq|T_n(v)|+c\varepsilon. 
$$
Thus $\limsup_n|T_n(w)|\leq c\varepsilon$ for all $\varepsilon>0$, which shows that the limit exists and is zero. 
