Weak convergence of unitary operators on a dense subset. Suppose $\mathcal{H}$ is a separable Hilbert space over $\mathbb{C}$ with scalar product $(\cdot,\cdot)$. Let $(U_n),\, n\in \mathbb{N},$ and $U$ be unitary operators on $\mathcal{H}$ and $D\subset \mathcal{H}$ be dense such that
$$ (x, U_n y) \longrightarrow (x,U y) \quad(n\to \infty)$$
for all $x,y\in D$. Can we already conclude that $U_n \to U$ weakly, that is, the above equation holds for all $x,y\in\mathcal{H}$? If not, is there a reasonable further assumption that would be sufficient to imply weak convergence?
I know this is not true if $U_n$ and $U$ are just bounded, but in all counter-examples I know of, $U$ is not unitary.
Edit: If it helps, $D$ is the linear span of an orthonormal system $(e_k)_{k\in\mathbb{N}}$.
 A: You only need the fact that $\|U_n\|$ is uniformly bounded. In our case $\|U\|\leq 1$ and $\|U_n\| \leq 1$ for all $n$.
It is obvious that the limit relation extends to the case when $x \in H$ and $y \in D$. 
Now let $x,y \in H$. Choose $y' \in D$ such that $\|y-y'\| <\epsilon$. Then $|(x,U_ny)-(x,Uy)| \leq |(x,U_ny)-(x,U_ny')|+|(x,U_ny')-(x,Uy')|+|(x,Uy')-(x,Uy)|.$ The middle term tends to $0$. The first and the last terms are both  bounded by $\|x\|\|y-y'\|<\epsilon \|x\|$. 
A: Since your set is dense, you can approximate $x,y$ via sequences $x_n,y_n$. Now note that:
$$|(x,U_ny)-(x,Uy)|^2=$$$$ |(x-x_n,U_ny)+(x_n,U_n(y-y_n))+(x_n,(U_n-U)y_n)+(x_n,U(y_n-y))+(x_n-x,Uy)+(x,Uy)-(x,Uy)|^2$$$$\leq |(x-x_n,U_ny)|^2+|(x_n,U_n(y-y_n))|^2+|(x_n,(U_n-U)y_n)|^2+|(x_n,U(y_n-y))|^2+|(x_n-x,Uy)|^2$$$$\leq||x-x_n||\cdot||U_ny||+||x_n||\cdot||U_n(y-y_n)||+||x_n||\cdot ||(U_n-U)y_n||+||x_n||\cdot||U(y_n-y)||+||x_n-x||\cdot||Uy||$$$$=||x-x_n||\cdot||y||+||x_n||\cdot||y-y_n||+||x_n||\cdot ||(U_n-U)y_n||+||x_n||\cdot||y_n-y||+||x_n-x||\cdot||y||$$
Where we have used (in order) the linearity of the inner product, the triangle inequality, the Cauchy-Schwarz inequality, and the fact that $U,U_n$ are unitary (and thus isometries). Now, each of the terms above gets smaller as $n\rightarrow \infty$, for a different reason.
