A question about the weak operator topology of $B(H)$

Question

Let $H$ be a Hilbert space and $V$ a dense subspace.

Let $A$ be a $*$-subalgebra of $B(H)$. The weak closure of $A$ is, by definition, the space of all $u\in B(H)$ satisfying:

For every $x_1,y_1,\ldots,x_n,y_n\in H$, there exists $a\in A$ with $|\langle (u-a)x_i,y_i\rangle|<\epsilon$ for all $i$.

Question: In the condition above, can we restrict ourselves to $x_1,y_1,\ldots,x_n,y_n\in V$?

I believe this should be true, but I can't prove. The problem is the condition (restricted to $V$) implying $u$ in the weak closure of $A$, because the condition does not give any nice bound for $\Vert a\Vert$.

Answer 1

Kaplansky's density theorem lets you choose $a $ with $\|a\|\leq\|u\|$.