I encounter the following problem in Murphy's book $C^*$ algebras and operator theory: show that relative weak topology on closed unit ball of $B(H)$ is metrisable where $H$ is a seperable Hilbert space with orthonormal basis $\{e_n\}$. The metric $d$ is defined as:
$$d(u,v)=\sum_{m,n = 1}^{\infty}{\frac{\left|\left\langle (u - v)e_n,e_m\right\rangle\right|}{2^{n+m}}}$$
To show that this metric on unit ball induces the weak topology I have to show that $u_n$ converges to $u$ weakly $\iff$ $d(u_n,u)$ converges to zero, where weak convergence means that $\langle u_ix,y\rangle\rightarrow\langle ux,y\rangle$ for all $x,y\in H$ as defined in Murphy's book. I know how to show that $u_n$ converges to $u$ weakly implies $d(u_n, u)$ converges to zero. But I can not show the other direction. Can anyone give a hint or solution to that?
Suppose that $d(u_n,0)\to0$. Fix $k,j$. Then $$ |\langle u_n e_j,e_k\rangle|\leq 2^{k+j} d(u_n,0)\to0. $$ Given $x\in H$, write $x=\sum_kx_ke_k$. Fix $\varepsilon>0$. Choose $k_0$ such that $\sum_{k\geq k_0}|x_k|^2<\varepsilon^2$. Note that $|\langle u_ne_j,e_k\rangle|\leq1$. Then \begin{align} |\langle u_nx,x\rangle| &\leq\sum_{k,j} x_j\,\overline{x_k}\,\langle u_ne_j,e_k\rangle\\[0.3cm] &=\sum_{k,j\leq k_0} x_j\,\overline{x_k}\,\langle u_ne_j,e_k\rangle +\sum_{k\ \text{ or } j\geq k_0} x_j\,\overline{x_k}\,\langle u_ne_j,e_k\rangle\\[0.3cm] &\leq\sum_{k,j\leq k_0} x_j\,\overline{x_k}\,\langle u_ne_j,e_k\rangle +2\left(\sum_{k\geq k_0} |x_j|^2\right)^{1/2}\,\|x\|\\[0.3cm]\\[0.3cm] &\leq \sum_{k,j\leq k_0} x_j\,\overline{x_k}\,\langle u_ne_j,e_k\rangle + 2\|x\|\,\varepsilon. \end{align} By the first part, we may choose $n_0$ such that for all $n\geq n_0$ we have $|\langle u_ne_j,e_k\rangle|\leq\varepsilon$ for all $k,j=1,\ldots,k_0$. Then $$ |\langle u_nx,x\rangle|\leq 2(1+\|x\|)\varepsilon. $$ As $\varepsilon$ was arbitrary, this shows that $\langle u_nx,x\rangle\to0$. Using polarization, we get that $\langle u_nx,y\rangle\to0$.