Weak convergence to uniform convergence Let $\mathscr{H}=\Bbb{C}^d$ , $d \in \Bbb{N}$, with the usual inner product. 
Prove that if a sequence of operators $\{T_n\}$ in $\mathscr{H}$ converges to zero weakly as $n \to \infty$, then it converges to zero in the operator norm. 
Proof ( My attempt)
Let $T_n \to 0$ weakly, then $(T_nx,y)=0$
By Riesz Representation Theorem 
There exists $f_n \in \mathscr{H}^*$ such that 
$f_n(y)=(T_nx,y)=0$ for all $y \in \mathscr{H}$.
$\Rightarrow f_n=0$ for all n. 
$\Rightarrow |f_n|=\|T_nx\|=0$  for all n.
$\Rightarrow \|T_n\|=0$
Hnece $T_n$ converges to zero in the operator norm.
Help: I feel this proof is wrong because it doesn't use the fact its finite dimensional or has an euclidean inner product. 
please help me as much as possible where i went wrong or how i would fix it.
Thanks
 A: You seem to go from the sequence $(T_nx,y)\rightarrow 0$ as $n\rightarrow\infty$ to then saying $(T_nx,y) = 0$ for all $n$. This is false. 
You should recall that you can represent $T_n$ as a matrix $(a^n_{i,j})$. Then try to show that for all $i,j$ we have that $a^n_{i,j}\rightarrow 0$ as $n\rightarrow\infty$. This is where you use weak convergence.
Then choose $N$ such that all the $a^N_{i.j}$ are sufficiently small, $\epsilon/d^2$ should work, and then you're basically done. 
A: I don't know how much functional analysis you have at your disposal, but a short argument works as follows: the operators on $\mathbb{C}^d$ form a finite dimensional topological vector space (isomorphic to the space $\mathbb{C}^{d\times d}$ of matrices). On finite dimensional spaces, all topologies which turn them into topological vector spaces (i.e. addition and scalar multiplication are continuous and the space is Hausdorff) are equivalent. Hence, convergence is the same with respect to any topology. The norm topology definitely turns the space of operators into a topological vector space, so all that remains to be shown is that there is a topology which induces the weak convergence. From the theory of locally convex spaces, we know this is true - just pick the weak topology with respect to the linear functionals $\phi_{x,y}:T\mapsto (Tx,y)$ for $x,y\in\mathbb{C}^d$. 
