Why the weak topology and the strong topology coincide? Why in any finite-dimensional Hilbert space the weak topology and the strong topology coincide?
 A: Let $\{e_1,\ldots,e_n\}$ be an orthonormal basis. Let 
$$
x_n=\sum_{j=1}^n x_{nj}e_j
$$
be a sequence that converges weakly to $0$. This means that $\langle x_n,y\rangle\to0$ for all $y$. In particular, $$x_{nj}=\langle x_n,e_j\rangle\xrightarrow[n]{}0.$$
Conversely, if $x_{nj}\to0$ for all $j$, 
$$
\langle x_n,y\rangle=\sum_{k=1}^nx_{nj}\bar y_j\to0. 
$$
So weak-convergence is precisely entrywise-convergence. 
Now, with $x_n$ as above, 
$$
\|x_n\|^2=\langle x_n,x_n\rangle=\sum_{j=1}^n |x_{nj}|^2\xrightarrow[n]{}0,
$$
since the sum is finite. Thus weak-convergence implies norm-convergence. 
A: You only have to look at neighborhoods of $0$ because of the translation invariance of the topologies.
Let $H$ be a finite-dimensional inner-product space with orthonormal basis $\{ e_{n}\}_{n=1}^{N}$ and inner-product $(\cdot,\cdot)$ and norm $\|\cdot\|$. It is an algebraic result that every $x^{\star}$ in the algebraic dual $X^{\star}$ can be written as $x^{\star}(x)=(x,y)$ for a unique $y \in H$. So, by the Cauchy-Schwarz inequality, $|x^{\star}(x)|\le \|y\|\|x\|=C\|x\|$, one sees that every $x^{\star}\in X^{\star}$ is continuous in the norm topology. The weak topology $\tau_{w}$ is then the weakest topology for which all $x^{\star} \in H^{\star}$ are continuous. So, $\tau_{w}$ must be weaker than $\tau_{s}$.
If $\mathcal{O}$ is an open neighborhood of $0$ in the norm topology, then there exists $\delta > 0$ such that
$$
         B_{\delta}(0)=\{ x \in H : \|x\| < \delta \} \subseteq \mathcal{O}.
$$
Let $x_{n}^{\star}(x)=(x,e_{n})$. Then $W_{n}=\{ x \in H : |x_{n}^{\star}(x)| < \delta/\sqrt{N}\}$ is a weak open neighborhood of $0$, and $\bigcap_{n=1}^{N}W_{n}\subseteq\mathcal{O}$ because $|(x,e_{n})| < \delta/\sqrt{N}$ for all $n$ implies
$$
   \|x\|^{2}=\sum_{n=1}^{N}|(x,e_{n})|^{2} < \delta^{2}.
$$
So the weak topology $\tau_{w}$ is stronger than the strong (norm) topology $\tau_{s}$. Therefore $\tau_{w}=\tau_{s}$.
