Weak convergence equal to coordinate-wise convergence 
Show for the Hilbert Space $\ell^2(\mathbb{N})$ weak convergence of
  a bounded sequence $(x_k)_{k \in \mathbb{N}}$ is equal to coordinate-wise
  convergence.
A sequence $(x_k)$ is weak convergent to $x$ $(x_k \rightharpoonup x)$
  in a Hilberspace E if $f(x_k) \rightarrow f(x)$ for every f in the
  Dualspace of E.

(=>) 
Let $(x_k)_{k \in \mathbb{N}}=(a^{(1)}_k,a^{(2)}_k,a^{(3)}_k,...)$ be weak convergent to the point $x=(a^{(1)},a^{(2)},a^{(3)},...)$. For $j \in \mathbb{N}$ we define a mapping f which maps a vektor $x \in \ell^2(\mathbb{N})$ to the j-Index. $f:(a^{(1)},a^{(2)},a^{(3)},...) \mapsto a^{(j)}$. f is linear and continuous, therefor f is in the Dualspace $(\ell^2(\mathbb{N}))'$. By definition of weak convergence: $x_k \rightharpoonup x$ in $\ell^2 $ $=> f(x_k)=f(x) => a^{(j)}_k=a^{(j)}$. $x_k$ is coordinate-wise convergent to $x$.
(<=) Can someone tell me how to show, that a coordinate-wise convergent sequence is weak convergent?  

Consider the set $B=\{e^j:j\in\mathbb{N}\}\subset (\ell^2(\mathbb{N}))^\prime$ defined as $e^j(x)=x^{(j)}$, where $x=(x^{(i)})_i\in\ell^2(\mathbb{N})$. $B$ is a basis of $\ell^2(\mathbb{N})$ in the sense that $\overline{B}=\ell^2(\mathbb{N})^\prime$, where the closure is of course taken with respect to norm convergence in $\ell^2(\mathbb{N})^\prime$ (you can see this easily using Riesz' representation theorem).
$B$ is the dual basis corresponding to the canonical orthonormal basis of $\ell^2(\mathbb{N})$, consisting of $e_j=(0,\dots,0,1,0,\dots)$ where the $1$ is at the $j$th position.
If now $(x_k)_k$ converges coordinate-wise to $x\in\ell^2(\mathbb{N})$, we have $e^j(x_k)=x_k^{(j)}\rightarrow x^{(j)}=e^j(x)$, therefore by linearity $f(x_k)\rightarrow f(x)$ for all $f\in B$. By density, $f(x_k)\rightarrow f(x)$ for all $f\in\ell^2(\mathbb{N})^\prime$. That last implication requires the sequences to be bounded in $\ell^2(\mathbb{N})$.
