Example of a net in $\mathcal{B}(\ell_2)$ that converges in the weak operator topology but not in the strong operator topology? I need to show that the Strong Operator topology is strictly stronger in $\ell_2$ (space of complex sequences that are square summable). I can show that convergence in the strong operator topology implies it in the weak sense, but cannot come up with an example of a weakly converging sequence of operators that does not converge strongly. 
 A: Convergence in strong operator topology implies convergence in weak operator topology. This result is valid for operators between arbitrary normed spaces.
Assume $(T_\lambda:\lambda\in\Lambda)\subset\mathcal{B}(E,F)$ is a net that converges to $T\in\mathcal{B}(E,F)$ in strong operator topology. Take arbitrary $x\in E$ and $y\in F^*$.  By assumption $\lim_{\lambda\in\Lambda} T_\lambda(x)=T(x)$, hence
$$
\lim\limits_{\lambda\in\Lambda}y((T_\lambda-T)(x))=
y(\lim\limits_{\lambda\in\Lambda}T_\lambda(x)-T(x))
=y(0)
=0
$$
Since $x\in E$ and $y\in F^*$ are arbitrary, then $(T_\lambda:\lambda\in\Lambda)$ converges to $T$ in the weak operator topology.
A: Your title question is answered by Norbert, but what you ask in the text of the question is the appropriate thing to ask.
The hint of the answer is given in the comments by minimalrho, but here it is as an answer. 
Let $U:l^2(\mathbb{N})\rightarrow l^2(\mathbb{N})$ be given by
$$ U(a_0, a_1, ...)=(0, a_0, a_1, a_2, ...) $$
Notice that $U$ is a (non-surjective) isometry so in particular $\|U(\xi)\|\not\rightarrow0$
for any nonzero $\xi$.
However let $\xi=(a_0, a_1, ...)$ and $\eta=(b_0, b_1, ...)$. Now note that 
$$|\langle U^n(\xi), \eta\rangle|=\left|\left\langle U^n(a_0, a_1, ...), (b_0, b_1, ...)\right\rangle\right|
=$$
$$|\langle (\underbrace{0, ..., 0}_{\text{n times}}, a_0, a_1, ...), (b_0, b_1, ...)\rangle| = $$
$$ |\langle (\underbrace{0, ..., 0}_{\text{n times}}, a_0, a_1, ...), (\underbrace{0, ..., 0}_{\text{n times}}, b_n, b_{n+1}, ...)\rangle| \leq $$
$$\|\xi\|\cdot\|(0, ..., 0, b_n, b_{n+1}, ...)\| $$
where the last inequality is by Cauchy-Schwarz. Then we just notice that
$$ \|(0, ..., 0, b_n, b_{n+1}, ...)\|\rightarrow 0 $$
in order to get that $U^n\rightarrow 0$ in the WOT. 
