If $H$ Hilbert and $(P_k)$ is a sequence of orthogonal projections in $B(H)$, then $0$ is in weak closure of $\{\sqrt{k}P_k:k\in\mathbb{N}\}$ Let $H$ be a separable Hilbert space with an orthonormal basis $(e_k)$. Let $P_k$ be the orthogonal projection onto $\{\lambda e_k:\lambda\in\mathbb{C}\}$. How do I show that $0$ (the zero operator) is in the weak closure of $\{\sqrt{k}P_k:k\in\mathbb{N}\}$. 
Recall: A net $(T_\iota)$ in $B(H)$ converges weakly to $T$ iff $\langle T_\iota(x),y\rangle\to\langle T(x),y\rangle$ for all $x,y\in H$. 
My idea was to assume that $0$ does not belong to the weak closure and construct a contradiction. Any help would be greatly appreciated!
 A: We can even show that $0$ belongs to the strong closure of this set that i will now on call $E$, i.e. we will construct a net in $E$ that converges strongly to $0$. Recall that a net $(T_\lambda)\subset B(H)$ converges strongly to $T\in B(H)$ when $T_\lambda x\to Tx$ for all $x\in H$. Obviously, if a net converges strongly to an operator, then it also converges weakly.
Suppose that $0$ does not belong to the strong closure of $E$. Then there exists a strong basic-neighborhood of $0$ that does not intersect $E$. 
Recall: A strong basic-neighborhood of $T\in B(H)$ is of the form
$$V(T,\varepsilon,x_1,\dots,x_n):=\{S\in B(H): \|Tx_i-Sx_i\|<\varepsilon\text{ for all i}\}$$
where $\varepsilon>0$, $x_1,\dots,x_n\in H$. 
So we have a strong-neighborhood of $0$ not intersecting with $E$. Suppose that this neighborhood is $V(0,\varepsilon,x_1,\dots,x_n)$ as above. Then the set
$$\{T\in B(H): \sum_{i=1}^n\|Tx_i\|^2<\varepsilon^2\}$$
is a subset of $V(0,\varepsilon,x_1,\dots,x_n)$ so it does not intersect $E$.
Therefore, for any $k\in\mathbb{N}$ it is $\sum_{i=1}^n\|\sqrt{k}P_kx_i\|^2\geq\varepsilon^2$, i.e. $\sum_{i=1}^n\|P_kx_i\|^2\geq\varepsilon^2/k$. But it is $P_kx_i=\langle x_i,e_k\rangle e_k$, so $\sum_{i=1}^n|\langle x_i,e_k\rangle|^2\geq\varepsilon^2/k$. This is true for any $k$.
But we have that $\sum_{k=1}^\infty|\langle x,e_k\rangle|^2=\|x\|^2$ for all $x\in H$ since $\{e_k\}$ is an orthonormal basis, so
$$\sum_{i=1}^n\|x_i\|^2=\sum_{i=1}^n\sum_{k=1}^\infty|\langle x_i,e_k\rangle|^2=\sum_{k=1}^\infty\sum_{i=1}^n|\langle x_i,e_k\rangle|^2\geq\sum_{k=1}^\infty\frac{\varepsilon^2}{k}=\infty,$$
a contradiction, since we only added up a finite number of norms of elements of $H$, so the sum cannot be infinity. 
The change of the two sums in the final equations is possible, since we add only positive quantities, so rearrangements do not affect summation (or, if you know measure theory, this is simply Tonelli's theorem)
