Are finite rank operators on Hilbert space $H$ dense in $B(H)$ in the weak operator topology?
Density of finite rank operators on Hilbert space
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2In the second part of my answer [here](http://math.stackexchange.com/a/126863) I show that the finite rank operators (it suffices to take those which are nilpotent of index 2) are strongly dense in $B(X)$ whenever $X$ is a Banach space. Since the weak topology is weaker than the strong topology on $B(H)$ it follows that the finite rank operators are weakly dense in $B(H)$ whether $H$ is separable or not. – 2012-06-28
2 Answers
Yes. You can check that a bounded operator is the WOT-limit of its "upper left corners" which are of course finite rank.
Or, if you know von Neumann algebras (more precisely, the Double Commutant Theorem), you can check that the commutant of the finite-rank operators is trivial, making their double commutant all of $B(H)$.
Edit (in case someone is interested in more detail): what I meant in the first paragraph is exactly what Norbert wrote in his solution, with the added (trivial) mention that if $p$ is finite-rank, then so is $pap$ (or even $pa$) for any $a\in B(H)$).
Regarding the second proof, suppose that $x\in B(H)$ commutes with all rank-one operators. Fix an orthonormal basis $\{\xi_j\}_{j\in J}$. Given $k,j\in J$, we define the operator $e_{kj}$ to be the map $e_{kj}\xi=\langle\xi,\xi_j\rangle\xi_k$ (the $\{e_{kj}\}$ are "matrix units", i.e. $e_{kj}e_{hl}=\delta_{jh}e_{kl}$, and $e_{kj}^*=e_{jk}$) . Now, for a given $k\in J$, we have $\tag{1} x\xi_k=xe_{kk}\xi_k=e_{kk}x\xi_k=\langle x\xi_k,\xi_k\rangle\xi_k $ and, for any $j\in J$, $\tag{2} \langle x\xi_k,\xi_k\rangle=\langle xe_{kk}\xi_k,\xi_k\rangle=\langle xe_{kj}e_{jk}\xi_k,\xi_k\rangle=\langle e_{kj}xe_{jk}\xi_k,\xi_k\rangle=\langle xe_{jk}\xi_k,e_{jk}\xi_k\rangle=\langle x\xi_j,\xi_j\rangle. $ If we write $\lambda=\langle x\xi_j,\xi_j\rangle$ (the equalities (2) shows that this number is independent of $j$), we have by (1) that $ x\xi_k=\lambda\xi_k, \ \ \ k\in J; $ that is, $x=\lambda I$.
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1@kaper: the proof that the commutant of the finite-rank operators (actually, rank-one are enough) is trivial does not use the dimension of $H$. – 2012-06-28
Let $\mathcal{P}(H)$ be a directed system of all finite rank orthogonal projections in $H$. We take $p\leq q$ in $\mathcal{P}(H)$ if $\mathrm{Im}(p)\subseteq\mathrm{Im}(q)$. One may show that for all $a\in\mathcal{B}(H)$ we have $ a=\lim\limits_{p\in\mathcal{P}(H)}pap\tag{1} $ in the weak operator topology. Indeed, take $x,y\in H$. Then for all orthogonal projections $p$ such that $\mathrm{Im}(p)\supseteq\mathrm{span}\{x,y,a(x)\}$ we will have $\langle (pap)(x),y\rangle=\langle a(x),y\rangle$. So we proved equality $(1)$.
Since $\{pap:p\in \mathcal{P}(H)\}\subseteq\mathcal{P}(H)$ we conslude that $\mathcal{P}(H)$ is dense in $\mathcal{B}(H)$ in the weak operator topology.