# Density of finite rank operators on Hilbert space

Are finite rank operators on Hilbert space $H$ dense in $B(H)$ in the weak operator topology?

• In the second part of my answer here 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. – t.b. Jun 28 '12 at 15:15

## 2 Answers

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.

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, $x=\lambda I$.

• Note: here H may be not separable. – kaper Jun 28 '12 at 14:46
• When $H$ is not separable, I don't think it is true. – kaper Jun 28 '12 at 15:03
• @Davide: compressions of the form $X\to PXP$ where $P$ is a finite-rank projection. – Martin Argerami Jun 28 '12 at 15:17
• @kaper: the proof that the commutant of the finite-rank operators (actually, rank-one are enough) is trivial does not use the dimension of $H$. – Martin Argerami Jun 28 '12 at 15:21