# Showing that the ideal of compact operators is not necessarily closed in the weak operator topology

For any Hilbert space $H$, one can show that the ideal of compact operators, is closed with respect to the strong operator topology by using that the unit ball is weakly compact or directly in the case when $H$ is separable using a diagonal argument.

The cantor diagonal argument fails when trying to see if the ideal of compact operators, is closed in $B(H)$ with respect to the weak topology. For this reason I have tried to find a counter example showing that $B(H)_{\text{compact}}$ is not closed in $B(H)$.

In the case when $H=:L_2(\mathbb{T})$, we have the sequence $T_n$ of elements in $B(H)_{\text{compact}})$ given by $T_n(f)=\mathscr{F}^{-1} \widehat{f(*+n)}$, i.e. the shift in fourier space. $T_n$,a sequence of compact operators, converges in the weak operator topology to $0$ by testing against all trigonometric polynomials.

The only problem with this example is that it is not a nonexample!: The zero map is a compact operator.

I am convinced however, that one can use an idea like this to show that the space of compact operators in $B(H)$ is not closed.

Question: How do I do it?

• It is not clear to me what you mean by "compact operators defined with respect to the strong operator topology". The "compact" is a compact operator is about the topology of $H$, not of $B(H)$. – Martin Argerami Jan 18 '17 at 3:53
• Yeah sorry about that. I edited the question. – user062295 Jan 18 '17 at 5:55
• I'm already lost at your first sentence, since the sot closure of $K (H)$ is $B (H)$. – Martin Argerami Jan 18 '17 at 8:08
• Thank you so much for spending your time here to help me. I must have some misconception. I thought that $\overline{K(H)}=K(H)$, in any topology that is at least as strong as the SOT. Is this wrong? I'll put my proof in the next comment. – user062295 Jan 18 '17 at 16:23
• Consider matrix units with respect to an orthonormal basis, and form $P_n-\sum_{k=1}^n E_{nn}$. Then $P_n\to I$ sot (now I see that Prahlad has already put this same example in his answer). And, more generally (like I said) the sot closure of $K(H)$ is $B(H)$; it follows very easily from von Neumann's Double Commutant Theorem. And, by the way, sot closure agrees with wot closure on convex sets (so, in particular, in subspaces/subalgebras). – Martin Argerami Jan 18 '17 at 16:42

It is very easy to see that the commutant of $K(H)$ is $\mathbb C I$. Then, using von Neumann's Double Commutant Theorem, we have $$\overline{K(H)}^{\rm sot}=K(H)''=(\mathbb C I)'=B(H).$$
The space $\mathcal{K}(H)$ of compact operators is not closed with respect to the strong operator topology (SOT) of $H$ is infinite dimensional. Take $H = \ell^2$, and let $\{e_n\}$ be the standard orthonormal basis of $H$. Then, for any $x\in H$, we may write $$x = \sum_{n=1}^{\infty} \langle x,e_n\rangle e_n$$ So if $P_n(x)$ denotes the $n^{th}$ partial sum of this series, then each such $P_n$ is bounded and has finite rank, and is thus compact. Also, $P_n(x) \to x$ for each $x\in H$, so $P_n \to I$ strongly, and $I\notin \mathcal{K}(H)$.
Take the limit of projector matrices with respect to the basis $e^{2\pi in}$ that act as the identity on the first basis elements and 0 on everything else. Then $T_n \rightharpoondown Id$ weakly. The identity matrix is not compact because the unit ball is not compact.