Let $X$ be a separable Banach space and consider $\mathcal{K}(X)$ the space of compact operators $K\colon X \rightarrow X$. Is it true that the space $\mathcal{K}(X)$ is separable? If yes, why? If no, where can I find a counterexample?

  • $\begingroup$ What topology do you choose on $\mathcal{K}(X)$? $\endgroup$ – YCor Dec 11 '17 at 15:00
  • $\begingroup$ @YCor, the operator norm. This is quite standard, isn't it? $\endgroup$ – Tomek Kania Dec 11 '17 at 17:47
  • $\begingroup$ @TomekKania it is standard but there are other standard topologies on the space of operators, e.g., the strong operator topology "SOT": $T_i\to T$ if $\|T_ix-Tx\|\to 0$ for every $x$: it is weaker than the norm topology (which requires the latter convergence to be uniform on the unit ball). When one defines continuous unitary representations, one uses SOT rather than the norm topology, which is too strong. $\endgroup$ – YCor Dec 11 '17 at 17:59

Let $X$ be a Banach space such that $X^*$ is separable. I claim that the space of compact operators $\mathcal K(X)$ on $X$ is separable.

It is enough to show that $nB_{\mathcal K(X)}$, the space of compact operators of norm at most $n$, is separable as then $\mathcal K(X) = \bigcup_{n=1}^\infty nB_{\mathcal K(X)}$ is separable too.

By Shauder's theorem, an operator is compact if and only if its adjoint is compact, and so $\mathcal K(X)$ isometric to the space $\mathcal K^{w^*}(X^*)$ of weak*-continuous compact operators on $X^*$. Take a compact operator $T$ on $X$ that has norm at most $n$. Then $T^*$ has the same norm. Moreover, it is uniquely determined by its restriction to the closed unit ball of $X^*$, hence the map $$T^*\mapsto T^*|_{B_{X^*}}\quad (T\in n B_{\mathcal{K}(X)})$$ is an isometry into the space of continuous functions $C(B_{X^*}, nB_{X^*})$ (indeed, these operators are weak*-to-norm continuous; this is where we employ compactness). The latter space is separable as the space of continuous functions between two compact, metrisable (hence second-countable) spaces.

  • 3
    $\begingroup$ Of course, separability of $X^*$ is necessary, because $X^*$ embeds isometrically in $\mathcal{K}(X)$ (consider rank-one operators). $\endgroup$ – Nate Eldredge Dec 11 '17 at 16:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.