Space of compact operators defined on separable Hilbert space

Question

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?

Answer 1

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.