What is a predual of the Banach space of compact operators on $\ell^2$?

Question

I am wondering if the space $K(\ell^2)$ of compact operators on $\ell^2$ can have a predual.

Thank you in advance for your help.

Answer 1

Here is an operator theorist's argument.

Assume $K(\ell_2)$ is a dual Banach space. Since $K(\ell_2)$ is a $C^*$-algebra, by Sakai's theorem we have that $K(\ell_2)$ is a von Neumann algebra. Take any $a\in K(\ell_2)$ with infinite dimensional image. Since $K(\ell_2)$ is a von Neumann algebra then there exists an orthogonal projection $p\in K(\ell_2)$ on $\operatorname{Im}(a)$. So $\operatorname{Im}(p)$ is infinite dimensional, which is impossible because $p$ is compact. Contradisction, so $K(\ell_2)$ is not a dual space.

Answer 2

This is the gist of the argument given in the linked paper in my comment above:

In said paper it is shown that if $H$ is a separable Hilbert space, then $K(H)$ is separable (look at finite-rank operators) and that $K(H)$ contains in isomorphic copy of $c_0$ (fix an orthonormal basis and look at the multiplication operators induced by elements of $c_0$).

$c_0$, however, does not embed in a separable dual space. See, e.g., Kalton and Albiac, Topics in Banach Space Theory, Theorem 6.3.7.

From this, it follows that $K(H)$ is not isomorphic to a dual space if $H$ is separable (in fact, it's not isomorphic to any subspace of a separable dual space).

Answer 3

Let me give an alternative approach.

Separable dual spaces have the Radon–Nikodym property. The algebra of compact operators on a separable, infinite-dimensional Hilbert space is of course separable. The closed unit ball of a space with the Radon–Nikodym property is the closed convex hull of its extreme points (using a fancy language: Radon–Nikodym property implies the Krein–Milman property).

But wait... an extreme point in the closed unit ball of a C*-algebra must be a partial isometry, so in particular the underlying C*-algebra must have a projection with infinite-dimensional range, which is not the case for $\mathscr{K}(H)$ – there are, thus, no extreme points in the ball of the compacts at all.