$K(H)^{**}= B(H)$

Question

Let $H$ be a Hilbert space. Is it true that we have an isometric linear isomorphism $$B(H) \cong K(H)^{**}$$

where $K(H)^{**}$ is the bidual of the compact operators on $H$.

I think I proved this but I could not find a reference on the internet. Is the result true?

Answer 1

Yes, this is completely standard and should appear in any operator theory book. One can easily show that $K(H)^*$ can be identified with $T(H)$ (the trace-class operators) via the duality $$ \langle T,S\rangle=\operatorname{Tr}(TS). $$ And similarly, $B(H)$ can be seen as the dual of $T(H)$ via the same duality.

To mention a few references:

• Conway, A Course in Operator Theory: Section 1.19

• Conway, A Course in Functional Analysis: Exercise IX.2.21

• Farenick, Functional Analysis, Theorems 10.99 and 10.102

• Reed-Simon, Methods of Mathematical Physics. Functional Analysis, Theorem VI.26

• Davidson, C$^*$-Algebras by Example, Exercises I.18 and I.20

• Murphy, C$^*$-Algebras and Operator Theory, Theorems 4.2.1 and 4.2.3

• Takesaki, Theory of Operator Algebras, Theorem II.1.8