Predual of $\mathcal{B}(K, H)$

Question

Is there a predual of $\mathcal{B}(K, H)$? So, what does the space $X$ look like, such that $X^*=\mathcal{B}(K, H)$.

Answer 1

if $H_1=H_2$ then $B(H)_*=L^1(H)$(trace class), so, in my opinion, $B(H_1,H_2)_*=L^1(H_1,H_2)$. takesaki shows that $K(H)^*=L^1(H) , K(H)^{**}=L^1(H)^*=B(H)$

Answer 2

The space $Y:=\mathcal{B}(K, H)$ have a predual $X$ which is the space of nuclear operators $\mathcal{N}(H, K)$, which in turn is nothing more than $H\otimes_{\pi} K^{cc}$.

If $\operatorname{hilb.dim}(H)=\operatorname{hilb.dim}(K)$, then $X\cong_1\mathcal{B}(H)$, so $Y$ is a von Neumann algebra and therefore predual $X$ is unique.