0

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

Norbert
  • 58,398
Reza
  • 9

2 Answers2

1

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.

Norbert
  • 58,398
  • Can you please explain how predual of B(K, H) is N(H,K), under which map? – Sushil May 01 '19 at 21:07
  • It is a non-trivial theorem that $i:\mathcal{B}(K,H)\to\mathcal{N}(H,K)^*:T\mapsto (S\mapsto Trace(TS))$ gives the desired isomorphism. – Norbert May 01 '19 at 21:27
  • Will norm on $\mathcal{N}(H, K)$ be infimum over $inf{\sum_{n=1}^{n=\infty} s_n: S = \sum_{n=1}^{n=\infty} s_n \langle , u_n\rangle v_n$ for some $u_n, v_n$ orthonormal and some $s_n$ such that $\sum_{n=1}^{n=\infty} s_n<\infty}$ – Sushil May 01 '19 at 21:56
  • Are you considering $H$ and $K$ to be hilbert space. What if these are general Banach Spaces, can we sat anything about Predual? – Sushil May 03 '19 at 13:37
  • Then, as far as I remember, it may not even exist. – Norbert May 03 '19 at 15:24
  • A question, is it true $\mathscr K(K, H)^* = \mathscr N(H, K)$, where $H, K$ are Hilbert spaces and $\mathscr K(K, H)$ denotes space of compact operators from $K$ to $H$. – Sushil Aug 07 '20 at 18:08
  • @Sushil, that is true – Norbert Aug 08 '20 at 16:25
  • @Norbert Do you have a reference for this? – Infimum Jul 28 '23 at 01:09
  • 1
    @Infimum Lectures and exercices on functional analysis. A.Ya. Helemskii, Chapter 3, Section 4, theorem 4, p. 229 – Norbert Jul 28 '23 at 14:44
0

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 $K(H)^*=L^1(H) , K(H)^{**}=L^1(H)^*=B(H)$

saeed
  • 489