Dualities on K(H), B(H)...

Notation. $H$ will denote a Hilbert space, $B\left(H\right)$ the algebra of the bounded operators on $H$, $K\left(H\right)$ the ideal of the compact operators, $H^{*}$ the dual of $H$.

$K\left(H\right)$ is identified with the injective tensor product $H^{*}\hat{\hat{\otimes }}H$. On the other hand, $B\left(H\right)$ is identified with the space ${\left(H^{*}\hat{\otimes }H\right)}^{*}$ via $<T,x^{*}\otimes y>=x^{*}\mathrm{Ty}$ (note that I permuted $H$ and $H^{*}$ from the talk on Oct 19 for simplicity). We have three mappings

1. $t:{K\left(H\right)}^{*}\to B\left(H\right)$ as the transpose of $H^{*}\hat{\otimes }H\to H^{*}\hat{\hat{\otimes }}H$.
2. $f:H^{*}\hat{\otimes }H\to B\left(H\right)$ as the composition of $H^{*}\hat{\otimes }H\to H^{*}\hat{\hat{\otimes }}H$ and the inclusion $K\left(H\right)\to B\left(H\right)$.
3. $g:H^{*}\hat{\otimes }H\to {K\left(H\right)}^{*}$ by $x^{*}\otimes y\to (u^{*}\otimes v\mapsto x^{*}v.u^{*}y)$.

We have $f=\mathrm{gt}$. By the theorem of Hahn-Banach, the injectivity of $g$ follows from the surjectivity of its tranpose $g^t:{K\left(H\right)}^{**}\to B\left(H\right)$. Now we can explicitly construct an inverse mapping to $g^t$: Let $h$ denote the composition of of $H^{*}\otimes H\to H^{*}\hat{\otimes }H$ and $g$. By the surfectivity of $g$ the image of $h$ is dense.

Lemma. For any $w\in H^{*}\otimes H$ and any $T\in B\left(H\right)$, we have $|<T,w>|\le \parallel h\left(w\right)\parallel .\parallel T\parallel$

Proof. $w$ can be written as ${\sum }_{i=0}^n{\left(x_i\right)}^{*}\otimes y_i$ for some ${x_i}^{*}\in H^{*}$ and $y_i\in H$. Let $P$ be the projection operator onto the finite dimensional space generated by the $y_i$ for $0\le i\le n$. Then $\mathrm{TP}$ is compact and we have $<T,w>=<\mathrm{TP},w>=<h\left(w\right),\mathrm{TP}>$, hence $|<T,w>|\le \parallel h\left(w\right)\parallel .\parallel \mathrm{TP}\parallel \le \parallel h\left(w\right)\parallel .\parallel T\parallel$.

From this lemma we get a continuos linear mapping $a:B\left(H\right)\to {K\left(H\right)}^{**}$. Finally we have $g^t\circ a={\mathrm{id}}_{B\left(H\right)}$ by $$<g^t\circ a\left(T\right),x^{*}\hat{\otimes }y>=<a\left(T\right),h(x^{*}\otimes y)>=<T,x^{*}\otimes y>.$$