Some formulae

For matricially normed spaces $X$ and $Y$, $m\in{\mathbb{N}}$, $y\in M_m(Y)$ and $T\in\mathit{CB}(X,Y)$ we have

$$\Vert y\Vert=\sup\left\{\Vert\Phi(y)\Vert\;\vert\;n\in{\mathbb{N}... ...athit{CB}(Y,{\mathbb{M}}_n),\;\Vert\Phi\Vert _{\mathrm{cb}} \leqslant 1\right\}$$

and

$$\Vert T\Vert _{\mathrm{cb}}=\sup\{\Vert\Phi^{(n)}\circ T\Vert _{\... ...i\in\mathit{CB}(Y,{\mathbb{M}}_n),\; \Vert\Phi\Vert _{\mathrm{cb}}\leqslant 1\}$$   .

A matrix $[T_{ij}]\in M_n(X^*)$ defines an operator

$$T:M_n(X) \to {\mathbb{C}}$$

$$\left[x_{ij}\right] \mapsto \sum_{i,j}T_{ij}x_{ij}$$

Thus we have an algebraic identification of $M_n(X^*)$ and $M_n(X)^*$ and further of $M_n(X^{**})$ and $M_n(X)^{**}$. The latter even is a complete isometry ([Ble92b, Cor. 2.14]):

$${\mathbb{M}}_n(X^{**})\stackrel{\mathrm{cb}}{=}{\mathbb{M}}_n(X)^{**}$$   .

The isometry on the first matrix level is shown in [Ble92a, Thm. 2.5]. This already implies¹¹the complete isometry. More generally we have¹²

$$\mathit{CB}(X^*,{\mathbb{M}}_n(Y))\stackrel{\mathrm{cb}}{=}\mathit{CB}({\mathbb{M}}_n(X)^*,Y)$$   .

$X$ is called reflexive, if $X \stackrel{\mathrm{cb}}{=}X^{**}$. An operator space $X$ is reflexive if and only if its first matrix level $M_1(X)$ is a reflexive Banach space.

Footnotes

... implies¹¹

$$M_k({\mathbb{M}}_n(X)^{**})=M_k(M_n(X))^{**}=M_{kn}(X)^{**}=M_{kn}(X^{**})=M_k({\mathbb{M}}_n(X^{**}))$$   .

... have¹²

This follows from ${\mathbb{M}}_n(X^{**})\stackrel{\mathrm{cb}}{=}{\mathbb{M}}_n(X)^{**}$ and the above mentioned formula

$$\Vert T\Vert _{\mathrm{cb}}=\sup\{\Vert\Phi^{(n)}\circ T\Vert _{\... ...i\in\mathit{CB}(Y,{\mathbb{M}}_n),\; \Vert\Phi\Vert _{\mathrm{cb}}\leqslant 1\}$$   .