Complete boundedness

For the definition of the completely bounded bilinear maps we need the amplification $\Phi^{(n)}$, the linearization $\tilde\Phi : X \otimes Y \rightarrow Z$. and the tensor matrix multiplication $x \odot y$ of operator matrices $x$, $y$. A bilinear mapping $\Phi: X \times Y \rightarrow Z$, $n\in{\mathbb{N}}$ is called completely bounded if

$$\Vert\Phi\Vert _\mathrm{cb}:= \sup \Vert \Phi^{(n)}(x, y)\Vert < \infty$$

where $n\in{\mathbb{N}}$, $x \in \mathrm{Ball}(M_{n}(X))$, $y \in \mathrm{Ball}(M_{n}(Y))$. The norm $\Vert\Phi\Vert _\mathrm{cb}$ equals the norm $\Vert\tilde\Phi\Vert _\mathrm{cb}$ of the linearization

$$\tilde\Phi : X \otimes_h Y \rightarrow Z$$

on the Haagerup tensor product. Furthermore, the norms $\Vert\Phi\Vert _n$ are obtained using the tensor matrix products of all ⁴⁹ rectangular matrices of $n$ rows resp. $n$ columns:

$$\Vert\Phi\Vert _n := \sup \Vert \Phi^{(n,l)}(x , y)\Vert$$

where $l \in {\mathbb{N}}$, $x \in \mathrm{Ball}(M_{n,l}(X))$, $y \in \mathrm{Ball}(M_{l,n}(Y))$. We have

$$\Vert\Phi\Vert _\mathrm{cb}= \sup \left\{ \Vert\Phi\Vert _n \ : \ n \in {\mathbb{N}}\right\}.$$

The norm $\Vert\Phi\Vert _n$ equals the norm $\Vert\tilde\Phi^{(n)}\Vert$ of the amplification of the linearization

$$\tilde\Phi^{(n)} : M_n(X \otimes_h Y) \rightarrow M_n(Z)$$

on the Haagerup tensor product. A bilinear form $\Phi$ is already seen to be completely bounded if $\Vert\Phi\Vert _1 < \infty$. Then we have $\Vert\Phi\Vert _\mathrm{cb}= \Vert\Phi\Vert _1$.⁵⁰ $\mathit{CB}(X \times Y; Z)$ denotes the operator space consisting of completely bounded bilinear maps. One obtains a norm on each matrix level using the identification

$$M_n(\mathit{CB}(X \times Y;Z)) = \mathit{CB}(X \times Y; {\mathbb{M}}_n(Z)).$$

Corresponding to the completely bounded bilinear maps we have the linear maps which are completely bounded on the Haagerup tensor product . The identification

$$\mathit{CB}(X \times Y; Z) \stackrel{\mathrm{cb}}{=}\mathit{CB}(X \otimes_h Y; Z)$$

holds completely isometrically. Completely bounded bilinear mappings are in particular jointly completely bounded. The embedding $\mathit{CB}(X \times Y; Z) \subset \mathit{JCB}(X \times Y;Z)$ is a complete contraction. The transpose $\Phi^t(y,x) := \Phi(x,y)$ of a completely bounded bilinear mapping $\Phi$ in general is not completely bounded.⁵¹ For completely bounded bilinear (and, more generally, multilinear) maps $\Phi \in \mathit{CB}(A \times B; B(\H))$ we have some generalizations of Stinespring's representation theorem.

Footnotes

... all⁴⁹

Note that the norm of the bilinear map

$$\Phi^{(n)}: M_n(x) \otimes M_n(Y) \rightarrow M_n(Z)$$

in general is smaller than the norm $\Vert\Phi\Vert _n$.

....⁵⁰

More generally, the equation $\Vert\Phi\Vert _\mathrm{cb}= \Vert\Phi\Vert _1$ holds for bilinear maps with values in a commutative C$^*$-algebra $A$ since every bounded linear map taking values in $A$ is automatically completely bounded and $\Vert\Phi\Vert _{\mathrm{cb}}=\Vert\Phi\Vert$ [Loe75, Lemma 1]. For bilinear maps $\Phi : X \times Y \rightarrow M_n(A)$ we have $\Vert\Phi\Vert _\mathrm{cb}= \Vert\Phi\Vert _n$.

...⁵¹

To this corresponds the fact that the Haagerup tensor product is not symmetric.