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Do completely bounded maps on an operator space have a completely contractive Banach algebra structure?

Let $X$ be an operator space and $CB(X)$ be the set of all completely bounded linear maps $f: X \to X$. Note that $CB(X)$ becomes a Banach algebra for the composition of operators.

Is the multiplication $$CB(X)\odot CB(X)\to CB(X): f\otimes g \mapsto f \circ g$$ completely contractive with respect to the projective tensor product norm on $CB(X)\odot CB(X)$? I.e. is $CB(X)$ a completely contractive Banach algebra?

Andromeda
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