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?

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? This is claimed in the paper New tensor products of C-algebras and characterization of type I C-algebras as rigidly symmetric C*-algebras.