The following is an abstraction of a more specific problem I've been grappling with, so I might give some extra unnecessary information. (The `bounded approximate
left identities' assumption is probably not necessary, but it is definitely
the case in the specific problem that this question is abstracted from).


Let $A$ and $B$ be Banach algebras, both having bounded approximate
left identities. We assume that we have non-degenerate, *isometric*
representations $\pi:A\to B(X)$ and $\rho:B\to B(Y)$ on some Banach
spaces $X$ and $Y$.


We can define the following algebra representation 
$$
\pi\otimes\rho:A\otimes B\to B(X\hat{\otimes}Y)
$$
(where $X\hat{\otimes}Y$ denotes the projective tensor product) in
the usual way, for all $a\in A$, $b\in B$, $x\in X$ and $y\in Y$,
by 
$$
\left(\pi\otimes\rho(a\otimes b)\right)x\otimes y:=\pi(a)x\otimes\rho(b)y.
$$
Now, since $\pi$ and $\rho$ were assumed to be isometric, we can
prove that the map $\pi\otimes\rho$ is injective and satisfies 
$$
||\pi\otimes\rho(a\otimes b)||\_{op} = ||a||\_A ||b||\_B
$$
for all elementary tensors $a\otimes b\in A\otimes B$, where the
norm on the left hand side denotes the operator
norm on $B(X\hat{\otimes}Y)$.


My question is the following: Does the map 
$$
\sum\_{i}a\_{i}\otimes b\_{i}\mapsto\left\Vert \pi\otimes\rho\left(\sum\_{i}a\_{i}\otimes b\_{i}\right)\right\Vert \_{\mbox{op}}
$$
define a reasonable cross norm on $A\otimes B$ (Ryan's book, Ch. 6)? Seeing that it is dominated by the projective tensor norm on $A\otimes B$ is easy, but is it bounded from below by the injective tensor norm on $A\otimes B$?


Note that we can extend bounded functionals on $A$ to bounded functionals
on $B(X)$ by using Hahn Banach and that the representations are isometric;
similarly with $B$. Still, using this, a proof still seems just out
of reach for want of being able to extend functionals on $B(X)\hat{\otimes}B(Y)$
to functionals on $B(X\hat{\otimes}Y)$ because the norms on these algebras
can't be favourably related (projective norm on the one, operator norm on the
other). 


Any ideas, helpful references or counterexamples showing how this might fail will be greatly appreciated.



Reasonable crossnorm on Banach algebra tensor product constructed from isometric non-degenerate representations
Let $D\_A\subseteq A^\*$ be the functionals of the form $\mu(\pi(\cdot)x)$, for $x\in X, \mu\in X^\*, \|x\|\leq 1, \|\mu\|\leq 1$. As $\pi$ is an isometry, Hahn-Banach shows that the convex hull of $X$ is weak$^\*$-dense in the closed ball of $A^\*$, say $A^\*\_{[1]}$.


Similarly for $D\_B$ using $\rho$. It's clear (\*) that


$$ |(\mu\_A\otimes\mu\_B)\tau| \leq \| (\pi\otimes\rho)(\tau) \|\_{op} \qquad (\tau\in A\otimes B, \mu\_A\in D\_A, \mu\_B\in D\_B). $$
However, this inequality is preserved by taking convex combinations of the elements of $D\_A$ and $D\_B$, and by taking weak$^\*$-closures. But that would then show that
$$ |(\mu\_A\otimes\mu\_B)\tau| \leq \| (\pi\otimes\rho)(\tau) \|\_{op} \qquad (\tau\in A\otimes B, \mu\_A\in A^\*\_{[1]}, \mu\_B\in B^\*\_{[1]}), $$
and that's all you need to show, I think?


Why is `(*)` true? Given $x\in X,\mu\in X^\*,y\in Y,\lambda\in Y^\*$, we have that $x\otimes y\in X\widehat\otimes Y$ of norm $\|x\|\|y\|$, and also $\mu\otimes\lambda \in (X\widehat\otimes Y)^\*$ (this is $w\otimes z\mapsto \mu(w)\lambda(z)$) has norm $\|\mu\| \|\lambda\|$. Then $(\mu\_A\otimes\mu\_B)\tau = (\mu\otimes\lambda)((\pi\otimes\rho)(\tau)(x\otimes y))$. Actually, this shows that the result would remain true if you replaced the projective tensor norm on $X\otimes Y$ be any reasonable cross-norm.


