Let $A$ and $B$ denote $C^{\ast}$-algebras. Let $\lVert\cdot\rVert\_h$ and $\lVert\cdot\rVert\_{\text {max}}$ denote the Haagerup norm and max $C^\*$-norms on $ A \otimes B$, respectively. I am looking for a reference/proof for the following result:



> 
> $\lVert\cdot\rVert\_{\text{max}} \leq \lVert\cdot\rVert\_h$.
> 
> 
> 



Need reference for: $\lVert\cdot\rVert_{\text{max}} \leq \lVert\cdot\rVert_h$
Let $v \in A \otimes B$, where $A\otimes B$ is the algebraic tensor product.


The Haagerup norm of $v$ is the infimum of the expressions $\sqrt{\|\sum\_{i=1}^{r} x\_{i} x\_{i}^{\ast}\|}\cdot \sqrt{\|\sum\_{i=1}^{r} y\_{i}^{\ast} y\_{i} \|}$, taken over all decompositions $v=\sum\_{i=1}^{r} x\_{i} \otimes y\_{i}$ ($r$ can be arbitrarily large).


On the other hand, the maximal norm is the supremum of the norm $\|\sum\_{i=1}^{r} \pi(x\_i) \sigma(y\_{i})\|\_{B(H)}$ over all pairs of representations $\pi: A \to B(H)$, $\sigma: B \to B(H)$ with commuting ranges. Note that $\sum\_{i=1}^{r} \pi(x\_i) \sigma(y\_{i})$ can be interpreted as the product of the row $(\pi(x\_1),\dots, \pi(x\_r))$ and the column $(\sigma(y\_1),\dots, \sigma(y\_r))^{T}$, so we get can estimate $\|\sum\_{i=1}^{r} \pi(x\_i) \sigma(y\_{i})\|\_{B(H)}$ by


$
\sqrt{\|\sum\_{i=1}^{r} \pi(x\_{i} x\_{i}^{\ast})\|\_{B(H)}} \cdot \sqrt{\|\sum\_{i=1}^{r} \sigma(y\_{i}^{\ast} y\_{i})\|\_{B(H)}}.
$


As representations of $C^{\ast}$-algebras are contractive, this is clearly not greater than the quantity defining the Haagerup norm.


It might be useful to note that we did not even have to use the fact the representations $\pi$ and $\sigma$ had commuting ranges.


