For which Banach space tensor products is the multiplication in a C*-algebra a linear map? It is well-known that there is a variety of tensor products on the category of Banach spaces, ranging from the injective tensor product to the projective tensor product. What I would like to know is: to which tensor product(s) $\otimes_\alpha$ does the multiplication map $A\times A \to A$ for a $C^*$-algebra $A$ — a bilinear map of Banach spaces satisfying $||xy|| \leq ||x||\,||y||$ — descend to give a linear map $A\otimes_\alpha A \to A$?
 A: I would say that the "optimal" tensor norm gives the $\gamma_2$ or Haagerup tensor product. That norm is given by
$$
  \|x\|_{h} = \inf \Big\{ \Big\| \sum_{j} a_j \, a_j^\ast \Big\|^\frac12
  \,  \Big\| \sum_{j} b_j^\ast b_j \Big\|^\frac12 : x = \sum_{j} a_j \otimes b_j \Big\},
$$
where the infimum runs through all representations of $x$ as a sum of simple tensors. The fact that multiplication for $C^\ast$ algebras lifts to this tensor norm can be seen by applying an operator valued Cauchy-Schwartz inequality. For more, see [Pi: Chapter 5] or other texts on operator space theory.
The projective tensor norm $\otimes_\pi$ works for the multiplication of $C^\ast$-algebras, as pointed out André S, but it is the largest norm and it works for every bilinear form in any Banach space. The natural question is what is the smallest possible norm that allows you to lift the multiplication on $C^\ast$-algebras. This has been studied in connection with the problem of finding an intrinsic characterization of "operator algebras" i.e. non-self adjoin, closed subalgebras of $B(H)$. Varopoulos proved that if the multiplication of a Banach algebra lifts to the tensor norm above (which funnily enough can be defined without using the involution) then the Banach algebra is an operator algebra. This gives you that $\otimes_h$ should be very closed to optimal since the multiplication lifting to that product implies that you can embed your algebra in $B(H)$ closely.
I do not know if that is optimal for Banach algebras (probably not), but it is surely optimal in the category of operator spaces.
[Pi] Pisier, Gilles, Introduction to operator space theory, London Mathematical Society Lecture Note Series 294. Cambridge: Cambridge University Press (ISBN 0-521-81165-1/pbk). vii, 478 p. (2003). ZBL1093.46001.
A: The projective norm does work: Recall, that it is defined on the algebraic tensor product $A \odot A$, as follows: 
$$
 \lVert x \rVert_\pi := \inf \left \{ \sum_{i=1}^n \lVert a_i \rVert \lVert b_i \rVert : x = \sum_{i=1}^n a_i \otimes b_i \right \}, \quad (x \in A \odot A).
$$
Now, let $x \in A \odot A$ and $\varepsilon > 0$. Find a representation $x = \sum_{i=1}^n a_i \otimes b_i$ such that 
$$
 \sum_{i=1}^n \lVert a_i \rVert \lVert b_i \rVert \leq \lVert x \rVert_\pi + \varepsilon.
$$
Then, with $\mu$ the multiplication $A \odot A \to A$, we get 
$$
 \lVert \mu(x)\rVert = \left \lVert \sum_{i=1}^n a_ib_i \right \rVert \leq \sum_{i=1}^n \lVert a_i \rVert \lVert b_i \rVert \leq \lVert x \rVert_\pi + \varepsilon.
$$
Hence, $\lVert \mu(x) \rVert \leq \lVert x \rVert_\pi$, i.e. $\mu$ is contractive on $A \odot A$ equipped with the projective norm.
