Let $p\in[1,\infty)$. An $L^p$ operator algebra refers to a Banach algebra that is isometrically isomorphic to a closed subalgebra of $B(L^p(X,\mu))$ for some ($\sigma$-finite) measure space $(X,\mu)$. I would like to know what is the "right" way to handle direct sums of two such algebras, and also unitizations of such algebras. More precisely, I would like to know what is the right norm to put on the direct sum and the unitization. In particular, can the direct sum or the unitization be realized as $L^p$ operator algebras?
Actually, I think that the $\ell_p$-direct sum of two such algebras should be represented on the $\ell_p$ direct sum of the two $L^p$ spaces that the two algebras are represented on. But there are probably details to be checked.
Just like for operator algebras (this is the case $p=2$), direct sum of two $L^p$-operator algebras is taken with respect to the supremum norm: if $A$ is represented on $L^p(\mu)$ and $B$ is represented on $L^p(\lambda)$, then $A\oplus B$, with the supremum norm, is represented as diagonal operators on the $L^p$-direct sum $L^p(\mu)\oplus L^p(\lambda)$. The point is that the norm of a diagonal operator on a direct sum is the supremum of the norms of the individual operators.
The situation with the unitization is much more subtle. In general, the unitization could have more than one norm which makes it into an $L^p$-operator algebra. For example, if $A$ is concretely represented on $L^p(\mu)$, you could represent $\widetilde{A}$ on the same $L^p$-space by sending the unit to the identity on $L^p(\mu)$. What is not clear (and probably not true), is that this definition of the norm on $\widetilde{A}$ is independent of the representation of $A$. If $A$ has a contractive approximate identity, then there is a canonical choice of a norm on $\widetilde{A}$ making it into an $L^p$-operator algebra, and this norm does not depend on the concrete representation of $A$.