Let $A$ be a non-separable reflexive Banach algebra. Every separable subspace of $A$ is contained in a separable 1-complemented subspace [Lindenstrauss,1966]. It is straightforward to show that every separable subalgebra is contained in a separable subalgebra $W$ of the form $$W=\bigcup_{n=1}^{\infty} E_n$$ where $(E_n)$ is an increasing sequence of separable 1-complemented subspaces of $A$.
Question: Is it true that every separable subalgebra of $A$ is contained in a separable complemented subalgebra?
Best Answer
I am inclined to say yes. This is because reflexive Banach spaces have projectional resolutions of the identity, which are increasing ordinal-indexed nets of contractive, commuting projections that exhaust the whole space and satisfy certain compatibility conditions. From this, it follows that separable subspaces of reflexive Banach space are contained in 1-complemented separable subspaces as you mentioned, but we got actually more.
Now, let $A$ be a separable subalgebra of a reflexive Banach algebra $B$. Denote by $\langle S \rangle$ denote the closed subalgebra generated by $S\subset B$. Take $A^1$ to be a separable subspace of $B$ containing $A$ that is 1-complemented by a projection $P^1$ from some fixed PRI and set $A_1 = \langle A^1 \rangle$. Let $A^2$ be a separable subspace of $B$ containing $A_1$ that is 1-complemented by a projection $P^2$ from the same PRI and set $A_2 = \langle A^2 \rangle$. Continue this process recursively so that you get intertwined sequences $A^1 \subset A_1 \subset A^2 \subset A_2 \ldots$. In particular, the unions of $(A_n)$ and $(A^n)$ have the same closure, call it $A_\omega$. Readily, $A_\omega$ is a closed subalgebra being the closure of an increasing chain of subalgebras. Since projections from a PRI commute, $(P^n)$ converge in the weak operator topology to a projection, which looks like a projection onto $A_\omega$.