# Separable subalgebras of non-separable reflexive Banach algebras

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?

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 $$A_\omega$$.