# Is the disk algebra a complemented subspace of the algebra of bounded analytic functions?

It is well known that the disk algebra (viewed as an algebra on the circle) is uncomplemented in $$C(\mathbb T)$$. What can be said about the pair $$(A(\mathbb D), H^\infty(\mathbb D))$$?

Call the smaller algebra $$A$$ and the larger algebra $$B$$ for convenience. Here is a ludicrously over-the-top way to prove that $$A$$ is not complemented in $$B$$: invoke Bourgain's result that $$B$$ is a Grothendieck space, which means that every bounded linear map from $$B$$ to any separable Banach space $$E$$ must be weakly compact. So if there were a continuous linear projection of $$B$$ onto $$A$$, it would be weakly compact, and composing this with the inclusion map $$A\hookrightarrow B$$ we could deduce that the identity map on $$A$$ is weakly compact. But this is impossible since $$A$$ is not reflexive (as it contains a closed copy of the non-reflexive Banach space $$c_0$$).
Of course this argument shows, more generally, that any complemented subspace of $$B$$ has to be non-separable.
• I wonder if it is possible to use some kind of averaging argument to show that if there is a bounded linear projection from $B$ onto $A$ then $A$ would be complemented as an $A$-module inside $B$, and then obtain a contradiction by looking at suitable Blaschke products – Yemon Choi Apr 7 '19 at 1:07