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 
$(A(\mathbb D), H^\infty(\mathbb D))$? 



Is the disk algebra a complemented subspace of the algebra of bounded analytic functions?
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*](https://en.wikipedia.org/wiki/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.


