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 $(A(\mathbb D), H^\infty(\mathbb D))$?
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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.
answered Apr 7, 2019 at 0:57
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$\begingroup$ 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 $\endgroup$ Commented Apr 7, 2019 at 1:07
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1$\begingroup$ I edited the wrong wikipedia item. so you may delete your two lines here. $\endgroup$– rayCommented Apr 9, 2019 at 15:58
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$\begingroup$ @ray Thank you! I have edited my answer $\endgroup$ Commented Apr 9, 2019 at 17:45