Let $A(\mathbb{D})$ denotes the disc algebra and $H^{\infty}(\mathbb{D})$ denotes the Banach space of bounded holomorphic functions on $\mathbb{D}.$ I have the following question;
is the subspace $A(\mathbb{D})+\{\varphi(z^2):\varphi\in H^{\infty}(\mathbb{D})\}$ equal to $H^{\infty}(\mathbb{D})$?
No, this is false. Let $f\in H^\infty(\mathbb{D})$ be a function that is continuous on $\partial D\setminus \{1\}$ and does not have a limit as at $z\to 1 $. (E.g., $f$ could be a conformal map onto a domain one with one nontrivial prime end.) If $f=g+\phi(z^2)$ with $g\in A(\mathbb{D})$, then $\phi$ must be discontinuous at $1$; but then $\phi(z^2)$ is also discontinuous at $-1$, where $f$ is continuous.