In the hope of completing the rich tapestry of complemented (or not) topological vector subspaces, I would like to know (maybe it is immediate for specialists) whether the space of analytic functions is complemented within the space of infinitely differentiable ones. I begin with the one-variable case

... and make this precise.

Let $\Omega\subset \mathbb{C}$ be an open subset. We consider

$$ H(\Omega)=C^\omega(\Omega;\mathbb{C})\subset C^\infty(\Omega;\mathbb{C}) $$ the large one being endowed with the standard topology defined by the seminorms $$ p_{\,n,B}=sup_{\ 0\leq |\alpha|\leq n\atop t\in B}|D^\alpha(f)[t]|\ . $$ where $n\in \mathbb{N}, \alpha\in \mathbb{N}^2$, $B$ is a relatively compact open subset of $\Omega$ and the bi-indexed derivative is $$ D^\alpha:=(\frac{\partial}{\partial x})^{\alpha[1]}(\frac{\partial}{\partial y})^{\alpha[2]}\ . $$ I know that the subspace $H(\Omega)=C^\omega(\Omega;\mathbb{C})$ is complete and then closed for this (standard) topology. My question is the followingQ) Is there a known closed complement of it i.e. a decomposition $$ C^\infty(\Omega;\mathbb{C})=C^\omega(\Omega;\mathbb{C})\oplus W=H(\Omega)\oplus W $$

where $W$ is closed ? (maybe the projector is an integro-differential operator ?) at least for some particular domains $\Omega$ ?

**Remark** i) This question is a reformulation of
this one
in MSE where it did not receive a complete answer.

ii) With the given topology, $C^\infty(\Omega;\mathbb{C})$ and $H(\Omega)=C^\omega(\Omega;\mathbb{C})$ are m-convex Fréchet algebras. Maybe (if possible) $W$ could have some algebraic structure (ideal ?).

real analyticfunctions (which is the usual convention) or do you mean the set of (complex valued) holomorphic functions, for which the usual notation would be $H(\Omega)$? If you mean the former, then there is no topological complement since the set is dense. And, if you really mean the usual topology of $C^\infty(\Omega)$, then $B\subset\Omega$ should be required to be relatively compact. Bounded generally gives a strictly stronger topology. $\endgroup$it is not. If e.g. $\Omega$ is the open unit disk, then $B=\Omega$ satisfies your requirement of being bounded open, and you get uniform convergence on $\Omega$ which is not the usual topology of $C^\infty(\Omega)$. I mean dense in your "large" space $C^\infty(\Omega)$. $\endgroup$