# Is this closed subspace of Fréchet space complemented

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 following

Q) 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 ?).

• To be precise, when you write $C^\omega(\Omega)$, do really mean the set of (real or complex?) valued real analytic functions (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.
– TaQ
Jul 29, 2017 at 20:36
• No, 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 $C^\infty(\Omega)$.
– TaQ
Jul 30, 2017 at 8:23
• O.K. thanks (I was thinking of $\Omega=\mathbb{C}$). Corrected ! Jul 30, 2017 at 12:49
• [If you mean the former, then there is no topological complement since the set is dense.]---> No, I meant the complex valued (i.e. holomorphic) functions, I made the ranges precise in the question. Jul 30, 2017 at 12:59

$H(\Omega)$ is not complemented in $C^\infty (\Omega)$ e.g. for the unit disc in $\mathbb C$. This follows from the structure theory of Frechet spaces: The space of smooth functions is isomorphic to $s^\mathbb N$ and has a certain property (DN$_{loc}$) of Vogt. If $H (\Omega)$ were complemented it would also have this property and hence even property (DN) because it has continuous norms. But this is not true for power series spaces of finite type.
• @JochenWengenroth [The space of smooth functions is isomorphic to $s^\mathbb N$]---> Sorry, but what is $s$ ? Jul 31, 2017 at 8:45
• @JochenWengenroth Well ... I feel sort of outsider. It seems, from the paper by P. Domanski and D. Vogt, that $s$ is the space of rapidly decreasing sequences. Aug 1, 2017 at 3:06