Complement of the space of analytic functions. Let $C^\infty(\mathbb{C})$ be the space of functions $\mathbb{C}\to \mathbb{C}$ with derivatives at all orders endowed with the topology of compact convergence of all the derivatives i.e. defined by the seminorms ($n\in \mathbb{N}, \alpha\in \mathbb{N}^2$ and $B$ a bounded open subset)
$$
p_{\,n,B}=sup_{\ 0\leq |\alpha|\leq n\atop t\in B}|D^\alpha(f)[t]|\ . 
$$
I know that the subspace $C^\omega(\mathbb{C})\subset C^\infty(\mathbb{C})$ (analytic functions on all $\mathbb{C}$, called entire functions) is complete and then closed. 

Q1) Is there a known closed complement of it i.e. a decomposition
  $$
C^\infty(\mathbb{C})=C^\omega(\mathbb{C})\oplus W
$$
  where $W$ is closed ? 
Q2) Can we replace the whole $\mathbb{C}$ by a domain $\Omega\subset \mathbb{C}$ ? 

 A: We consider the $\sup$ norm.


*

*By convolution with the Gaussian kernel $g_n(x,y)=n^2 e^{-\pi n^2(x^2+y^2)}$ the bounded real analytic functions on $|x+iy| < r$ are dense in the continuous functions on $|x+iy | \le r$, thus the polynomials in $x,y$ are dense too.

*A polynomial $f(x+iy) = \sum_{n+m \le N} a_{n,m} x^n y^m$ is complex analytic iff $$\partial_x f = \sum_{n+m \le N,n\ge 1} a_{n,m} n x^{n-1} y^m = -i\partial_y f = -i\sum_{n+m \le N,m\ge 1} a_{n,m} x^{n} m y^{m-1}$$ ie. iff $n a_{n,m-1} = -i m a_{n-1,m}$ ie. $a_{k-m,m} = a_{k,0} {k \choose m}i^m$

*The polynomial $$h(x+iy) = \sum_{k\le N} (x+iy)^k  \frac{1}{k+1}\sum_{m=0}^{k} \frac{a_{k-m,m}}{ {k \choose m}i^m}$$
is complex-analytic and if $f$ was complex analytic then $f=h$.

*The map $T : f \mapsto h$ is $\mathbb{C}$ linear and it is a projection : the polynomials are the direct sum of $\text{Im}(T)$
(complex analytic part) and $\text{ker}(T)$ (the non-complex analytic part).

*The complex analytic functions are closed for the uniform convergence on compacts, so taking the closure for some norms or semi-norms you can extend this decomposition for the continuous functions or the $C^\infty$ functions.
