I am trying to make sense of some infinite dimensional topology.
Let $\mathcal{A}_r$ to be the set of all real-analytic functions $\mathbb{C}^2\to\mathbb{C}^2$, whose radius of convergence is at least $r$ and let $\mathcal{M}_r=\{f\in\mathcal{A}_r:\det f'\equiv 1 \}$.
I have a functional $\Theta:\mathcal{M}_r\to\mathbb{C}$ such that for any finite analytic parametric family $\mathcal{F}:\mathbb{C}^n\to \mathcal{M}_r$ the function $\Theta \circ \mathcal{F} : \mathbb{C}^n \to \mathbb{C}$ is analytic in the usual sense.
What I would like to say is that $\Theta$ is actually analytic on $\mathcal{M}_r$ or even a subset of it.
As far as I understand, if $\mathcal{M}_r$ was a Banach space, this would have been solved. But since it is not, I am trying to understand what I can get out of this situation.
So, does the supremum norm turn $\mathcal{A}_r$ into a Banach space?
I am not sure what will be the most convenient topology on $\mathcal{A}_r$ for this.
Is $\mathcal{M}_r$ an analytic manifold?
The answer seems to be yes by the implicit function theorem, but I don't trust my knowledge enough to claim that.
Is $\Theta$ an analytic functional on $\mathcal{M}_r$?
Again the answer seems to be yes, even though I don't really know exactly why. But I guess I should say that there is probably no way to extend $\Theta$ on $\mathcal{A}_r$. At least none that I can think of anyway.