# metric on the space of real analytic functions

Hello, this question may be simple but I couldn't find a reference.

Let $E$,$F$ be real Banach spaces and $\Omega\subset E$ be a bounded domain and let $C_b^{\omega}(\Omega,F)$ be the vector space of bounded real analytic functions from $\Omega$ to $F$. Now I would like to know if there is a natural way to define a metric on $C_b^\omega(\Omega,F)$, that makes the space complete.

Concretely, I have a series of real analytic functions that converge uniformly as well as their Frechet derivatives and now I would like to know if their limit is analytic again.

My first idea was to show that $C_b^\omega(\Omega,F)$ is a closed subspace of $C_b^\infty(\Omega,F)$. Here $C_{b}^{\infty}(\Omega,F)$ is the set of infinitely Frechet-differentiable functions equipped with the usual set of seminorms (i.e. $\Vert f\Vert_k:=\Vert D^k f\Vert_\infty$) defining a Frechet space and thus a metric $d(f,g):=\sum_{k=1}^\infty 2^{-k}\frac{\Vert f-g\Vert_k}{1+\Vert f-g\Vert_k}$ which makes $C_{b}^{\infty}(\Omega,F)$ a complete metric space.

Actually, I have no idea if this is true at all. Could someone please confirm if this is the right thing to prove or not?

Regards, Mirko

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Think in terms of the complex extensions of your real analytic functions. Convergence to an analytic limit means that the domain of complex extension does not shrink to $\Omega$. Then I feel what you are looking for reduces to uniform convergence on compact subsets of this extension domain (at least if E is finite dimensional). If you want to express this convergence only in terms of the traces at $\Omega$, I'm afraid you will need to put some a priori bound on the growth of rate of derivatives (like $\|D^k f\|_\infty \le C^k k!$, $C$ depending on the neighbourhood of the point) –  Piero D'Ancona Jan 28 '11 at 0:16
It's not a closed subspace of smooth functions (indeed, it's dense in it), and moreover it isn't a metrisable space as it is a union of a countable family of Banach spaces, not a projective limit. I'm pretty sure that it isn't complete either (since an arbitrary analytic function should be a limit of bounded ones, but I'd need to think a bit harder about the topologies involved to be sure). –  Loop Space Jan 28 '11 at 11:39

The problem is nontrivial already in the finite dimensional case $E= \mathbb R^d$, $F=\mathbb R$. The space $C^{\omega}(\Omega)$ of real-valued real analytic functions on the open bounded set $\Omega\subset \mathbb R^d$ does not have any obvious or natural metric which would make it a Fréchet space.
The good news is that there is a "canonical" topology which renders $C^{\omega}(\Omega)$ as a complete (reflexive nuclear separable) space. In fact, it is natural to endow $C^{\omega}(\Omega)$ with either an inductive limit or a projective limit topology but these two are equivalent on $C^{\omega}(\Omega)$ as was shown by Martineau in 1966.
For practical purposes, the topology can be described following the suggestion of Piero D'Ancona in his comment above. Let $\{U_j\}_{j\in\mathbb N}$ be a monotonically decreasing sequence of open sets of $\mathbb C^d$ such that $\Omega=\bigcap U_j$. Let $\{h_j\}_{j\in\mathbb N}$ be a sequence of bounded holomorphic functions $h_j:U_j\to\mathbb C$ such that $h_j|_{U_k}=h_k$ for $k\geq j$. Then a subbase element of the topology on $C^{\omega}(\Omega)$ has the form $$\mathcal V_{j, K}=\left\{f\mbox{ is real analytic on }\Omega:\ \sup_{x\in K} \left|\partial^{\alpha} f\right|\leq C_j[\delta_j(K)]^{-|\alpha|}\ \mbox{ for every }\alpha\in\mathbb N^{d}_{0}\right\},$$ where the set $K\subset\Omega$ is compact, $\delta_j(K)=\mbox{dist}\{K,\partial U_{j+1}\}$ and $C_j$ is a constant which depends on the supremum of $h_j$ on $U_{j+1}$.
[EDIT. Concerning your specific question whether the limit of a sequence of real analytic functions is itself an analytic function. Let $f\in C^\infty(\mathbb T)$ be a periodic smooth but non-analytic function. Then the partial Fourier sums $S_N f$ converge to $f$ in the uniform metric with all their derivatives.]