MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
up vote 9 down vote accepted

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 $U_{j+1}$.

A sketch of the construction in the finite dimensional setting can be found, for instance, in A Primer of Real Analytic Functions by Krantz and Parks. Hopefully, it generalizes to the case of Banach spaces in a straightforward way.

[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.]

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.