metric on the space of real analytic functions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T12:19:24Z http://mathoverflow.net/feeds/question/53557 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/53557/metric-on-the-space-of-real-analytic-functions metric on the space of real analytic functions Mirko 2011-01-27T23:42:17Z 2011-01-31T20:32:31Z <p>Hello, this question may be simple but I couldn't find a reference.</p> <p>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.</p> <p>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.</p> <p>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.</p> <p>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?</p> <p>Regards, Mirko</p> http://mathoverflow.net/questions/53557/metric-on-the-space-of-real-analytic-functions/53623#53623 Answer by Andrey Rekalo for metric on the space of real analytic functions Andrey Rekalo 2011-01-28T15:26:27Z 2011-01-29T11:54:44Z <p>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. </p> <p>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. </p> <p>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 $U_{j+1}$.</p> <p>A sketch of the construction in the finite dimensional setting can be found, for instance, in <a href="http://books.google.com/books?id=i4vw2STJl2QC&amp;printsec=frontcover&amp;dq=a+primer+on+real+analytic&amp;hl=en&amp;ei=VdJCTaTOFcTsOYqp7ZUC&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=1&amp;ved=0CCoQ6AEwAA#v=onepage&amp;q&amp;f=false" rel="nofollow"><em>A Primer of Real Analytic Functions</em></a> by Krantz and Parks. Hopefully, it generalizes to the case of Banach spaces in a straightforward way. </p> <p>[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.]</p>