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$\def\bbR{\mathbb R}\def\bbC{\mathbb C}\def\scrT{\mathscr T}\def\ssp{\kern.4mm} $More specifically, I ask whether $S$ be a Borel set in the topological space $(\Omega,\scrT)$ in the following situation.

Let $\Omega$ be the set of all real analytic functions $x:\bbR\to\bbR$, and let $X$ be the abstract real vector space obtained when $\Omega$ is equipped with the "obvious" pointwise operations. Let $S$ be the set of all $x\in\Omega$ having a holomorphic extension $\bbC\to\bbC$. For any open set $U$ in $\bbC$ with $\bbR\subseteq U$, let ${\rm F}\,U$ be the real Fréchet space of analytic functions $x:\bbR\to\bbR$ possessing a holomorphic extension $U\to\bbC$, equipped with the topology induced from the compact-open topology of the space of holomorphic functions $U\to\bbC$. Now let $\scrT$ be the strongest locally convex topology for $X$ such that for $F=(X,\scrT)$ and for all the above sets $U$ the identity is a continuous linear map ${\rm F}\,U\to F$.

Letting ${\rm F}_\bbC\,U$ denote the similar real spaces of complex valued real analytic functions, and using Runge's theorem in conjunction with the open mapping theorem applied to the linear map ${\rm F}\,U\times{\rm F}\,U\to{\rm F}_\bbC\,U$ given by $(x,y)\mapsto x + {\rm i}\,y$, if I am not mistaken, one can show that $S$ is a dense linear subspace in $F$.

Added. By Robert Israel's answer we have $S=\bigcap\ssp\big\{\,\bigcup\ssp\big\{\,\bigcap\kern1mm\{\,\{\,x\in\Omega:|\,a_n(x)\,|^{\,n^{\ssp-1}} < j^{\ssp-1}\,\} : k < n \in\mathbb Z\kern2mm\} : k\in\mathbb N_0 \,\} : j\in\mathbb N\kern2mm\} \kern1mm$, a Borel set since every $x\mapsto a_n(x)=(\ssp n\ssp!\ssp)^{\ssp-1}\,{\rm D}^{\,n\ssp}x\ssp(0)$ is a continuous linear functional on $F\,$.

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  • $\begingroup$ What is $k$ in the $\bigcap\bigcup\bigcap$ expression ? $\endgroup$
    – Todd Leason
    Commented Dec 12, 2015 at 10:18
  • $\begingroup$ @Todd Leason: I do not understand you question. It is just a bound variable like $j$, $n$ and $x$, too. $\endgroup$
    – TaQ
    Commented Dec 12, 2015 at 23:24
  • $\begingroup$ Oh, I see that $k$ is used to express that $|a_n(x)|^{n-1} < j^{-1}$ should hold for almost all $n$ (for fixed $j$). Now it's clear. Thanks. $\endgroup$
    – Todd Leason
    Commented Dec 13, 2015 at 1:33

1 Answer 1

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$f$ has an entire extension iff the radius of convergence of the Maclaurin series of $f$ is $+\infty$, i.e. iff $\limsup_{n \to \infty} |a_n|^{1/n} = 0$ where $a_n$ are the coefficients of that series. The coefficients can be defined as limits of expressions defined in terms of point evaluations (finite-difference formulas), and those expressions are continuous functions on $\Omega$. So yes, it is a Borel set, in fact an $F_{\sigma\delta\sigma\delta}$.

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  • $\begingroup$ Thank you very much for the answer. I had in my mind a vague idea that something related to the convergence radius could be used but I couldn't formulate it explicitly. $\endgroup$
    – TaQ
    Commented Dec 12, 2015 at 3:21

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