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I have been confused with this problem for weeks now. Suppose I have Banach spaces $E$ and $F$ and a sequence of functions $f_{n}: U \subset E \to F$, where $U$ is open and nonempty. Let $x \in U$ be fixed and let $\Omega \subset \mathbb{C}$ be a neighborhood of the origin such that $\lambda x \in U$ for every $\lambda \in \Omega$. Suppose that the functions $\Omega \ni \lambda \mapsto f_{n}(\lambda x)$ are all smooth and consider their Taylor expansions: $$f_{n}(\lambda x) = \sum_{k=0}^{\infty}\frac{\lambda^{k}}{k!}f_{n}^{(k)}(x) \tag{1}\label{1}$$ with: $$f_{n}^{(k)}(x) = \frac{d^{k}}{d\lambda^{k}}f_{n}(\lambda x)\bigg{|}_{\lambda = 0}.$$ What I want to understand is the following. Suppose that the derivatives of these functions are all uniformly bounded, that is, there exists a constant $C>0$ which is independent of $n$ such that: $$|f_{n}^{(k)}(x)| \le C k! \tag{2}\label{2}$$ Under these conditions, can I conclude anything about the existence of the limit $\lim_{n\to \infty}f_{n}(x)$?

The reason why I am asking this question is that my intuition says that (\ref{2}) implies that the radius of convergence of (\ref{1}) does not depend on $n$, so in particular it does not shrink to zero as $n$ grows. This suggests me that the limit could be taken. But of course, this also looks a bit strong and I haven't found anything in the literature that supports this idea.

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  • $\begingroup$ This expansion is a bit strange. You want $x\in U$ to be fixed, but then $\lambda x$ only varies in a nbd of $\lambda$, so you have essentially maps $\Omega\subset\mathbb C\to F$. Shouldn't you consider $$f_{n}( x+u) = \sum_{k=0}^{\infty}\frac{1}{k!}D^kf_{n} (x)[u^k]$$ ($D^kf_n(x)\in L^k_{sym}(E^k;F)$ is the $k$-th order differential as a symmetric $k$-linear map) $\endgroup$
    – Pietro Majer
    Commented Nov 12 at 11:13
  • $\begingroup$ @PietroMajer I really meant my Taylor expansion. In fact, I am considering a function $\Omega \ni \lambda \mapsto f_{n}(\lambda x)$, where $\Omega \subset \mathbb{C}$ is small enough so that $\lambda x \in U$ for all $\lambda \in \Omega$. $\endgroup$
    – InMathweTrust
    Commented Nov 12 at 12:15
  • $\begingroup$ I edited the post. Hope it is better explained now. $\endgroup$
    – InMathweTrust
    Commented Nov 12 at 12:19
  • $\begingroup$ If $1$ and $0$ are not in the same connected component of $\Omega$ (in particular if $x$ and $0$ lie in different components of $U$), you can define $f_n=0$ near $0$ and $f_n=n$ near $x$. In that case I fail to see why there should exist a link between what happens near $x$ and near $0$. $\endgroup$
    – Loïc Teyssier
    Commented Nov 12 at 14:54
  • $\begingroup$ @LoïcTeyssier thanks for your comment. I am trying to mimic the definition usually used for directional derivatives. In the case of real Banach spaces, this would be $(-\epsilon, \epsilon) \ni \lambda \mapsto f(\lambda x)$. Of course, in this case there is only one connected component. So maybe I should require $\Omega$ to be connected in addition (or treat the underlying Banach spaces as defined over $\mathbb{R}$). $\endgroup$
    – InMathweTrust
    Commented Nov 12 at 15:04

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