A question about converging derivatives

Solution 1:

You can apply the following theorem from A theorem on analytic functions of a real variable by R. P. Boas, Jr. The proof is an application of the Baire category theorem and I am copying my quote from a previous question.

Let $f(x)$ be a function of class $C^\infty$ on $a\leqq x\leqq b.$ At each point $x$ of $[a, b]$ we form the formal Taylor series of $f(x),$ $$\sum\limits_{k=0}^\infty \frac{f^{(k)}(x)}{k!}(t-x)^k.$$

This series has a definite radius of convergence, $\rho(x),$ zero, finite, or infinite, given by $1/\rho(x)=\overline{\lim}_{k\to\infty}|f^{(k)}(x)/k!|^{1/k}.$ The function $f(x)$ is said to be analytic at the point $x$ if the Taylor development [I believe this just means Taylor series - Dap] of $f(x)$ about $x$ converges to $f(t)$ over a neighborhood $|x-t|<c,$ $c>0,$ of the point; $f(x)$ is analytic in an interval if it is analytic at every point of the interval. [...]

THEOREM A. If there exists a number $\delta>0$ such that $\rho(x)\geqq\delta$ for $a\leqq x\leqq b,$ $f(x)$ is analytic in $[a, b].$

Given your condition on $f,$ for each $x,$ the sequence $f^{(n)}(x)$ is bounded so the Taylor series has infinite radius of convergence. By Theorem A, this is enough to force $f$ to be analytic and in fact entire. We then have bound the convergence uniformly on compact sets, e.g. for $|x|\leq M$ and $n,m\geq N,$

\begin{align*} |f^{(n)}(x)-f^{(m)}(x)| &\leq \sum_{k}\tfrac{1}{k!}M^k|f^{(n+k)}(0)-f^{(m+k)}(0)|\\ &\leq e^M\sup_{n\geq N}|f^{(n)}(0)-g(0)|\\ &\to 0. \end{align*}