# A question about converging derivatives

Suppose $$f \in C^{\infty}(\mathbb{R})$$ and $$\forall x \in \mathbb{R} \text{ } \exists \lim_{n \to \infty} f^{(n)}(x) = g(x)$$. Does this mean that $$\exists a \in \mathbb{R} \forall x \in \mathbb{R} \text{ } g(x) = ae^x?$$

If $$f^{(n)}$$ converge to $$g$$ uniformly, then it does, as $$\forall x_0 \in \mathbb{R} \implies \begin{split} g’(x_0) &= \lim_{x \to x_0} \frac{g(x) - g(x_0)}{x - x_0} \\ &= \lim_{x \to x_0} \lim_{n \to \infty} \frac{f^{(n)}(x) - f^{(n)}(x_0)}{x - x_0} \\ &=\lim_{n \to \infty} \lim_{x \to x_0} \frac{f^{(n)}(x) - f^{(n)}(x_0)}{x - x_0} \\ & = \lim_{n \to \infty} f^{(n + 1)}(x_0) \\ & = \lim_{n \to \infty} f^{(n)}(x_0) \\ & = g(x_0) \end{split}$$ and all solutions of a differential equation $$g’(x) = g(x)$$ have the form $$ae^x$$ for some $$a \in \mathbb{R}$$.

However, this proof does not work in case, when $$f^{(n)}$$ does not converge uniformly, as in this case $$\lim_{x \to x_0} \lim_{n \to \infty} \frac{f^{(n)}(x) - f^{(n)}(x_0)}{x - x_0}\text{ not necessarily equals }\lim_{n \to \infty} \lim_{x \to x_0} \frac{f^{(n)}(x) - f^{(n)}(x_0)}{x - x_0}.$$ And I do not know how to proceed in this case.

Any help will be appreciated.

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>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*}