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.


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*}


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