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.