Pointwise limits of the sequence of derivates of a fixed function =? Assume you are given a function $f:\mathbb R\to \mathbb R$ with the following properties:


*

*All derivatives $f^{(k)}$ exist everywhere

*The sequence $(f^{(k)})_k$ converges pointwise to a function $h:\mathbb R \to \mathbb R$


What can be said about the function $h$ ?
I looked at a few examples and came up with functions like
$f(x) = 3e^x+7e^{-\frac x 2} +42\sin(x/7)+x^{32}$.
In this special example the limit funtion is
$h(x) = 3e^x$.
What I would like to know: Is it always true that if you start with a function $f$ satisfying the assumptions above, that you end up with a function $h(x)$ which is just a scalar multiple of $e^x$ ?
 A: [Disclaimer: in this answer, I assume that the Taylor series for $f$, which converges everywhere, actually converges to $f$. I believe that, given the conditions stated in the question, this must be true, but I can't prove it. This shouldn't be regarded as the "final" answer to the OP's question.]
Since the sequence $f^{\left(k\right)}\left(0\right)$ converges to $h\left(0\right)$, the Taylor series for $f$ is: $$\sum_{k=0}^\infty \frac{f^{\left(k\right)}\left(0\right)}{k!}x^k=\sum_{k=0}^\infty \frac{h\left(0\right)}{k!}x^k+\sum_{k=0}^\infty \frac{\varphi_k}{k!}x^k$$
and $\varphi_k$ goes to zero. Clearly, both of the power series on the right-hand side converge everywhere, and we know that the $h(0)$ part is the exponential: $$f\left(x\right)=h\left(0\right)e^x + \sum_{k=0}^\infty \frac{\varphi_k}{k!}x^k$$
Now, differentiating this equation $n$ times: $$f^{\left(n\right)}\left(x\right)=h(0)e^x + \sum_{k=n}^\infty \frac{\varphi_k}{\left(k-n\right)!}x^{k-n}=h(0)e^x+\sum_{k=0}^\infty \frac{\varphi_{k+n}}{k!}x^k$$
Since $\varphi_k\to 0$, we know that: $$\left|\sum_{k=0}^\infty \frac{\varphi_{k+n}}{k!}x^k\right|\leqslant\sum_{k=0}^\infty \frac{\left|\varphi_{k+n}\right|}{k!}\left|x^k\right|\leqslant\varepsilon\sum_{k=0}^\infty \frac{\left|x^k\right|}{k!}=\varepsilon e^{\left|x\right|}$$
and we can make $\varepsilon$ be as small as we want by choosing $n$ big enough. This means that: $$\lim_{n\to\infty}\left|\sum_{k=0}^\infty \frac{\varphi_{k+n}}{k!}x^k\right|=0$$ pointwise, and therefore: $$\lim_{n\to\infty}f^{\left(n\right)}\left(x\right)=h(0)e^x$$
so that the answer to your question is positive.
