Let $f$ be a function defined everywhere on the real line, which is infinitely differentiable everywhere, in other words, $f$ is everywhere smooth. I define the $\omega$-th derivative, where $\omega$ is the first infinite ordinal, to be the pointwise limit of the sequence of $n$-th derivatives. And then, the $\omega + 1$-th derivative would be the derivative of the $\omega$-th derivative, and so on through all the ordinals, like $\omega *2$, $\omega^2$, and even higher ordinals, where at limit ordinals you take the limit of the ordinal-indexed sequence of derivatives that you have so far.
Let me give some examples. The $\omega$-th derivative of any polynomial is the constant $0$ function. The $\omega$-th derivative of $e^x$ is $e^x$. Somewhat more interestingly, the $\omega$-th derivative of $2^x$ is $0$. However, I have noticed that in each case where it exists, the $\omega$-th derivative is a constant multiple of the exponential function $e^x$, which forces all further ordinal derivatives to be the same as the $\omega$-th derivative. But is that really the case?
To make my question precise, if $f$ is an everywhere smooth function, whose sequence of $n$-th derivatives converges pointwise to an everywhere defined function $g$, must $g$ be a constant multiple of the exponential function $e^x$? Note, we are not making any apriori assumptions that $g$ is differentiable or even continuous.
I have asked questions similar to this before, but in all previous times, people did not really understand the question. I have now given a very through explanation, in the hopes that people can finally understand it, and give an answer.
Edit: If the question is not solved as stated, can it at least be solved under the hypothesis that $g$ is everywhere differentiable? After all, that is precisely the case that interests me, because only then can you take the derivative of $g$.
