Switching a derivative and a limit is sometimes possible, sometimes not...

[nomadreid]

TL;DR Summary

Switching a derivative and a limit is sometimes possible, sometimes not. I have a case that I am not sure about.

Here is a cute calculation about which I have my doubts:

Treating the derivative as a limit makes the first step a case of switching the order of limits. One cannot automatically do this, as for example for the sequence of functions:

More precisely, that one should be able to switch limits iff at least one of the limits is uniformly convergent and the other one at least point-wise convergent (Moore-Osgood Theorem).

So my problem is that it appears to me that this condition is met here, but I am not certain.

Any pointers? Thanks.

 

[fresh_42]

I'm not convinced that Moore-Osgood can be applied, at least not from what I saw on Wikipedia,
https://en.wikipedia.org/wiki/Iterated_limit#Interchanging_limits_of_functions,
and it causes my suspicion that they didn't mention differentiation among their corollaries.

What we have here is the situation
$$
\left(e^x\right)' = \dfrac{d}{dx}(\lim_{n \to \infty}f_n(x))\stackrel{!}{=}\lim_{n \to \infty}\dfrac{d}{dx}f_n(x)=e^x
$$
$$
\lim_{n \to \infty}f_n(x)=f(x)\stackrel{?}{\Longrightarrow } \lim_{n \to \infty}f'_n(x)=f'(x)
$$
My textbook proves this for continuously differentiable $\mathbf{f_n}$, pointwise convergence of $\mathbf{f_n}$ and uniform convergence of $\mathbf{f_n\,'}.$ Uniform convergence of $f_n$ is not sufficient. The proof uses the interchangeability of limit and integral.

So what remains to show is indeed whether
$$
\left(x\longmapsto \left(1+\dfrac{x}{n}\right)^{n-1}\right)_{n\in \mathbb{N}}
$$
converges uniformly. I have my doubts, too, that this is uniform.

My book uses the functional equation of the exponential function and differentiability at $x=0$ and proves directly
\begin{align*}
\dfrac{d}{dx}e^x&=\lim_{h \to 0}\dfrac{e^{h+x}-e^x}{h}=e^x\lim_{h \to 0}\dfrac{e^h-1}{h}=e^x
\end{align*}
This avoids uniform convergence and uses the functional equation instead. That means we have to answer the question of whether the functional equation implies the uniform convergence of that specific sequence of functions. I don't think so. Isn't the functional equation the reason why it cannot be done uniformly?

 

[nomadreid]

Thanks, fresh42. Gives me food for thought.