This is a fairly basic analysis question. Consider a continuous function $f: \mathbb{R} \to \mathbb{R}$ which is twice differentiable at a point $x$. If necessary, also assume that $f \in C^2(\mathbb{R})$. Chasing through the definitions, we have
$$f''(x) := \lim_{h\to 0} \frac{f'(x+h) - f'(x)}{h} = \lim_{h\to 0}\left(\lim_{k\to 0} \frac{f(x+h+k) - 2f(x+k) + f(x)}{hk}\right).$$
We have two limit processes - one with $h$ and one with $k$, but perhaps we can sort of "set $h=k$" and combine these limit processes, as in:
$$f''(x) \overset{?}= \lim_{h\to 0}\frac{f(x+2h) - 2f(x+h) +f(x)}{h^2}.$$
I am trying to convert this into a Baby Rudin-type problem as follows:
Define (continuous) functions $F_n(k)$ for $k\in[0,a]$ by $$F_n(k) = \begin{cases} \frac{f(x+\frac1n+k) - 2f(x+k) + f(x)}{k/n}&k \in(0,a]\\ \lim_{k\to 0}\frac{f(x+\frac1n+k) - 2f(x+k) + f(x)}{k/n}&k=0\end{cases}.$$
Showing that the $F_n$ converge uniformly to $F_{\infty} := \lim_{n\to \infty}F_n$ for some $a >0$ seems to suffice for our problem. (How to prove this is my question.)
With uniform convergence, then by Baby Rudin Theorem 7.11, $F_{\infty}$ is continuous at $0$ so that we may find a $\delta$ such that if $k < \delta$, then $|F_{\infty}(0) - F_{\infty}(k)| < \varepsilon/2$, and also there is an $M$ such that for $n\geq M$ we have $|F_n(k) - F_{\infty}(k)| < \varepsilon/2$ for all $k$.
So we have $|F_n(k) - F_{\infty}(0)| < \varepsilon$ when both $k < \delta$ and $n \geq M$. We can find $N$ (such as $\max(M,\lceil\delta^{-1}\rceil)$ I think) for which we can set $k = \frac1n$ if $n \geq N$, which is essentially what we wanted.
So my questions is: How can we show that the $F_n$ converge uniformly to $F_{\infty}$, or if we need some extra conditions, then what are they? I would also be interested in easier (still rigorous) ways of concluding that we may "set $h=k$", or obviously any mistakes in my approach.
Finally, I should mention that I became interested while on the page for the somewhat related question Second derivative "formula derivation" in which the current issue was brought up in a comment by Peter S. to an answer provided by Mhenni Benghorbal.
