Problem Statement:
Let $f:[a,b]\to \mathbb{R}$ be a continuous function such that the following condition holds true. $$\lim_{h\to 0}{\frac{f(x+h)+f(x-h)-2f(x)}{h^2}}=0\ \ \forall \ \ x\in[a,b]$$ Prove that $f(x)$ is linear on $[a,b]$, i.e., $f(x)=cx+d$ for some $c,d \in \mathbb{R}$
Because of the pesky $f(x-h)$ term, and the fact that there is no given information about the differentiability of $f$, and hence no information about the existence of the first derivative, I was unable to start the solution...as common tricks like L'Hospital's rule don't work...
I am looking for hints to this problem, as well as some good resources to practice more problems of this kind..
Also, I read an article on Wikipedia that this limit is precisely the expression for the second symmetric derivative of $f$, but is not equal to the actual second derivative in general....and any proof of the statement assumes the existence of the first derivative (I also read this on an answer on MSE)..which does not seem to be helpful in this problem...
Links:
The Wikipedia article: https://en.wikipedia.org/wiki/Second_derivative#Limit
The answer on MSE: Can the limit $\lim_{h \to 0}\frac{f(x + h) - 2f(x) + f(x - h)}{h^2}$ exist if $f'(x)$ does not exist at $x$?
Thanks for any answers!!
It's possible to do this elementarily, without appealing to calculus; here's a hint.
Define $g(x,h) = \frac{f(x+h)-f(x)}{h}$. So we're given that $$\lim_{h \to 0} \frac{g(x+h,h)-g(x,h)}{h} = 0. \; \; \; \;(*)$$
Take $h = \frac{b-a}{n}$ (i.e. cut up the interval $[a, b]$ into $n$ parts), for some $n \in \mathbb N$. Note that $$f(a+kh) - f(a) = \sum_{i=0}^{k-1} \left(f\left(x+(i+1)h, h \right) - f\left(x+ih, h \right)\right) = h\sum_{i=0}^k g(x+ih,h).$$ Taking $k$ such that $kh$ is roughly fixed as $n$ increases, and using $(*)$, show that $g(x+kh, h) - g(x,h)$ goes to $0$ as $n \to \infty$; this means that the function can be arbitrarily well approximated by linears, whence you conclude by continuity.