Does a weaker condition than vanishing derivative imply a function being constant?     I learned this question from math.stackexchange, which is equivalent to ask that if $f:[0,1]\to \mathbb{R}$ is a continuous function with bounded variation, does 
$$g(x):=\lim_{\epsilon\to 0}\frac{f(x+\epsilon)-f(x-\epsilon)}{2\epsilon}$$
exist for every $x\in[0,1]$ imply that $f$ is absolutely continuous?
Moreover, if we do not know whether $f$ is of bounded variation or not, what can we say about the differentiability of $f$? For example, if $g\equiv 0$, will $f$ be a constant?
Any help is appreciated. 
Edit: I think the original question cited from math.stackexchange has been solved by the asker himself there, which is based on the Vitali covering theorem for Radon measures on $\mathbb{R}^n$. For my own question, where $f$ is not assumed to be of bounded variation a priori, Jack Huizenga's answer is good enough for me.
 A: *

*For continuous $f$, there is a completely elementary proof that $g=0$
implies that $f$ is constant.


By uniform continuity,
for every $\delta>0$ and for every $x$ there exists $\epsilon>0$
such that the oscillation of $f$ on $[x-\epsilon,x+\epsilon]$ is
at most $2\epsilon\delta$. (Oscillation is $\sup$ minus $\inf$).
Now use the following Lemma: if a closed bounded set $E$ on the real
line is covered by intervals such that every point is the center of some
interval, then there exists a subcovering such that the intervals intersect
at most by two. (No triple of intervals has non-empty intersection).
From this we deduce that the total oscillation of $f$ is at most $4\delta$.
As $\delta$ is arbitrary, this implies that $f$ is constant.
2 . For discontinuous $f$ the statement is not true.
Let $E$ be a finite set on $(0,1)$, $f(x)=1$ for $x\in (0,1)\backslash E$,
and arbitrary values on $E$. Then $g=0$.
A: $f$ is a continuous function thus can be considered as a distribution. For $\phi\in C^1_c(0,1)$, we have
$$
\langle f',\phi\rangle=-\int_{\mathbb R} f(x) \phi'(x) dx=\lim_{\epsilon \rightarrow 0}
\int_{\mathbb R} f(x) \frac{\phi(x-\epsilon)-\phi(x+\epsilon)}{2\epsilon} dx,
$$
$$=\lim_{\epsilon \rightarrow 0}
\int_{\mathbb R} \phi(x) \frac{f(x+\epsilon)-f(x-\epsilon)}{2\epsilon} dx=\int \phi(x) g(x) dx,
$$
if we know some domination for some $\epsilon_0>0$
$$
\sup_{0<\epsilon<\epsilon_0}\bigl\vert\frac{f(x+\epsilon)-f(x-\epsilon)}{2\epsilon}\bigr\vert \le h(x),\quad h\in L^1.
$$
In that case, we get $f'=0$ in the distribution sense, so that $f$ is constant on the connected (0,1).
A: The ordinary proof that 0 derivative implies constant rests on the mean value theorem.  Your function $g(x)$ is the "symmetric derivative" of $f$, so we should look for a "quasi-mean value theorem" for symmetric derivatives.  A quick google search came up with the paper 
http://www1.au.edu.tw/ox_view/edu/tojms/j_paper/Full_text/Vol-27/No-3/27%283%298-5%28279-301%29.pdf
Here it is shown (Theorem 2) that if $f$ is continuous on $[a,b]$ and symmetric differentiable on $(a,b)$ with symmetric derivative $f^s$, then there are points $\xi,\eta\in (a,b)$ with 
$$f^s(\eta) \leq \frac{f(b)-f(a)}{b-a} \leq f^s(\xi).$$
Clearly then if $f^s = 0$ we must have that $f$ is constant.
