# 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.

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The existence of $g$ for every $x \in [0,1]$ is saying that $f$ is differentiable everywhere, right? I mean, this is not exactly the definition, but close enough, maybe? –  Daniel Spector Dec 10 '12 at 8:55
I cannot see why they are close enough, because when $f$ is differentiable everywhere and of bounded variation, I know $f$ is absolutely continuous. –  qianzhang Dec 10 '12 at 9:36
@Daniel Spector : take for $f$ any even function, so that $g(0)=0$ is well-defined. Yet it happens that some even functions are not differentiable at 0. You can elaborate easily on this example to show that there exists continuous functions $f$ nowhere differentiable for which $g$ exists everywhere. You can also extend the construction of Alexander Eremenko given below: if you take for $f$ the indicator function of the rational numbers then $g$ is identically zero on the rational numbers, since both $x+\eps$ and $x-\eps$ are irrational or rationa at the smae time. –  Loïc Teyssier Dec 10 '12 at 9:48
OK. Thanks for the differentiable correction :). So my thinking is the requirement that $f$ is $BV$ is enough to get rid of these kind of pathologies and regain the fundamental theorem, if $g$ is locally integrable (since the fundamental theorem implies absolute continuity). –  Daniel Spector Dec 10 '12 at 11:46
You can show the following. If $g$ exists everywhere, then $f$ is absolutely continuous outside of a closed nowhere dense set. This is the best you can do in the general case, as given any closed nowehere dense set $S$ there exists a differentiable function which is absolutely continuous outside of $S$ but has infinite variation on any neighbourhood of each point of $S$. However, if $g$ exists everywhere and is locally integrable, then $f$ is absolutely continuous. You can prove this using the Baire category theorem. I don't have time to post a full answer right now, so just leaving a comment. –  George Lowther Dec 13 '12 at 6:51

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.

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Thank you for your answer. Do you have any idea about the absolute continuity of $f$? An immediate corollary of the "quasi-mean value theorem" is that if $f^s$ is bounded, then $f$ is Lipschitz, but I still have no idea about the general situation. –  qianzhang Dec 10 '12 at 3:10
1. 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 $g=0$.

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Sorry, I cannot see why your statement in the second paragraph is true. –  qianzhang Dec 10 '12 at 3:03

$f$ 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).

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Bazin, thank you for your answer. I think the dominated convergence assumption is too strong for me, and I am only interested in pointwise convergence without further condition. I am sorry for not excluding this situation in my question. –  qianzhang Dec 9 '12 at 19:24