TITLE: Suppose $f$ is continuous on R such that $\lim_{h\to 0} \frac{f(x+h)-f(x-h)}{h} = 0$ for all $x\in\mathbb R$. Prove that $f$ is constant.
QUESTION [5 upvotes]: I'm having a bit of trouble with the title problem out of Davidson and Donsig's Real Analysis. I'll state it again:
Suppose $f$ is continuous on $\mathbb{R}$ such that $$\lim_{h\to 0} \frac{f(x+h)-f(x-h)}{h} = 0\  \forall x\in\mathbb{R}.$$ Prove that $f$ is constant.
They provide the following hint which I have been trying to apply.
HINT: Fix $\epsilon > 0$. For each $x$, find a $\delta > 0$ so that $|f(x+h)-f(x-h)| \leq \epsilon h$ for $0\leq h \leq \delta$. Let $\Delta$ be the supremum of all such $\delta$. Show that $\Delta = \infty$.
Here's how I've started. Fix $\epsilon > 0,\ x\in\mathbb{R}$. By the definition of the limit, 
$$(\forall x\in\mathbb{R})(\forall\epsilon>0)(\exists\delta>0)(0<|h|\leq\delta\implies\Bigg|\frac{f(x+h)-f(x-h)}{h}\Bigg|<\epsilon).$$ 
Therefore, we immediately get a $\delta>0$ for our $x,\epsilon$ such that $$|f(x+h)-f(x-h)|\leq\epsilon h,\ 0\leq h \leq\delta.$$
I don't know how to proceed from here. I'm not even sure conceptually how showing that $\Delta=\infty$ would give us that $f$ is constant. Any help would be appreciated!

REPLY [6 votes]: I don't understand the hint in your book. But here is an alternative way to do it. 
The limit $$SDf(x):=\lim_{h\to 0}\frac{f(x+h)-f(x-h)}{2h},$$ if exists, is called the symmetric derivative of $f$ at $x$. In Brian Thomson's Symmetric Properties of Real Functions, it is shown that

Theorem. A continuous function is necessarily increasing in any interval in which its symmetric derivative exists and is positive. 

As a corollary one can deduce that continuous functions with zero symmetric derivatives are constant:
For any $\epsilon>0$, $SD(f(x))=0$ means that $f(x)+\epsilon x$ has a positive symmetric derivative and so, by the theorem, must be increasing; similarly $-f(x)+\epsilon x$ has a positive symmetric derivative and must also be increasing. As $\epsilon>0$ is arbitrary $f$ must be constant as required.

Here is the proof of the theorem in Thomson's book: