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. 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!
 A: 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:




A: We are going to show that the set of points where $f$ is differentiable is dense. Let $(a,b)$ an open interval, since $f$ is continuous on $[a,b]$, it attains its maximal at $x_0$, suppose that $x_0\in (a,b)$
$\frac{f(x_0+h)-f(x)}{h}\geq 0$ and $\frac{f(x_0-h)-f(x_0)}{-h}\geq 0$, this implies that $\frac{f(x_0+h)-f(x_0-h)}{h}\geq \frac{f(x_0+h)-f(x_0)}{h}\geq 0$, simce $lim_{h\rightarrow 0}\frac{f(x_0+h)-f(x_0)}{h}=0$ and $f$ is derivable at $x_0$, if the maximum is $f(a)$ or $f(b)$, we consider all the intervals $(c,d)\subset (a,b)$ and if for every $(c,d)\subset (a,b)$ the maximum is $f(c)$ or $f(d)$, $f$ is monotone, we deduce that the set of points where $f$ is differentiable on $(a,b)$ is dense.
Monotone+continuous but not differentiable
Suppose $f$ is differentiable at $x$
$$\lim_{h\to 0} \frac{f(x+h)-f(x-h)}{h} = 0=2f'(x)\  \forall x\in\mathbb{R}.$$
since $\frac{f(x+h)-f(x-h)}{h}=$ $\frac{f(x+h)-f(x)}{h}$+ $\frac{f(x-h)-f(x)}{-h}$
so $f'(x)=0$ implies that $f$ is constant.
Uses of $\lim \limits_{h\to 0} \frac{f(x+h)-f(x-h)}{2h}$
A: Consider a compact interval $[a,b]$
Fix  $\epsilon>0$ For each $x \in [a,b]$, there exist $\delta(x)$, such that $|f(x+h)-f(x-h)|\le h\epsilon$  whenever $|h|<\delta(x)$
Consider the open intervals $\left\{I_x=(x-\delta(x),x+\delta(x))
\right\}_{x\in[a,b]}$, they clearly cover $[a,b]$ which is compact, hence there exists a finite sub-cover$\left\{I_{x_n}\right\}_{n=1}^{N}$ for $[a,b]$ 
Assume $x_1<x_2<...<x_N$ and choose a sequence of points $y_n$ such that $y_n$ lies in the intersection of $I_{x_n}$ and $I_{x_{n+1}}$ so that
$|f(y_{n+1})-f(y_n)| <\delta(x_n)\epsilon$,
and $|f(y_{n+2})-f(y_{n+1})| <\delta(x_{n+1})\epsilon$ 
Let
$y_0= a$ and $y_{N+1}=b$
Now 
$|F(b)-F(a)|= \left|\sum_{i=0}^{N}f(y_{i+1})-f(y_i)\right|\le\sum_{i=0}^{N}\left|f(y_{i+1})-f(y_i)\right|\le\sum_{i=0}^{N}\delta(x_i)\epsilon\le(b-a)\epsilon$
The last inequality holds since we can choose any $\eta(x)<\delta(x)$  and the  first inequality will still hold. 
Finally since $\epsilon$ is arbitrary $|f(b)-f(a)|=0$, hence $f(b)=f(a)$, letting $b$ vary, we conclude that $f(x)$ is constant.
Note: Anybody is welcome to edit the answer to get a better format and possiby a better way of writing the proof
