If $f$ is differentiable and $\forall x \in \mathbb R $ and $\forall h >0 ( \mid {f(x+h)-f(x-h)}\mid If $f$ is differentiable ,and $ \forall x \in \mathbb R  $ and $\forall h >0 $ hold $\mid {f(x+h)-f(x-h)}\mid <h^2$.Prove that $f$ is constant.I tried to use Lagrange theorem.
 A: Hint: divide the inequality by $h$, can you recognize the left hand side? Remember also that this "new" inequality still holds for every $h > 0$.
A: We get $\mid {f(x+h)-f(x-h)}\mid <h^2\iff\left|\frac{ {f(x+h)-f(x)}}{h}+\frac{f(x-h)-f(x)}{-h}\right|<h$ for all $x\in\mathbb{R}$ and $h>0$, which means that
$2|f'(x)|=\left|\lim_{h\downarrow 0}\frac{ {f(x+h)-f(x)}}{h}+\lim_{h\uparrow 0}\frac{{f(x+h)-f(x)}}{h}\right|=\left|\lim_{h\downarrow 0}\frac{ {f(x+h)-f(x)}}{h}+\lim_{h\downarrow 0}\frac{f(x-h)-f(x)}{-h}\right|=\lim_{h\downarrow 0}\left|\frac{ {f(x+h)-f(x)}}{h}+\frac{f(x-h)-f(x)}{-h}\right|\leq \lim_{h\downarrow 0}h=0$ because by differentiability this right limit equals the left limit. So $f'(x)=0$ for all $x\in\mathbb{R}$, which means that $f$ is constant.
A: A more direct proof.
Take $x_1$ and $x_2 > x_1$ arbitrarily. Let $h = (x_2 - x_1)/(2N)$ and divide the interval $(x_1 , x_2)$ into $N $ equally long intervals. We then have 
$$ |f(x_1) - f(x_2 )| < N * h^2 = (x_2 - x_1)/ (4N ) $$
for arbitrary $ N$. As $N \rightarrow \infty $, we get $f(x_1 ) = f(x_2 )$.
