TITLE: Prove that $f(x)$ be a constant
QUESTION [12 upvotes]: Let $f(x)$ be continuous function in R,note
$$ h_{n}(x)=2^{n}\left[f\left(x+\frac{1}{2^{n}}\right)-f(x)\right]$$
with
$$ |h_{n}(x)|\leq M \qquad (x\in R,n\in N)$$
and 
$$ h_{n}(x)\rightarrow 0  \qquad (n\rightarrow\infty)$$
Show that $f(x)$ is a constant funciton.
I tired to prove that for any $x,h \in R$,
$$ |f(x+h)-f(x)|=0 $$
but I don't know how to deal with the condition.

REPLY [5 votes]: The condition $|h_n(x)|\le M$ is needless.
Given $\epsilon>0$, let $f_\epsilon(x)=f(x)+\epsilon x$ on $\mathbb{R}$. From 
$$\lim_{n\to\infty}2^n(f_\epsilon(x+2^{-n})-f_\epsilon(x))=\epsilon>0,\quad \forall x\in\mathbb{R},$$
it is easy to see that $f_\epsilon$ is strictly increasing on $\mathbb{R}$. Letting $\epsilon\to 0$, it follows that $f$ is non-decreasing on $\mathbb{R}$. A similar argument also shows that $f$ is non-increasing on $\mathbb{R}$, which completes the proof.