Prove that $f(x)$ be a constant 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.
 A: 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.
A: Take any $a, b \in \mathbb{R}$ such that $a < b$.
Since $f$ is continuous over $\mathbb{R}$, each  $h_n$ is Lebesgue integrable over $[a,b]$.
Since $h_n$ converges pointwise to $0$ and all $|h_n|$ are bounded by a constant
$M$ (which is trivially Lebesgue integrable) over $[a,b]$. By 
Lebesgue's dominated convergence theorem,
$$\lim_{n\to\infty} \int_{a}^{b} h_n(x) dx = \int_{a}^{b} \lim_{n\to\infty} h_n(x) dx = \int_{a}^b 0\; dx = 0$$
Notice $$\int_{a}^{b} h_n(x) dx = 2^{n} \int_{a}^{b}\left( f(x+\frac{1}{2^n}) - f(x) \right) dx = 2^{n} \left( \int_{b}^{b+2^{-n}} f(x) dx - \int_{a}^{a+2^{-n}} f(x) dx \right)$$
and by continuity of $f$ at $a$ and $b$, we have:
$$\lim_{n\to\infty} 2^n \int_{a}^{a+2^{-n}} f(x)dx = f(a) \quad\text{ and }\quad
\lim_{n\to\infty} 2^n \int_{b}^{b+2^{-n}} f(x)dx = f(b) 
$$
This implies 
$$f(b) - f(a) = \lim_{n\to\infty} \int_{a}^{b}h_n(x)dx = 0$$
and hence $f$ is a constant.
A: Assume to the contrary that there is a function $f$ that satisfies the conditions that is not constant, so there are some $x_0,a,b\in \mathbb{R}$ with $a>0, b\ne0$ such that
$$ f(x_0+a)=f(x_0)+b
$$
Since $f$ is continuous there is a $\delta>0$ such that $|f(x_0+a+d)-f(x_0+a)|<|b|/2$ for all $d\in(-\delta,\delta)$.
Choosing $N$ such that $2^{-N}<\delta$ we can find $D\in[0,\delta)$ so that $2^N(a+D)=K\in \mathbb{Z}$, i.e. so that we can divide the interval $[x_0,x_0+a+D]$ into $K$ equal parts of length $2^{-N}$.
Then
$$
|f(x_0+a+D)-f(x_0)|>|b|/2\\
|f(x_0+K 2^{-N}) - f(x_0)|>|b|/2
$$
so considering the changes in $f$ in each interval $[x_0+j2^{-N},x_0+(j+1)2^{-N}]$ for $j=0,1,\ldots,K-1$ there must be a $j$ that satisfies
$$
|f(x_0+(j+1)2^{-N})-f(x_0+j2^{-N})|>\frac{|b|}{2K}
$$
Similarly, for any $n>N$ we have $2^n(a+D)=2^{n-N}K$ and there must be a $j\in\{0,1,\ldots,2^{n-N}K-1\}$ that satisfies
$$
|f(x_0+(j+1)2^{-n})-f(x_0+j2^{-n})|>\frac{|b|}{2^{n-N+1}K}
$$
that is, writing $X=x_0+j2^{-n}$,
$$
h_n(X) = 2^n|f(X+2^{-n})-f(X)|>\frac{2^{N-1}|b|}{K}
$$
But this is impossible, since the right side is a fixed positive number whereas we must have $h_n(X)\rightarrow 0$ as $n\rightarrow \infty$. Hence our initial assumption must be incorrect, and there cannot be a non-constant $f$ with this condition on $h$.
