Show $f$ is constant if $|f(x)-f(y)|\leq (x-y)^2$. Problem: Let $f$ be defined for all real $x$, and suppose that 
$$|f(x)-f(y)|\le (x-y)^2$$
for all real $x$ and $y$. Prove $f$ is constant.
Source: W. Rudin, Principles of Mathematical Analysis, Chapter 5, exercise 1. 
 A: For any $x\in\mathbb{R}$,
$$
\begin{align}
|f'(x)|
&=\lim_{h\to0}\frac{|f(x+h)-f(x)|}{|h|}\\
&\le\lim_{h\to0}\frac{h^2}{|h|}\\
&=0
\end{align}
$$
Therefore, $f$ is constant.
A: Here's a proof more elementary.
Let $c=f(0)$, we have to prove that $f(x)=c$ whenever $x\neq0$. Supposing that $n$ is an arbitrary positive integer, we have
$$\left|f\left(\frac{m+1}nx\right)-f\left(\frac mnx\right)\right|\le\left(\frac{m+1}nx-\frac mnx\right)^2=\frac{x^2}{n^2}$$
Hence
\begin{align*}
|f(x)-f(0)|
\;&=\;\left|\,\sum_{m=0}^{n-1}\left(f\left(\frac{m+1}nx\right)-f\left(\frac mnx\right)\right)\,\right|\\
&\le\;\sum_{m=0}^{n-1}\,\left|f\left(\frac{m+1}nx\right)-f\left(\frac mnx\right)\right|\\
&\le\;\frac{x^2}n
\end{align*}
Let $n\to\infty$, we have $|f(x)-f(0)|=0$, thus $f(x)=c$.
A: It suffices to show that $f'(x)=0$ for all $x\in\mathbb{R}$. We see that the given condition implies
$$\left| \frac{f(x)-f(y)}{x-y} \right| \le |x-y|.$$
So in a $\delta$-neighborhood of $x$, the quotient in definition of the derivative is less than $\delta$. So the limit is 0, and we are done. 
