Fix rationals $a<b$ . For all positive integers $n$, we have $|f(a)-f(b)|=|(f(x_{0})-f(x_{1}))+(f(x_{1})-f(x_{2}))+...+(f(x_{n-1})-f(x_{n}))|\leq \sum_{i=0}^{n-1}|f(x_{i})-f(x_{i+1})|\leq n\cdot\frac{(b-a)^{2}}{n^{2}}\to 0$, where $x_{0}=a<x_{1}<...<x_{n-1}<x_{n}=b$ and $x_{i+1}-x_{i}=\frac{b-a}{n}$. Therefore $f(a)=f(b)$. Done!

This problem has been posed with hint such as N.T.TUAN did for entrance examination of Waseda University/Politics & Economics 1998.

In notes of pbornsztein and other, has following : Find all continuous functions $f: \mathbb{R}\to \mathbb{R}$ such that $|f(x)-f(q)|\leq 7(x-q)^{2}\;\forall x\in\mathbb{R}\;\forall q\in\mathbb{Q}$. Solution is same as above.

It's easy to see that $f$ is continuous on $\mathbb Q$ so it can be extended uniquely (because $\mathbb Q$ is dense in $\mathbb R$) to a function $g: \mathbb R\to\mathbb R$ so that $g|_{\mathbb Q}\equiv f$. It is easy to see that $g$ satisfies the inequality $|g(x)-g(y)|\le(x-y)^{2},\forall x,y\in\mathbb R$ (just use the fact that $\mathbb Q$ is dense in $\mathbb R$ and that $g$ satisfies the ineq for rationals). If we multiply the inequality with $|x-y|^{-1}$ and let $y\to x$ we get that $g'(x)=0,\forall x\in \mathbb R$, so $g(x)=c\in\mathbb R$ hence $f$ is constant. Instead of $(x-y)^{2}$ we could have $(x-y)^{\alpha}, \alpha>1$ and the conclusion still holds.

mateivld wrote: Let $f: \mathbb{Q}\rightarrow \mathbb{R}$ be a function such that \[|f(x)-f(y)|\leq (x-y)^{2}\]for all $x,y \in\mathbb{Q}$. Prove that $f$ is constant. I really enjoyed this problem