TITLE: About the derivative of a function defined on rational numbers
QUESTION [15 upvotes]: I have found this problem:

Let $f : \mathbb{Q} → \mathbb{R}$ with property: $$|f(x) − f(y)| \le (x − y)^2 \tag1$$ for all $x, y \in \mathbb{Q}$. Prove $f$ is constant.


My idea is to consider the formal derivative of $f$ like this:
$$f'(x) = \lim_{h \rightarrow 0} \frac {f(x + h) - f(x)}{h}, h \in \mathbb{Q} $$
Using (1) it's easy to prove the limit exists and is equals $0$ for all $x \in \mathbb{Q}$. So $f$ has derivative and $f'(x)=0$ for all $x \in \mathbb{Q}$. It follows that $f$ is constant. 
Unfortunately, it's not that simple because: $$f'\equiv 0 \implies \ f  \ constant \tag 2$$ is a consequence of  Mean Value Theorem which is valid only on real intervals.

My question: is (2) valid for $f: \mathbb{Q} \rightarrow \mathbb{R}$?
Obs. I don't need a proof for the problem.

REPLY [16 votes]: The desired version of the Mean Value Theorem is not generally true for functions $g:\mathbb Q\to \mathbb R$.  To see that consider the function:  $$g(x) =
\begin{cases}
0,  & \text{if $x<\pi$} \\
1, & \text{if $x>\pi$}
\end{cases}$$
That function has derivative $0$ everywhere (as it is locally constant) but it is not a constant.
To your problem:
We have $$|f(x)-f(y)|=|f(x)-f\left(\frac {x+y}2\right)+f\left(\frac {x+y}2\right)-f(y)|≤|f(x)-f\left(\frac {x+y}2\right)|+|f\left(\frac {x+y}2\right)-f(y)|$$ $$≤\frac {(x-y)^2}2$$  
Repeat this using the stronger estimate to see that $$|f(x)-f(y)|≤\frac {(x-y)^2}4$$  and iterate to see that ($\forall n\in \mathbb N$) $$|f(x)-f(y)|≤\frac {(x-y)^2}{2^n}$$ 
And as $n\to \infty$ we see that $f(x)=f(y)$.