Constant function

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a function such that $\left | f(x)-f(y) \right |\leq \left | x -y \right |^2 , \;\; \forall x, y \in \mathbb{R}$. Prove that $f$ is constant.

Solution

The exercise is much more easier if $f$ is given to be differentiable. (actually you may prove this and finish the exercise right away but we won't in this case). Now , that we only have the continuity (that is easily extracted by the given condition) the exercise is a little more difficult. What we will do here is to invoke a partition of the $[x, y]$ interval. Indeed:

We split the $[x, y]$ interval in $N$ equal parts and we use the given condition in each of them. Hence:

$$\begin{align*}
\left | f(x)-f(y) \right | &\leq \left| f\left( x\right) -f\left( x + \frac {y-x}{N}\right) \right|+\cdots + \left| f\left( x + (N-2)\frac {y-x}{N}\right ) -f\left( x + (N-1) \frac {y-x}{N}\right) \right|\\
& \quad \quad + \left| f\left( x + (N-1)\frac {y-x}{N}\right ) -f\left( y\right) \right|   \\
 &\leq \left|\frac {y-x}{N}\right|^2+ \cdots + \left|\frac {y-x}{N}\right|^2  \\
 &=N \left|\frac {y-x}{N}\right|^2  \\
 &=\frac {(y-x)^2}{N}
\end{align*}$$

Taking limits as $N \rightarrow +\infty$ we get that $\left | f(x)- f(y) \right |\leq 0$. Now the result follows and $f$ is constant.

Notes: 

1.  We can drop the differentiability condition since $\mathbb{R}$ is known to be Archimedean. As a matter of fact:
   

   If $\mathbb{F}$ is an ordered field such that: every differentiable function $f:\mathbb{F} \rightarrow \mathbb{F}$ with a derivative identically to zero is constant then $\mathbb{F}$ is isomorphic to $\mathbb{R}$. 

   The above holds because the continuity axiom of Dedekind is valid. Except of the above theorem it also possible for someone to prove that if an ordered field satisfies the Rolle's theorem, Bolzano's theorem , maximum value theorem of Weierstrass or Lagrange's mean value theorem then $\mathbb{F}$ is isomorphic to $\mathbb{R}$. This has to do with the structure of $\mathbb{R}$ and it also reveals that analysis cannot be done relied upon the "visual" angle.
    
2.  As strange as it sounds
   
   

   Let $\mathbb{F}$ be an ordered field such that: every function $f:\mathbb{R} \rightarrow \mathbb{R}$ satisfying $\left | f(x)-f(y) \right |\leq \left | x -y \right |^2 , \;\; \forall x, y \in \mathbb{R}$ is constant.

   $\mathbb{F}$ is not necessarily Archimedean. To be more precise there exists a non Archimedean ordered field that contains $\mathbb{R}$ in which the only functions $f:\mathbb{F} \rightarrow \mathbb{F}$ satisfying $\left | f(x)-f(y) \right |\leq \left | x -y \right |^2 , \;\; \forall x, y \in \mathbb{R}$ are the constant ones. Of course such a field must be weird. We are not giving an answer to the question here but the interested reader will undoubtely find more information in the Non Standard Analysis of Abraham Robinson about the tools needed to answer it. Then the result stated above will come as a theorem and can be found in the book Henle και Kleinberg - Infinitesimal Calculus. 
3. We get the same result if we also work over $\mathbb{Q}$. That is $f$ is also constant if:
   
   $$|f(x)-f(y)| \leq (x-y)^2 , \;\; \forall x, y \in \mathbb{Q}$$
   
   

The exercise can also be found in mathematica.gr