Find all function satisfying a given inequality Find all functions $ F : R \rightarrow R $ having the property that for any $x_1$ and $x_2$ the following inequality holds:
$ F(x_1) - F(x_2) \le (x_1 - x_2)^2 $
My attempt:
Observe that 
$ -(x_1 - x_2)^2 \le F(x_1) - F(x_2) \le (x_1 - x_2)^2  $
Assume WLOG that $ x_1 - x_2 >0 $, then we have 
$ -(x_1 - x_2) \le (F(x_1) - F(x_2))/(x_1 - x_2) \le x_1 - x_2 $ 
As $ x_1 $ tends to $x_2$, we have
$ 0 \le dF/dx \le 0 $ 
hence 
$ dF/dx = 0 $
Therefore $ F(x) = $ constant
Is this solution correct? (probably not because the problem didn't say nothing about the function be differentiable)
Any tip will be great, thanks.  
 A: I like your approach. My only remark is that you should give some more details on this inequality that you used:
$$\tag{1}
|F(x_1)-F(x_2)|\le (x_1-x_2)^2.$$ 
This is a consequence of the one given in the text, namely 
$$\tag{2}
F(x_1)-F(x_2)\le (x_1-x_2)^2,$$
because in this inequality the right-hand side is invariant under permutation of $x_1, x_2$ while the left-hand side is not. So we can upgrade (2) to 
$$\tag{3} \max\{ F(x_1)-F(x_2), F(x_2)-F(x_1)\} \le (x_1-x_2)^2, $$ 
and this is exactly (1).
P.S. The upgrading from (2) to (3) is an example of the amplification technique, in the words of Terry Tao. 
A: We have $$|F(x+nh)-F(x)| \le \sum_{k=1}^n|F(x+kh)-F(x+(k-1)h)|\le n h^2.$$ Let $x=x_1$ and $h=(x_2-x_1)/n$ so $|F(x_2)-F(x_1)|\le (x_2-x_1)^2/n$. Now let $n\to\infty$.
A: You are correct, and as a point of interest this problem is one of the coffin problems. Problem 2 on This link
A: I think you wanted to say abs(F(x1)-F(x2))<(x1-x2)^2 . Anyway,  if you divide by abs(x1-x2) , and then tend x1 to x2 you can prove that F is differentiable in x1
A: I agree with user Aymane Gr. An alternative solution is to notice that, for arbitrary $x$
$$F(x) - F(0) = \sum_{i=1}^n F\left(i\frac{x}{n}\right)-F\left((i-1)\frac{x}{n}\right) \leq n \left(\frac{x}{n}\right)^2 = \frac {x^2}{n} \quad \forall n,$$
so $F(x) = F(0)$.
