Find All Functions $F(x)$ with $(x-y)^2$ Inequality

[Fallen Angel]

Find all functions $F(x):\Bbb{R}\longrightarrow \Bbb{R}$ such that

$F(x)-F(y)\leq (x-y)^2$ for all $x,y\in \Bbb{R}$
Edited for correct a typo.

 

[Amer]

Something wrong $x,y \in R^2 $ but $F: R \rightarrow R$ how we can define $F(x), F(y)$ then ?

 

[Fallen Angel]

Oh sorry, it was a typo, it should read

$F(x)-F(y)\leq (x-y)^2, \ \forall x,y \in \Bbb{R}$

 

[Opalg]

Find all functions $F(x):\Bbb{R}\longrightarrow \Bbb{R}$ such that

$F(x)-F(y)\leq (x-y)^2$ for all $x,y\in \Bbb{R}$

[sp]
Interchange $x$ and $y$ to see that $|F(x) - F(y)| \leqslant (x-y)^2$ for all $x,y\in \Bbb{R}$.

Divide the interval $[0,x]$ into $n$ subintervals $\bigl[\frac{k-1}nx,\frac knx\bigr]$. Then $$|F(x) - F(0)| = \Bigl|\sum_{k=1}^n F\bigl(\tfrac knx\bigr) - F\bigl(\tfrac{k-1}nx\bigr) \Bigr| \leqslant \sum_{k=1}^n \bigl| F\bigl(\tfrac knx\bigr) - F\bigl(\tfrac{k-1}nx\bigr) \bigr| \leqslant \sum_{k=1}^n \bigl(\tfrac1nx\bigr)^2 = \tfrac1nx^2.$$ Now let $n\to\infty$ to see that $|F(x) - F(0)| = 0$. In other words, $F$ must be a constant function.[/sp]

 

[Fallen Angel]

Well done Opalg.

Another way

Spoiler

Divide by $|x-y|$ to get
$\dfrac{F(x)-F(y)}{|x-y|}\leq |x-y|$, and now letting $x-y \to 0$ this implies the derivative of $F$ exists and equals zero, by Fundamental theorem of Calculus, $F$ is constant