Let $f:\Bbb{R}^2\to \Bbb{R}$ be such that for all $(x,y) \in (\Bbb{R}^2)^2$ we have that $|f(x)-f(y)|\leq \|x-y\|^2$. I need to show that $f$ is constant.
First, by hypothesis we have that $\lim_{x\to y}\frac{|f(x)-f(y)|}{\|x-y\|}=0$ which implies that $f$ is differentiable (Taylor expension of order 1) and that $Df(y)=0 \ \forall y \in \Bbb{R}^2$. Now by mean value theorem considering $[x,y]$ we have that there is $c\in ]x,y[$ such that $f(x)-f(y)=Df(c)(x-y)=0\implies f(x)=f(y)$. As $x$ and $y$ are arbitrary we conclude that $f$ is constant. Is it correct? Thank you in advance.