# $|f(x)-f(y)|\leq \|x-y\|^2$ implies $f$ is constant

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.

• It is correct. More generally, if $|f(x)-f(y)|\le\|x-y\|^{1+\alpha}$ for some $\alpha>0$ the same conclusion holds. Apr 19 at 15:44
• @GReyes Why not an official answer? Apr 24 at 14:39
• @PaulFrost Done. Apr 24 at 20:08

If you take $$x=y+tu$$ where $$u$$ is a unit vector, you get $$|f(y+tu)-f(y)|\le t^2$$ Dividing by $$t$$ and taking the limit $$t\to 0$$ you conclude that (the directional derivatives exist and) $$D_uf(y)=0$$ for any direction $$u$$ and point $$y$$. Therefore for any two points $$x$$ and $$y$$, if you consider the restriction of $$f$$ to the segment joining the points, $$f$$ is constant along that segment and $$f(x)=f(y)$$. You can join any point in $$\mathbb{R}^2$$ with the origin. Therefore $$f(x)=f(0)$$ is a constant.
If you replace the square on the right hand side by $$1+\alpha$$ with $$\alpha>0$$ , the conclusion is the same.