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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.

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

1 Answer 1

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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.

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