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.

  • 6
    $\begingroup$ 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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.