Let $f:[0,1]\to \mathbb R$ s.t. there is $a\in (0,1)$ s.t. $$\forall x,y\in [0,1], |x-y|\geq a\implies |f(x)-f(y)|<|x-y|.$$

Show there is $C\in [0,1)$ s.t. $$\forall x,y\in [0,1], |x-y|\geq a\implies |f(x)-f(x)|\leq C|x-y|.$$

As Hint I have that : Set $E=\{x,y\in [0,1]\mid |x-y|\geq a\}$. Prove that $$F:E\to \mathbb R, (x,y)\longmapsto \left|\frac{f(x)-f(y)}{x-y}\right|,$$ is continuous. I really don't understand why $F$ is continuous... so

Q1) Why is $F$ continuous ?


I tried to do an other way. Suppose that for all $n$, there is $x_n,y_n\in [0,1]$ s.t. $$|x_n-y_n|\geq a\quad \text{and}\quad |f(x_n)-f(x_n)|\geq \left(1-\frac{1}{n}\right)|x_n-y_n|.$$

I suppose WLOG that $(x_n)$ and $(y_n)$ converge to $x,y\in [0,1]$. Then $|x-y|\geq a$ and since $$|x_n-y_n|>|f(x_n)-f(y_n)|\geq \left(1-\frac{1}{n}\right)|x_n-y_n|,$$ we get $$\lim_{n\to \infty }|f(x_n)-f(y_n)|=|x-y|\geq a,$$ but I don't see in what it's a contradiction.


1 Answer 1


I think we need more assumptions on $f$ (such as continuity), because $F$ is not necessarily continuous.

Otherwise, if $F: E\rightarrow \mathbb{R}$ $$F(x,y) = \left| \frac{f(x) - f(y)}{x-y}\right|$$ is continuous, note that its domain $E$ is a closed subset of the unit square thus compact, so $F$ attains a maximum. Lets call it $C:= \max_E F(x,y)$, then for some $(x^*, y^*)$ the maximum is attained, and together with our assumption, we have
$$C = F(x^*, y^*) = \left| \frac{f(x^*) - f(y^*)}{x^*-y^*}\right| < 1.$$ This $C$ will satisfy the constant we are trying to find in the question.

  • $\begingroup$ Thank you. They probably forgot to say that $f$ is continuous. $\endgroup$
    – user386627
    Jul 21, 2018 at 21:40

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