Show that $f$ is constant.

Question

Im trying to prove that if $f\colon \mathbb{R}\to\mathbb{R}$ such that for all $x,y\in\mathbb{R}$ we have $\lvert f(x)-f(y)\rvert \leq (x-y)^2$ then $f$ is constant.

I have shown that if $f$ is differentiable, then the statement is true. But without the differentiability condition, I'm having a hard time.

Then show first that $f$ is differentiable ... – Martin R May 25 '18 at 08:58 — Martin R, May 25 '18 at 08:58
Another one: https://math.stackexchange.com/q/164804/42969. – Martin R May 25 '18 at 09:01 — Martin R, May 25 '18 at 09:01

score 4 · Accepted Answer · answered May 25 '18 at 09:00

4

Hint: $$0 \leq \lim_{x \to y} \frac{\lvert f(x)-f(y) \rvert}{\lvert x-y \rvert} \leq \lim_{x \to y} \lvert x-y \rvert.$$

answered May 25 '18 at 09:00

Gibbs

8,230

score 0 · Answer 2 · answered May 25 '18 at 09:00

0

For every $x_0\in\mathbb{R}$ we get $0\leq{}|\frac{f(x)-f(x_0)}{x-x_0}|\leq{}|x-x_0|$ and thus the derivative exists and $f’(x_0)=0$ everywhere.

answered May 25 '18 at 09:00

Μάρκος Καραμέρης

1,748
11
22

Show that $f$ is constant.

2 Answers2