Is this a valid way to prove that this function is constant for some $C$?

Question

I think it is mostly valid but wanted to know if it was too "lazy" and not rigorous enough. The question is suppose $f: \mathbb{R} \to \mathbb{R}$ where $|f(x) - f(y)| \leq |x-y|^2$ for all x and y. Now, I want to show that f(x) = C for some constant C and this is what I have:

Dividing both sides by $|x – y|$, we get $|\frac{f(x)-f(y)}{x-y}| < C|x – y|$. If we let $x$ approach $y$, we get that the function is differentiable at any point of $y$ given that $y > x$ and $f’(y) = C$. Finally, applying the MVT, we conclude that $f(x) = C$. Honestly, something feels off and I feel like I skipped a very important step.

$|\frac{f(x)-f(y)}{x-y}| < C|x – y|$ should be $|\frac{f(x)-f(y)}{x-y}| \le |x – y|$, and the conclusion should be $f'(y) = 0$, otherwise it is fine. — Martin R, Nov 23 '21 at 19:28
Ah yes that is where I messed up I think. I was unsure how to handle the constant. — TreausreDragon, Nov 23 '21 at 19:29
You can find the same proof e.g. here: https://math.stackexchange.com/q/537216/42969. — Martin R, Nov 23 '21 at 19:30
Here is another way https://math.stackexchange.com/a/1119422/27978. — copper.hat, Nov 23 '21 at 19:47

score 1 · Answer 1 · answered Nov 23 '21 at 19:37

$|f(x) - f(y)| \leq |x-y|^2$

$|\frac{f(x)-f(y)}{x-y}| \le |x – y| $ $(for , x \neq y)$

$|{lim_{x\to y}{\frac{f(x)-f(y)}{x-y}}|}\le 0$

Hence, $f'(y) =0$

Since, $y $ is arbitrary implies $f'(y) =0$ forall $y\in \mathbb{R}$ And as $\mathbb{R}$ is connected together implies $f$ is constant.

Is this a valid way to prove that this function is constant for some $C$?

1 Answers1