For the $f:\mathbb{R} \rightarrow \mathbb{R}$ for every $x,y$ real numbers is true, that $|f(x)-f(y)|\le(x-y)^2$
-I think that a good way would be to show, that $f'(x)$ is a constant.
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You need to show that $f'$ is zero, not just any constant. – dxiv Nov 16 '16 at 20:19
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@dxiv You're right. I have misread the question. – mfl Nov 16 '16 at 20:25
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Hint: let $x=y+h$ and write the inequality as: $$0 \le \left| \frac{f(x+h)-f(x)}{h}\right| \le |h|$$
Then think the squeeze theorem.
dxiv
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and how can I finish the proof with the squeeze theorem? I still can't see it – Zauberkerl Nov 16 '16 at 22:02
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@Zauberkerl For a fixed $x$ let $h \to 0$ then the inequalities prove that $f'(x)$ exists and is $0$. – dxiv Nov 16 '16 at 22:41