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if $f: \mathbb{R}\to \mathbb{R}$ satisfies $$|f(x)-f(y)|\le |x-y|^{\sqrt{2}}$$ for all $x,y\in \mathbb{R}$ ,then is f increasing ,decreasing or constant?
in my view ,it is clear that $|f(x)-f(y)|$ is not zero so it is not constant. And since the rate of change is positive so the function would be increasing.Is my solution correct or wrong?

Mayank Deora
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  • Your solution is incorrect, in particular the part you say is clearly non-zero. –  Jan 21 '16 at 01:12
  • So what is the solution? Would you please answer it? – Mayank Deora Jan 21 '16 at 01:17
  • For starters: Why do you say it's clear? Can you name a non-constant function that satisfies the inequality? As a hint for the solution, try computing the derivative of $f$. –  Jan 21 '16 at 01:18
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    There is a solution avoiding differentiability, available on several pages of the site. (Hint: Decompose the interval $(x,y)$ into $n$ subintervals, apply the hypothesis to each of these subintervals, let $n\to\infty$, collect the result that $f(x)=f(y)$.) – Did Jan 21 '16 at 01:51
  • See http://math.stackexchange.com/a/1119422/27978 for example :-). – copper.hat Jan 21 '16 at 02:34

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Consider the more general case: $|f(x)-f(y)|\le |x-y|^{1+\varepsilon}$, with $\varepsilon>0$.

Then $f$ is differentiable with $f'=0$ because $$ \lim_{x\to x_0} \left|\frac{f(x)-f(x_0)}{x-x_0}\right| \le \lim_{x\to x_0}|x-x_0|^{\varepsilon} = 0 $$

So, $\sqrt2$ is a red herring. Its only relevant property is $\sqrt2>1$.

lhf
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