$|f(x)-f(y)|\leqslant C|x-y|^{\alpha},\forall x,y\in(a,b)$ Prove that $f$ is constant in $(a,b)$.

Asked Mar 23 '21 at 08:02

Active Mar 23 '21 at 08:06

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Let $f:(a,b)\to\mathbb{R}$ and assume that there is $C>0$ and $\alpha>1$ such that $|f(x)-f(y)|\leqslant C|x-y|^{\alpha},\forall x,y\in(a,b)$

Prove that $f$ is constant in $(a,b)$.

Prove that $\underset{x\to y}{\lim}\frac{f(x)-f(y)}{x-y}=0$

Ie $(\forall\epsilon>0)(\exists\delta>0):|x-y|<\delta\implies|\frac{f(x)-f(y)}{x-y}|<\epsilon$

Suppose that $|x-y|<1$.

Then by the stated inequality, $\frac{|f(x)-f(y)|}{|x-y|}\leqslant\frac{|f(x)-f(y)|}{|x-y|^{\alpha}}\leqslant C\iff\frac{|f(x)-f(y)|}{|x-y|}\leqslant C$.

I need to show that this is smaller than $\epsilon$ but I'm not too sure where to go from here.

edited Mar 23 '21 at 08:06

asked Mar 23 '21 at 08:02

expl0it3r

3

What are the hypothesis on $\alpha$? – mathcounterexamples.net Mar 23 '21 at 08:04
what's ur $\alpha$? Is it a real number ?, a positive number ?, greater that $1$? – Someone Mar 23 '21 at 08:04
Sorry, added hypothesis on $\alpha$ – expl0it3r Mar 23 '21 at 08:07
Hint: Note that $ \left| {\frac{{f(x) - f(y)}}{{x - y}}} \right| \le \left| {x - y} \right|^{\alpha - 1} $ and $\alpha-1>0$ (by hypothesis). – Gary Mar 23 '21 at 08:09
More similar questions: https://math.stackexchange.com/questions/linked/361400 – Martin R Mar 23 '21 at 08:10

$|f(x)-f(y)|\leqslant C|x-y|^{\alpha},\forall x,y\in(a,b)$ Prove that $f$ is constant in $(a,b)$.

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