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This question already has an answer here:

Let $\alpha > 1$ and $M \geq 0$. Suppose $f: \mathbb{R} \longrightarrow \mathbb{R}$ satisfies $|f(x)-f(y)|\leq M|x-y|^\alpha$ for all $x, y\in \mathbb{R}$.

How can we prove that $f$ is a constant function? I don't even know where to start.

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marked as duplicate by Did real-analysis Sep 13 '15 at 22:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Fix $x\in\mathbb R$. For any $y\in\mathbb R$ distinct from $x$, we have $0\leq |\frac{f(x)-f(y)}{x-y}|\leq M|x-y|^{\alpha -1}$. Let $y$ approach $x$ and use the squeeze theorem to conclude that $f'(x)=0$.

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Hint: divide by $|x-y|$. What does this tell you about the derivative of $f$?

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Divide [x,y] into the n partitions and use equation of the problem then use triangle inequalities to reach this equation : $$|f(x) - f(y)| \leq \frac{M|x - y|^\alpha}{n^{\alpha-1}}$$ when $n\rightarrow \infty $ we have : $$\rightarrow f(x) = f(y)$$

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Consider using the fact that $|f(x)-f(y)|\leq \sup_{x<a<y} (|(f'(a)|) (x-y)$.

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  • $\begingroup$ How do you know $f(x)$ is differentiable? $\endgroup$ – user223391 Sep 13 '15 at 22:45

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