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This is my question :

Let $f$ be defined on an interval $I$, and suppose there exists an $M>0$ and $\alpha>0$ such that $$ |f(x) - f(y)| \leq M|x -y|^\alpha, $$ for $x,y \in I$. Prove that $f$ is uniformly continuous on $I$. If $\alpha>1$, prove that $f$ is constant on $I$.

Should I use the mean value theorem for this problem?

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How do you hope to use the mean value theorem? Where are you going to get a differentiable function from? – Chris Eagle Nov 22 '12 at 20:56
For the first part: Given $\epsilon>0$, find a $\delta>0$ such that $M\delta^\alpha<\epsilon$ and see if it helps. – user3533 Nov 22 '12 at 21:04
For the second part: I'll just interpret Chris Eagle's comment as I see it: There is a differentiable function here. Why is it differentiable? – user3533 Nov 22 '12 at 21:12

I'll write my comments as an answer.

  1. For the first part: Given $\epsilon>0$, find a $\delta>0$ such that $M\delta^\alpha<\epsilon$ and see if it helps.

  2. Prove that if $\alpha>1$ then $f$ is differentiable by definition of being differentiable. You'll also get the value of the derivative. What do you expect it to be?

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Given $\epsilon>0$, let $\delta = \left( \frac \epsilon M\right)^{\frac1\alpha}$. Then $|x-y|<\delta$ implies $$|f(x)-f(y)|\le M |x-y|^\alpha< M\delta^\alpha = M\frac \epsilon M=\epsilon$$ as was to be shown.

Assume $\alpha>1$ and $f(x)\ne f(y)$. For $n\in \mathbb N$ let $x_k=x+\frac kn (y-x)$, $0\le k\le n$ (so that $x_0=x$, $x_n=y$). Then $$|f(x_{k+1})-f(x_k)|\le M|x_{k-1}-x_k|^\alpha=M\left(\frac{|y-x|}n\right)^\alpha$$ for $0\le k<n$ and hence $$\tag1|f(y)-f(x)|\le \sum_{k=0}^{n-1}|f(x_{k+1})-f(x_k)|\le n M \left(\frac{|y-x|}n\right)^\alpha= M|y-x|^\alpha n^{1-\alpha}.$$ As $1-\alpha<0$, the factor $n^{1-\alpha}$ gets arbitrarily small as $n\to\infty$, especially there exists $n$ such that $$\tag2n^{1-\alpha}<\frac{|f(x)-f(y)|}{M|y-x|^\alpha}.$$ (More explicitly, take $n>\left(\frac{M|y-x|^\alpha}{|f(x)-f(y)|}\right)^{\frac1{\alpha-1}}$). But (1) and (2) together imply $|f(y)-f(x)|<|f(y)-f(x)|$, a contradiction. Hence $f(x)\ne f(y)$ is not possible if $\alpha>1$.

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