Thread: analysis

1. analysis

Problem Statement:Suppose that $f : \Re \rightarrow \Re$ is such that, for some $\epsilon > 0$, if $x,y \in \re$, then $|f(x)-f(y)|<|x-y|^{1+\epsilon}$. Prove that $f$ is constant.

I dont see how this is true. for instance, $f(x)=x/100$. Is there something I am missing?

2. I think it is something with the derivative... f'(x)=|f(x)-f(y)|/|x-y|=0 when epsilon is very small.

f'(x)=0 for all x ==> f(x) constant for all x.

Hmmm...

3. Originally Posted by Also sprach Zarathustra
I think it is something with the derivative... f'(x)=|f(x)-f(y)|/|x-y|=0 when epsilon is very small.

f'(x)=0 for all x ==> f(x) constant for all x.

Hmmm...
You could state this a little bit more rigorously.

1) Let $x_0\in\mathbb{R}$ then the above says that $\left|\frac{f(x_0)-f(y)}{x_0-y}\right|\leqslant |x_0-y|^{\varepsilon}$. Thus, since $\varepsilon>0$ we have that $\displaystyle \lim_{y\to x_0}|x_0-y|^{\varepsilon}=|x_0-x_0|^{\varepsilon}=0$ (by continuity of $f(y)=|x_0-y|^{\varepsilon}$) and thus by the squeeze theorem $f$ is differentiable at $x_0$ and $\displaystyle f'(x_0)=\lim_{y\to x_0}\frac{f(x_0)-f(y)}{x_0-y}=0$. Since $x_0$ was arbitrary it follows that $f'(x)=0$ for all $x\in\mathbb{R}$

2) Since $\mathbb{R}$ is an interval we may conclude that $f(x)=c$.

Just a comment though.