# Function on $[a,b]$ that satisifies a Hölder condition of order $\alpha > 1$ is constant

I want to show that if a function $f:[a,b]\rightarrow \mathbb R$ satisfies a Hölder condition of order $\alpha > 1$ then it is constant.

The way I think of it is as follows:

$$|f(x) - f(y)| < K|x-y|^\alpha$$

$$\frac{|f(x) - f(y)|} {|x-y]} < K|x-y|^{\alpha -1}$$

$$\lim_{y\rightarrow x} \frac{|f(x) - f(y)|} {|x-y]} \le \lim_{y\rightarrow x} K|x-y|^{\alpha -1} =0$$ As the limit is $0$, we can remove the modulus, so we get:

$$\lim_{y\rightarrow x} \frac{f(x) - f(y)} {x-y} = 0$$ So $f$ is derivable and $f'(x) = 0$ for all $x$ in $[a,b]$. Note that the reason we can add the limit $y\rightarrow x$ is because $[a,b]$ is closed in $\mathbb R$.

However, the question gives as a hint using the mean value theorem. I am not sure why one should do that. You would first have to prove that $f$ is derivable in a similar manner to what I did, and then prove that $f$ is constant. Or is there a simpler way to do it and I am missing it?

Also please inform me of any mistakes I did in the proof (if any)/ Thank you!

• The mean value theorem just allows you to conclude that $f'(x) = 0$ on $[a,b]$ means the function is constant there. – copper.hat Apr 14 '13 at 16:22
• @copper.hat Oh, OK. I just assumed that I well known. Thanks. – elaRosca Apr 14 '13 at 16:23
• Btw, this is called the $\alpha$-Hölder condition. The Lipschitz condition is only the special case where $\alpha=1$. – Yoni Rozenshein Apr 14 '13 at 16:24

For every $x\ne y$, split the interval $[x,y]$ into $n$ subintervals of length $\frac1n\cdot|x-y|$. The Hölder condition on each interval yields a bound $K\cdot \left(\frac1n\cdot|x-y|\right)^\alpha$. By the triangular inequality used $n-1$ times, $$|f(x)-f(y)|\leqslant n\cdot K\cdot \left(\tfrac1n\cdot|x-y|\right)^\alpha=K\cdot|x-y|^\alpha\cdot \frac1{n^{\alpha-1}}.$$ Now, consider the limit $n\to\infty$.
The mean value theorem is the usual way to show that if $f'$ on an interval, then $f$ is constant on the interval. If there were two points $a,b$ in the interval for which $f(a)\ne f(b)$, i.e. if $f$ were not constant on the interval, then there would be some point in the interval at which $f'$ is equal to $(f(a)-f(b)/(a-b)\ne0$.
However, I think the stated result can be proved without showing that $f'=0$ everywhere on the interval and without the mean value theorem so that, for example, it's true of functions whose domain is $\mathbb Q$ as well as those whose domain is $\mathbb R$.