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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!

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

This was asked on the site before and requires no differentiability argument nor the mean value theorem, nor anything but the hypothesis, really...

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 $n\to\infty$.

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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 $\mathbb R$.

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  • $\begingroup$ I see now. I just assumed that is well known - I did not even use that in my answer. Thank you! $\endgroup$ – elaRosca Apr 14 '13 at 16:25
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    $\begingroup$ It is well known, in any reasonable sense of the term, and that is how it is proved. $\endgroup$ – Michael Hardy Apr 14 '13 at 16:27

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