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I want to show that $$f: \mathbb{R}_{\geq0} \to \mathbb{R}_{\geq0}$$ $$x \mapsto x^s$$ is Holder continuous with Holder exponent $s \in \mathbb{R}$, where $0<s \leq 1$. So what I want to show is that $\exists \hspace{2 mm} C \in \mathbb{R}_{\geq0}$ sucht that for all $ x,y \in \mathbb{R}_{\geq0}, $
$$|x^s -y^s| \leq C|x-y|^s$$
and therefore, assuming, wlog $\hspace{1mm} x>y$ $$(x^s -y^s) \leq C(x-y)^s.$$

I thought about Bernoulli's inequality but couldn't make that work.I thought about the binomial theorem, but didn't know how to handle the fact that $s \in \mathbb{R}$.

ghthorpe
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    the function is concave. – HyJu Dec 24 '16 at 16:52
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    To motivate a general solution one might tackle the case $s = \frac{1}{2}$ as typical of those where $s \in (0,1)$. Certainly the case $s = 1$ poses no difficulty. – hardmath Dec 24 '16 at 16:56

2 Answers2

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Assume $x>y$.

$$x^s-y^s \leq C(x-y)^s$$

$$\Leftrightarrow x^s \leq C(x-y)^s+y^s$$

Claim: this holds for all $(x,y)$ when $C=1$. Proof: Because $0<s\leq 1$, for all $a,b \geq 0$

$$1 \leq (\frac{a}{a+b})^s+(\frac{b}{a+b})^s$$

$$\Leftrightarrow (a+b)^s \leq a^s+b^s $$

Now set $a=x-y$ and $b=y$.

fes
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    Why is $(\frac{a}{a+b)})^s+(\frac{b}{a+b)})^s\geq1$ given that $s\in \mathbb{R}$? – ghthorpe Dec 24 '16 at 17:52
  • The derivative of $z^s$ is $s\cdot z^{s-1}$, then that derivative is continuous in $0<z\le1$, but if we take the limit as $z$ approaches $1$, then that derivative approaches $s$, then, if we define $f(z) = z^s - z$, then $f'(x) = s\cdot z^{s-1} - 1$ for all $z$ in which $0<z<=1$, then taking limits in both sides, we'd have $\lim_{x\to1}f'(x) = s-1$, and with the fact that $f'(x)$ is continuous, we'd have a certain interval containing 1 which intersection with $(0,1]$ we'd have $f'(x) \le 0$ (since $s-1\le0$), so we wouldn't have $f'(x)>0$ in all of this range, is there any mistake in all this? – Geovane Coelho Oct 03 '20 at 02:22
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    You can solve $z^s-z=0$ to find the zero points $z=0$ and $z=1$. $z^s-z > 0$ e.g. when $z=\frac{1}{2}$. By continuity $z^s-z \geq 0$ for all $z \in [0,1]$. – fes Oct 03 '20 at 06:05
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    For future people looking at this: use the above remark to find that $z^s \geq z$, let $z = a/(a+b)$ and $b/(a+b)$ and add the resulting inequalities to obtain the mysterious inequality in the answer – yoshi May 05 '21 at 11:53
  • @fes Hi fes, sorry to comment on this old post. I have a question. We can easily see that $z=0$ and $z=1$ are zero points of the equation $z^s-z=0$ for $0<s<1$. But how we can prove that there is not any other zero point which is neither $0$ nor $1$? Thanks. – Sam Wong Mar 05 '24 at 09:31
  • @yoshi The crucial step is to prove the zero points of $z^s-z=0$ for $0<s<1$ are only $s=0$ and $s=1$. To be honest, your comment is a very trivial step. – Sam Wong Mar 05 '24 at 09:33
  • @SamWong $z^s-z=z(z^{s-1}-1)$. $z^s-z=0$ iff $z=0$ or $z^{s-1}-1=0$. $z^{s-1}-1=0$ iff $z=1$. – fes Mar 05 '24 at 13:58
  • @fes Why it is true that $z^{s-1}-1=0$ iff $z=1$? This is the step that I could not prove yesterday. Thanks. – Sam Wong Mar 06 '24 at 16:08
  • @SamWong To solve $z^{s-1}=1$, when $0<s<1$, e.g. apply $^{\frac{1}{s-1}}$ on both sides. The case $s=1$ is trivial. – fes Mar 06 '24 at 17:15
  • @fes I can understand the argument but I still feel like it is a bit strange. Here is the reason why I think it is strange. I know $s\in (0,1)$, but let us consider the case $z^2=1$ where $s-1=2$ here. When we apply $^{1/2}$ on both sides, we get $z=1$. We miss the solution $z=-1$. Our $s$ is in $(0,1)$ and cannot be $3$, but my example still makes me not convinced. Can you tell what is the difference between the case $s\in(0,1)$ and my case $s=3$? And why our argument for $s\in(0,1)$ does not miss any solution? Thanks. – Sam Wong Mar 06 '24 at 17:46
  • @SamWong Hmm. perhaps it is clearer if we take $log$ instead (in our case $z>0$). We obtain $(s-1)\log(z)=0$ – fes Mar 06 '24 at 18:30
  • @fes Yea it makes more sense right now. Thanks! – Sam Wong Mar 06 '24 at 19:46
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$$|x^s - y^s||x^{1-s} + x^{1-2s}y^s + ... + y^{1-2s}| = |x - y|$$

$\forall{x,y},\, \exists{m\gt0}\;\text{such that} \;m\le|x^{1-s} + x^{1-2s}y^s + ... + y^{1-2s}|$

If $|x - y|\lt 1$, $|x-y|\le |x-y|^s$.

Hence,

$$|x^s-y^s|\le\frac{1}{m}|x-y|\le\frac{1}{m}|x-y|^s$$

If, however, $|x-y|\gt1$, then $|x-y|\ge |x-y|^s$.

By the Archimede's theorem (or the Archimedean property of $\mathbb{R}$), $\exists{k}\;\text{such that}\;k|x-y|^s\ge|x-y|$.

Then, $$|x^s-y^s|\le\frac{1}{m}|x-y|\le\frac{k}{m}|x-y|^s$$

Notice that $k\ge1$, so $\frac{k}{m}|x-y|^s\ge\frac{1}{m}|x-y|^s$

Therefore, $$\forall{x,y \in \mathbb{R}_{\geq0}},\;|x^s-y^s|\le\frac{k}{m}|x-y|^s$$.

If we let $C=\frac{k}{m}$, then this becomes $$|x^s-y^s|\le C|x-y|^s$$

Hence proven.

Divide1918
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