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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 $s \in \mathbb{R}$.

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    $\begingroup$ the function is concave. $\endgroup$
    – HyJu
    Commented Dec 24, 2016 at 16:52
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    $\begingroup$ 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. $\endgroup$
    – hardmath
    Commented Dec 24, 2016 at 16:56

2 Answers 2

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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 $b=y$.

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    $\begingroup$ Why is $(\frac{a}{a+b)})^s+(\frac{b}{a+b)})^s\geq1$ given that $s\in \mathbb{R}$? $\endgroup$
    – ghthorpe
    Commented Dec 24, 2016 at 17:52
  • $\begingroup$ 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? $\endgroup$ Commented Oct 3, 2020 at 2:22
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    $\begingroup$ 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 $z \in [0,1]$. $\endgroup$
    – fes
    Commented Oct 3, 2020 at 6:05
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    $\begingroup$ 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 $\endgroup$
    – yoshi
    Commented May 5, 2021 at 11:53
  • $\begingroup$ @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. $\endgroup$
    – Sam Wong
    Commented Mar 5 at 9:31
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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.

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