# proving that $f(x) = x^s$ is holder continuous with holder exponent s

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 $s \in \mathbb{R}$.

• the function is concave.
– HyJu
Commented Dec 24, 2016 at 16:52
• 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. Commented Dec 24, 2016 at 16:56

## 2 Answers

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 $$0, for all $$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 $$b=y$$.

• Why is $(\frac{a}{a+b)})^s+(\frac{b}{a+b)})^s\geq1$ given that $s\in \mathbb{R}$? Commented Dec 24, 2016 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? Commented Oct 3, 2020 at 2:22
• 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
Commented Oct 3, 2020 at 6:05
• 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 Commented May 5, 2021 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. Commented Mar 5 at 9:31

$$|x^s - y^s||x^{1-s} + x^{1-2s}y^s + ... + y^{1-2s}| = |x - y|$$

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

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.