Proving that $f(x) = \vert x \vert^{\alpha}$ is Holder continuous, inequality help

Question

The definition of $\alpha$-Holder continuity for a function $f(x)$ at the point $x_0$ is that there exist a constant $L$ such that for all $x \in D$ such that

\begin{equation} \vert f(x) - f(x_0) \vert \leq L \vert x - x_0 \vert^\alpha \end{equation}

The function $f(x) = \vert x \vert^\alpha$ is cited as the canonical example of an $\alpha$-holder continuous function. I'm trying to verify this fact. So far I have \begin{equation} \vert f(x) - f(x_0) \vert = \vert \vert x \vert^\alpha - \vert x_0 \vert^\alpha \vert \leq \vert x^\alpha - x_0^\alpha \vert \end{equation} I would like to get from $\vert x^\alpha - x_0^\alpha \vert$ to $L \vert x - x_0 \vert^\alpha$ to complete the proof, but I'm not sure what inequality allows me to do this. I was looking at the concavity of the $x^\alpha$ function, but I'm not sure if that's useful in this case.

Answer 1

On any set which is bounded away from $0$, $f$ is actually Lipschitz, as you can check by bounding its derivative. So on such sets $f$ is trivially $\alpha$-Holder continuous, since it is actually $1$-Holder continuous. The issue is all near $0$.

So to finish, identify the bigger number as $x+h$ and the smaller one as $x$, you get $||x+h|^\alpha - |x|^\alpha| \leq \alpha |x|^{\alpha-1} h$ if $x>0$ and otherwise $h^\alpha$. Divide by $h^\alpha$ and send $h \to 0$. In the first case you get $0$, in the second you get $1$, either way you get something that is at most $1$.

Answer 2

For $x_0 = x$, $x=0$ or $x_0=0$ the inequality is obvious. W.l.o.g. let $x > x_0$. Then

$$ \frac{\left|x^\alpha-x_0^\alpha\right|}{|x-x_0|^\alpha} =\frac{\left|1-\left(\frac{x_0}{x}\right)^\alpha\right|}{|1-\left(\frac{x_0}{x}\right)|^\alpha} $$

Since $0 < 1- x_0/x < 1$ and $0<\alpha\leq 1$we get $|1-\left(\frac{x_0}{x}\right)|^\alpha \geq |1-\left(\frac{x_0}{x}\right)|$. Similarly $\left|1-\left(\frac{x_0}{x}\right)^\alpha\right| \leq \left|1-\left(\frac{x_0}{x}\right)\right|$. So

$$ \frac{\left|x^\alpha-x_0^\alpha\right|}{|x-x_0|^\alpha} \leq \frac{\left|1-\left(\frac{x_0}{x}\right)\right|}{|1-\left(\frac{x_0}{x}\right)|} = 1 $$