# $x^\alpha$ as an example of an $\alpha$-Hölder continuous function

I saw the following statement by user Mark Joshi in response to the question : Non-trivial exemple of Hölder continuous function.

$$x^\alpha$$ for $$x > 0$$ and $$0$$ otherwise for $$0 < \alpha < 1$$ is Holder continuous of order $$\alpha$$

I cannot seem to prove this statement. How do I proceed to show that the function $$f(x) = x^{\alpha}$$ is Holder continuous of order $$\alpha<1$$, i.e., $$|f(x_1) - f(x_2)| \leq c |x_1-x_2|^{\alpha}$$ for all $$x_1, x_2 \in (0, \infty)$$, and some $$c>0$$.

Take $$x_{0} \geq 0$$ and, for $$x > x_{0}$$, define $$g(x) = x^{\alpha} - x_{0}^{\alpha} - (x-x_{0})^{\alpha}.$$ Then $$g'(x) = \alpha x^{\alpha-1} - \alpha(x-x_{0})^{\alpha-1}.$$ Since $$\alpha - 1 < 0$$ and $$x > x_{0}$$, $$g'(x) \le 0$$. So, $$g'$$ is decreasing with $$g(x_{0}) = 0$$, for $$x > x_{0}$$. Can you continue?
• Yes, I think I can continue. Since $g'$ is decreasing for $x>x_0$, with $g(x_0)=0$, it implies that $g'\leq 0$ for $x>x_0$. Using this we can conclude that $g(x) \leq g(x_0) = 0$ for $x>x_0$. I think that should imply the result with $c=1$. Apr 26 '19 at 1:23