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?

  • 2
    $\begingroup$ 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 $c=1$. $\endgroup$
    – panini
    Apr 26 '19 at 1:23
  • $\begingroup$ @panini your conclusion is right! $\endgroup$
    – Corrêa
    Apr 26 '19 at 1:31

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