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I am trying to show that $f(x) = x^{{1}/{3}}$ is Hölder continous on $\mathbb{R}$, but I have not been able to make much progress.

To show that this is indeed true we would need to show that

$$\left|x^{1/3}-y^{1/3}\right| \leq C|x-y|^\alpha.$$ I think $C = 1$ and $\alpha = 1/3$, but I can't seen to make the inequalities work.

Math1000
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fosho
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  • I'm assuming you mean $[0,\infty)$ instead of $\mathbb R$ since $x\mapsto x^{\frac13}$ is not real-valued for negative $x$. – Math1000 Jul 23 '16 at 08:15
  • Yes, sorry -- you are right. – fosho Jul 23 '16 at 08:16
  • @Math1000: What do you mean? The function $x\mapsto x^{3}$ is bijective (look at the graph), so why isn't $x\mapsto x^{1/3}$ just the inverse? – Will R Jul 23 '16 at 09:39
  • @WillR How do you define $(-1)^{1/3}$? It is true that $(-1)^3 = 1$, but also true that $$\left(e^{2\pi i/3}\right)^3=\left(e^{4\pi i/3}\right)^3=1.$$ – Math1000 Jul 23 '16 at 10:41
  • @Math1000: $(-1)^{3}=-1$, not $1$. While there are non-real numbers that give $-1$ as their cube, that doesn't change the fact that $-1$ is a perfectly valid real cube root for $-1.$ Similarly, $-(a^{1/3})$ is a perfectly valid real cube root for $-a$ whenever $a\geq0$ is real. – Will R Jul 23 '16 at 10:58
  • Whoops, typo. Anyway, I suppose it is a moot point - the phrasing "Hölder continous on $\mathbb R$" would imply that the function $f$ is defined for real numbers. – Math1000 Jul 23 '16 at 11:07

2 Answers2

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Take $0<p<1$ and use $|u+v|^p\leq |u|^p+|v|^p$ (look here). By letting $u=x-y$, $v=y$, you find that $|x|^p-|y|^p\leq |x-y|^p$. By symmetry you get the other part.

Community
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Robert Z
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Here is a direct proof. For $x=0$ or $y=0$, the inequality holds obviously so assume $x,y \in (0, \infty)$ and without loss of generality $x \geq y$. Now the inequality holds for $x$ and $y$ if and only if $$(\sqrt[3]{x} - \sqrt[3]{y})^3 \leq x-y.$$ Expanding the left hand side gives the inequality holds if and only if $$\sqrt[3]{xy^2} - \sqrt[3]{x^2y} \leq 0.$$ This is true if and only if $$xy^2 \leq x^2y$$ which is equivalent to $x \geq y$, which is true by assumption.

M. Van
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