Is $u:B_1(0)\rightarrow \mathbb{R}, \beta$-Hölder continuous given by $u(x) =|x|^\beta$?

Is Is $u:B_1(0)\rightarrow \mathbb{R}, \beta$-Hölder continuous given by $u(x) =|x|^\beta$? i.e $$\sup \left \{ \dfrac{||x|^\beta-|y|^{\beta}|}{|x-y|^{\beta}} : \begin{matrix}|x|<1 \\ |y|<1 \end{matrix}\right \} \le C <\infty$$ Assume $0< \beta <1$.

-
Do you really think that it was necessary to create a new tag regularity for this question? And if you think that this tag would be useful, you should add description of the tag in the tag-wiki and tag-excerpt; since the word regularity is used in many different meanings in mathematics. (I don't think that such tag is needed.) –  Martin Sleziak Jul 2 '12 at 4:21
Really, none of the answers you received on 7 questions is satisfying? I am asking because this is the message your 0% accept rate is sending. –  Did Jul 2 '12 at 6:22
If you don't know what did is talking about in his comment, you can read more here: How do I accept an answer? and Why should we accept answers?. –  Martin Sleziak Jul 2 '12 at 10:04
@Martin: I agree. Re-tagged as holder-spaces. (Sorry, but tags don't support umlauts...) –  Willie Wong Jul 2 '12 at 11:24
Since we know that $|a+b|^\beta\le |a|^\beta + |b|^\beta$ (see e.g. here), we get for $a=x-y$, $b=y$ $$|x|^\beta \le |x-y|^\beta + |y|^\beta$$ which is equivalent to $$|x|^\beta - |y|^\beta \le |x-y|^\beta.$$ By symmetry we also have $|y|^\beta - |x|^\beta \le |x-y|^\beta$, which together gives $$||x|^\beta - |y|^\beta| \le |x-y|^\beta.$$