Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
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
add comment

1 Answer 1

up vote 4 down vote accepted

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 $$||x|^\beta - |y|^\beta| \le |x-y|^\beta.$$

This is basically the same trick as in the proof of this form of triangle inequality.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.