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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 $0< \beta <1$.

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    @Martin: I agree. Re-tagged as holder-spaces. (Sorry, but tags don't support umlauts...)2012-07-02

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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.