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Show that $f(x):=\sqrt{\lvert x\rvert}$ belongs to $C^{0,\frac{1}{2}}(\mathbb{R})$.

Hello, when I got it right, I have to show four things:

(1) $f\in C(\mathbb{R})$

(2) $f\in C(\overline{B_R(0)})$ for all $R>0$

(3) $$ \sup\limits_{x,y\in\overline{B_R(0)}), x\neq y}\left\{\frac{\lvert f(x)-f(y)\rvert}{\lvert x-y\rvert^{\frac{1}{2}}}\right\}<\infty $$

(4) $f$ uniform continious in $B_R(0)$ for all $R>0$

Am I right?

  • That depends on which definition of $C^{0,\frac12}(\mathbb{R})$ you are working with. Note, however, that proving (3) for all $R > 0$ suffices, the other points are consequences of that. – Daniel Fischer Oct 25 '13 at 17:21
  • We had so many different definition, but I believe here it is: $\Omega\subset\mathbb{R}^n$ a domain which does not have to be limited, $\lambda\in (0,1], k\in\mathbb{N}_0$, then $C^{k,\lambda}(\Omega):=\left{u\in C^k(\Omega)|\forall\overline{B}_R(0): u\in C^{k,\lambda}(\overline{B}_R(x))\right}$. And for a limited domain, we defined $C^{k,\lambda}(\overline{\Omega}):=\left{u\in C^k(\overline{\Omega})|\forall\alpha\in\mathbb{N}_0^n\text{ with }\lvert\alpha\rvert\leq k: sup... <\infty\right}$ So I do need this two definitions and the definition of $C^k(\overline{\Omega})$? –  Oct 25 '13 at 17:30
  • When I see it right, (3) $\Rightarrow$ (4), too. So it reduces to show (3)... –  Oct 25 '13 at 17:45
  • Right. (3) implies (4) directly for the same $R$. (3) for $R$ implies (2) for all $R' < R$. (3) for all $R$ implies (1). – Daniel Fischer Oct 25 '13 at 17:49
  • What I do not understand is, why one defines so many different sets then? –  Oct 25 '13 at 17:52
  • Don't ask me, I didn't invent that definition. – Daniel Fischer Oct 25 '13 at 17:57
  • Then I try to show (3): Consider any $R>0$, then $\frac{\lvert\sqrt{\lvert x\rvert}-\sqrt{\lvert y\rvert}\rvert}{\lvert x-y\rvert^{1/2}}\leq\frac{\lvert\lvert x\rvert-\lvert y\rvert\rvert}{\lvert x-y\rvert^{1/2}}\leq\frac{\lvert x-y\rvert}{\lvert x-y\rvert^{1/2}}=\lvert x-y\rvert^{1/2}$ so the supremum over all $x,y\in \overline{B}_R(0), x\neq y$ is $\sqrt{2R}<\infty$? –  Oct 25 '13 at 18:21
  • You don't generally have $\lvert \sqrt{\lvert x\rvert} - \sqrt{\lvert y\rvert}\rvert \leqslant \lvert \lvert x\rvert - \lvert y\rvert \rvert$. Consider the case $0 \leqslant y < x$, and then argue that all other cases follow from that. – Daniel Fischer Oct 25 '13 at 18:25
  • When is $\lvert\sqrt{\lvert x\rvert}-\sqrt{\lvert y\rvert}\rvert\leq\lvert\lvert x\rvert-\lvert y\rvert\rvert$ not fullfilled? –  Oct 25 '13 at 18:35
  • $y = 0 < x < 1$, for example. – Daniel Fischer Oct 25 '13 at 18:40
  • Ok, yes. I do not know what you meant with: "Consider the case $0\leq y<x$ and argue that all...". –  Oct 25 '13 at 18:43

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$$\frac{\left||x|^{1/2}-|y|^{1/2}\right|}{|x-y|^{1/2}}=\frac{||x|-|y||}{|x-y|^{1/2}\left(|x|^{1/2}+|y|^{1/2}\right)}\leq\frac{|x-y|^{1/2}}{|x|^{1/2}+|y|^{1/2}}=\sqrt{\frac{|x-y|}{|x|+|y|+2|x|^{1/2}|y|^{1/2}}}\leq\sqrt{\frac{|x|+|y|}{|x|+|y|}}=1.$$ This gives that $\sqrt{|x|}$ is uniformly Holder-continuous with exponent $\frac{1}{2}$. (1),(2) and (4) follow.

Jack D'Aurizio
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