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Let $f=\sqrt{|x|} \in \text{Lip}_\alpha(T)$, where $\text{Lip}_\alpha(T)$ is the set of Lipschitz function with Lipschitz constant $\alpha=1/2$ on the unit circle $T$. What is $$ \|f\|=\sup_{t\in T,h \ne 0} \frac{|f(t+h)-f(t)|}{|h|^\alpha} \text{ ?} $$

I need to evaluate this supremum to show that translation is not continuous in $\text{Lip}_\alpha(T)$, i.e., $$ \lim_{c \to 0} \|f(\cdot)-f(\cdot+c)\| \ne 0. $$

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    Is $T$ the unit circle?2012-09-24
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    What is the meaning of $\sqrt{|x|}$ if $x\in T$?2012-09-24
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    @DavideGiraudo Yes, $T$ is the unit circle.2012-09-24
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    @did I want $|x|$ to be the absolute value of x, for example $|-\pi|=\pi$. Then the map $x \mapsto \sqrt{|x|}$ is well-defined on $T$.2012-09-24
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    Then it seems $\|f\|=1$, or am I wrong?2012-09-24
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    @did I don't see how you get $\|f\|=1$. I have $\|f\| \le \frac{\sqrt{|h|}}{|\sqrt{x+h}+\sqrt{|x|}|}$.2012-09-24
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    Precisely. The norm $\|f\|$ is the supremum of the RHS of your last comment, which is $\leqslant1$ for every $x$ and $=1$ when $x=0$.2012-09-24
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    @did Can you help me with the translation of continuity? I try to work $\frac{|f(t+h)-f(t+c+h)-f(t)+f(t+c)|}{|h|^\alpha}$, but I get nowhere.2012-09-24

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Notice that any translated copy of square root is differentiable at $0$. Then prove the following more general statement: if $g$ is a function differentiable at $0$, then $\lim_{h\to0 } \frac{(\sqrt{|h|}+g(h)) -g(0) }{\sqrt{|h|} }=1$. Deduce that the $Lip_{1/2}$ norm of $\sqrt{|\cdot |}+g$ is at least $1$.