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I would like to know how to show that the supremum $$\sup_S\frac{(x-y)^2 + (t-s)^2}{(|x-y|^2 + |t-s|)^{\epsilon}}$$ is less than infinity, where $0<\epsilon \leq 1$ and $$S = \{ (x,t), (y,s) \in [0,2\pi] \times[0,T] \mid (x,t) \neq (y,s)\}$$ where $0 < T \leq 1$. What can I use?

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up vote 2 down vote accepted

We have \begin{align} \frac{(x-y)^2+(t-s)^2}{((x-y)^2+|t-s|)^{\varepsilon}}&\leq ((x-y)^2+|t-s|)^{1-\varepsilon}+\frac{|t-s|(1+|t-s|)}{((x-y)^2+|t-s|)^{\varepsilon}}\\ &\leq ((x-y)^2+|t-s|)^{1-\varepsilon}+|t-s|^{1-\varepsilon}(1+T), \end{align} and using the fact that $s\mapsto x^s$ is increasing for $s\in (0,1)$, we get $$\sup_S\frac{(x-y)^2+(t-s)^2}{((x-y)^2+|t-s|)^{\varepsilon}}\leq (4\pi^2+T^2)^{1-\varepsilon}+(1+T)T^{1-\varepsilon}.$$

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Thank you!!!$$$$ – TagWoh Aug 4 '12 at 17:11

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