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I have a function $f(x)$ in the space $H^{2,-s}(\mathbb{R}^3)$; have this limit sense $$\lim_{|x-y|\to 0} f(x)$$ ? ($y$ is a fixed point) If i have $f$ in $H^2$ I can say that $$\lim_{|x-y|\to 0} f(x)=f(y)$$ because $H^2(\mathbb{R}^3)\subset C(\mathbb{R}^3)$, but in the space $H^{2,-s}$? In general I can say nothing, isn't it?

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Not quite sure. If you are in $H^{2-s}(\mathbb R^3)$ you can embed in $W^{1,\frac{6}{1+2s}}(\mathbb R^3)$, and from there you can get into $C^{0,\alpha}({\mathbb R}^3)$ by Morrey's inequality, for all $\alpha,s\in (0,1)$ such that $s+\alpha=\frac12$. So, I would say there's plenty of space for you beneath $H^2(\mathbb R^3)$ to still get a continuous function in the end.

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Can you write me your definition of $C^{0,\alpha}$? – Sue Mar 12 '13 at 21:15
The same used here , from where I've taken the second embedding (I can never remember the embeddings by heart). – Delio Mugnolo Mar 12 '13 at 21:27
So the limit that I considered in my first answer exists? From your reasoning I deduce that I can find a continuous representative for my function in SH^{2,-s}$, isn't it? – Sue Mar 12 '13 at 21:28
Yes, so I believe. – Delio Mugnolo Mar 12 '13 at 22:36

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