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Nov 6, 2013 at 11:07 comment added Jean Van Schaftingen $H^{1/2} (\mathbb{R}^n_+) = B^{1/2, 2}_2 (\R^n_+) \supsetne B^{1/2, 2}_1 (\R^n_+)$
Nov 6, 2013 at 11:06 history edited Jean Van Schaftingen CC BY-SA 3.0
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Nov 5, 2013 at 20:26 comment added Delio Mugnolo sorry, i do not have triebel's book at my disposal right now. does his result hold for all $p$? i am trying to make sense of how it relates to the assertion of lions-magenes i quote. i guess $B^{\frac12,2}_1(\mathbb R^n_+)$ does not easily compare to $H^{\frac12}(\mathbb R^n_+)$, right?
Nov 5, 2013 at 16:23 comment added Jean Van Schaftingen According to Triebel "There does not exist a linear extension operator from $L^p (\partial \mathbb{R}^n_+)$ to $B^{1/p, p}_1 (\mathbb{R}^n_+)$", that is, there is no linear operator $\operatorname{ext}: L^p (\partial \Omega) \to B^{1/p, p}_1 (\Omega)$ such that $\operatorname{ext} \circ \operatorname{tr} = \operatorname{id}$.
Nov 5, 2013 at 11:53 comment added Delio Mugnolo what do you mean by "the linear extension operator [...] cannot be linear?
Nov 5, 2013 at 9:10 history answered Jean Van Schaftingen CC BY-SA 3.0