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Update: After some more thinking and asking I've come to the conclusion that there is no reasonable way to achieve this for all possible $\varphi$ because of the mixed terms. I believe something useful can only be said under additional assumptions on the behaviour on the boundary both of $g$ and the $\varphi$, something I can do in my original problem, so I've decided to move on.

Let $\Omega \subset \mathbb{R}^n$ be a bounded Lipschitz domain (or at most $C^{1, 1}$) with boundary $\Gamma$ split in open, disjoint subsets $\Gamma_1, \ldots, \Gamma_k$ each with Lipschitz boundary as well. Let $g \in H^{- 1 / 2} ( \Gamma)$, the dual of the Hilbert space $H^{1 / 2} ( \Gamma) = W^{1 / 2, 2} ( \Gamma) = \{ v \in L^2 ( \Gamma) : ( v, v)_{1 / 2} < \infty \}$ where the scalar product is given by

$$ ( u, v)_{1 / 2} := \int_{\Gamma} uv + \int_{\Gamma} \int_{\Gamma} \frac{u ( x) v ( y)}{| x - y |^n} d x d y. $$

Question: Under what assumptions may I split the action of $g$ in a way like the following? $$ \langle g, \varphi \rangle_{- 1 / 2, \Gamma} \overset{!}{=} \sum_{i = 1}^k \langle g_i, \varphi_{| \Gamma_i} \rangle_{- 1 / 2, \Gamma_i}, $$ where the $g_i$ are somehow the "restrictions" of $g$ and the $\langle \cdot, \cdot \rangle_{- 1 / 2, \Gamma_i}$ are the duality pairings of $H^{- 1 / 2} ( \Gamma_i) \times H^{1 / 2} ( \Gamma_i)$.

Note: I'm starting to suspect this is not possible without changing the subspaces where the $g_i$ live (but how?).

On the one hand, using the representation of $g$ with the scalar product of $H^{1 / 2} ( \Gamma)$ has led me nowhere because of the double integral, where mixed terms appear. On the other hand, because the $\Gamma_i$ are Lipschitz, there exist continuous embeddings $H^{1 / 2} ( \Gamma_i) \overset{E_i}{\hookrightarrow} H^{1 / 2} ( \Gamma)$ which allow to define for example

$$ \langle g_i, \varphi_{| \Gamma_i} \rangle_{- 1 / 2, \Gamma_i} := \langle g, E_i ( \varphi_{| \Gamma_i}) \rangle_{- 1 / 2, \Gamma}, $$

but this doesn't achieve much because the extensions with $E_i$ aren't extensions by zero.

Help anyone? Thanks!

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Where do you get this definition of $W^{\frac{1}{2},2}$? I need to ruminate on it a bit. – Daniel Spector Feb 21 '13 at 7:55
Hi, thanks for your interest. I guess you mean the scalar product, right? See for instance these lecture notes, proposition 4.4: . I used this scalar product because the one that came immediately to mind to me from the Gagliardo norm was more complicated, involving differences of the arguments. I haven't actually checked whether they are equivalent, but it seems plausible enough. I've also left out the details about the surface measures using the charts for the boundary $\Gamma$, etc., thinking that was non-essential. – Miguel Feb 21 '13 at 9:09
I don't have enough points to edit, but the exponent should be $N+1$, since here $\lambda=\frac{1}{2}$ and $p=2$. That was part of the question on my mind. Thanks for the notes. I will have a look. – Daniel Spector Feb 21 '13 at 9:22
But $n$ is the dimension of the space and the integral is on the boundary so that's one dimension less... – Miguel Feb 21 '13 at 10:23
Okay. So it is n-1+1. Got it. So the question is if it is possible to write a linear functional as the sum of linear functionals along pieces of the boundary then? – Daniel Spector Feb 21 '13 at 11:56

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