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 $\varphi$.

Let $\Omega \subset \mathbb{R}^n$ be a bounded Lipschitz domain (or at most $C^{1, 1}$) with boundary 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 $g$, $\varphi$ may be taken in $C^{\infty} ( \Gamma)$ and the $\langle \cdot, \cdot \rangle_{- 1 / 2, \Gamma_i}$ are the duality pairings of $H^{- 1 / 2} ( \Gamma_i) \times H^{1 / 2} ( \Gamma_i)$.

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.

I know that the $\Gamma_i$ being 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 I think this doesn't achieve much because the extensions with $E_i$ aren't extensions by zero. This somehow reflects the same situation as with the "mixed terms" in the double integral.

Help anyone? Thanks!



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