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In lecture we had the following theorem:

Let $\Omega \subseteq \subseteq \mathbb{R}^n$. Then $$H^1(\Omega) = H^1_0(\Omega) \oplus \{u \in H^1 : \Delta u = 0\}$$ where $\Delta u$ is understood in the distributional sense. Moreover, if $\Omega$ is of class $C^1$, $u\vert_{\partial \Omega}$ does admit a unique harmonic extension.

Now the proof of the above is straight forward, however, the moreover part is where I stumble: In the notes it is phrased that this is a direct consequence of the decomposition, but I do not see this. I mean, we do have $$u\vert_{\partial \Omega} = u_0\vert_{\partial\Omega} + u_1 \vert_{\partial \Omega} = u_1\vert_{\partial \Omega}$$ when $u = u_0 + u_1$ and hence $u_1$ is a harmonic extension of $u\vert_{\partial \Omega}$, but why is this extension unique? The reason why I am asking this is because I need to proof that $$H^1_0(\Omega) = \{u \in H^1(\Omega) : u\vert_{\partial \Omega} = 0\}.$$

TheGeekGreek
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    The last equation can be shown independently of the above using the trace operator. If this would be an option for you, too. – Jonas Lenz Aug 05 '18 at 18:25
  • @JonasLenz I am aware of this fact, but in the oral exam I should reproduce the proof given in the lecture. Sigh... – TheGeekGreek Aug 05 '18 at 18:26
  • that's unfortunate. – Jonas Lenz Aug 05 '18 at 18:29
  • I don't see why you need the trace characterization here; the uniqueness amounts to saying that if $u\in H^1_0(\Omega)$ satisfies $\Delta u=0$ in the distributional sense, then $u\equiv 0$. A proof of that amounts to justifying $\int |\nabla u|^2 = -\int u \Delta u$. –  Aug 05 '18 at 20:34
  • https://en.wikipedia.org/wiki/Dirichlet_problem - we need a Green function for the Dirichlet problem, if the existence is part of the problem... – dan_fulea Aug 05 '18 at 21:16

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