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Let $n\in\mathbb{N}$ and $\Omega \subseteq \mathbb{R}^n$ sufficiently smooth. Then we have the Hilbert space $H^1(\Omega)$ and the trace operator $\operatorname{tr}: H^1(\Omega) \to L^2(\partial \Omega)$. The continuity of $\operatorname{tr}$ implies that $\operatorname{ker} \operatorname{tr}$ is closed in $H^1(\Omega)$. Hence, \begin{align} H^{1/2}(\partial \Omega) := \operatorname{ran} \operatorname{tr} \quad \text{with} \quad \|\phi\|_{H^{1/2}} := \inf \{\|f\|_{H^1} : \operatorname{tr} f = \phi\} \end{align} is a Banach space because it is isomporphic to quotient space $H^1(\Omega)\big/\operatorname{ker}\operatorname{tr}$. But why is it even a Hilbert space?

I would appreciate some literature about it. Furthermore, I think sometimes I even saw $H^{1/2}(\Omega)$ instead of $H^{1/2}(\partial\Omega)$. Is there a reason for that?

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    $\begingroup$ Quotient space of a Hilbert space is again a Hilbert space (naturally isomorphic to the orthogonal complement of the subspace you are factoring over). $\endgroup$ Sep 6, 2018 at 9:16
  • $\begingroup$ This is a really simple answer to my question. Thank you. I feel kind of stupid now^^ $\endgroup$ Sep 6, 2018 at 9:18
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    $\begingroup$ $H^{1/2}(\Omega)$ is a different space from $H^{1/2}(\partial\Omega)$. $\endgroup$ Sep 6, 2018 at 10:15
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    $\begingroup$ Without teaching Grandma to suck eggs, if you are convinced that $H^{1/2}$ is a Hilbert space, then identifiying an inner product on that space allows you to define $H^{-1/2}$ as Hilbert also... $\endgroup$
    – asymptotic
    Sep 6, 2018 at 10:22

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