# Why is $H^{1/2}$ a Hilbert space?

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?

• Quotient space of a Hilbert space is again a Hilbert space (naturally isomorphic to the orthogonal complement of the subspace you are factoring over). Sep 6, 2018 at 9:16
• This is a really simple answer to my question. Thank you. I feel kind of stupid now^^ Sep 6, 2018 at 9:18
• $H^{1/2}(\Omega)$ is a different space from $H^{1/2}(\partial\Omega)$. Sep 6, 2018 at 10:15
• 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... Sep 6, 2018 at 10:22