The Sobolev Space $H^{1/2}$

This is a very stupid question. In my course on linear PDEs, the professor used $H^{1/2}$ without defining it, and I have been looking on google trying to find a definition, but the only related thing I found was $H^{-1/2}$ as being the dual space to $H^{1/2}$ which does not really help. Plugging in the one half in the defintion of the standard Sobolev spaces $H^m$ does not make any sense. Could someone quickly help me out there?

Thank you.

• $f\in H^r(\mathbb R^n)$ means that $(1+|\xi|^2)^{r/2}\hat f(\xi)$ (as a function of the variable $\xi\in\mathbb R^n$, where $\hat f$ is the Fourier transform of $f$) is in $L^2(\mathbb R^n)$. – user8268 Jan 7 '15 at 14:17
• @user8268 So for $r=1/2$ it would be a power of one fourth? That is a somewhat strange definition. Where does it come from? And why does it suddenly appear in trace theory? – Löwe Simon Jan 7 '15 at 14:21
• In my point of view, if you just use it in PDE class, maybe replace $H^{1/2}$ by $L^2(\partial\Omega)$ will be better. It just declare the boundary value of your solution. – spatially Jan 8 '15 at 1:34

There are multiple definitions of $H^{1/2}(\partial Ω)$ which are equivalent if the boundary is regular enough. The most intuitive is probably as the range of the trace operator $tr\colon H^1(Ω) \to L^2(\partial Ω)$: $$H^{1/2}(\partial Ω) = \{ u ∈ L^2(\partial Ω) \;|\; ∃ \tilde u ∈ H^1(Ω)\colon u = tr(\tilde u) \}, \quad \| u \|_{H^{1/2}(\partial Ω)} = \inf \{ \| \tilde u \|_{H^1(Ω)} \;|\; tr(\tilde u) = u \}.$$

Intrinsic definitions of $H^{1/2}(\partial Ω)$ are quite technical in detail as $\partial Ω$ is a $(n-1)$-dimensional manifold. In case $\partial Ω$ is a plane you have $\partial Ω \cong \mathbb R^{n-1}$ and you end up having to define $H^{1/2}(Ω')$ for $Ω' \subset \mathbb R^{n-1}$. For a general Lipschitz boundary you can "straighten" your boundary locally to look like a plane (this is a general technique while working with manifolds) and in the end you ask for a transformation of your boundary function to be in $H^{1/2}(Ω')$. (See [1] for details.)

All in all, you end up having to define $H^{1/2}(Ω')$. There are multiple ways for doing that, one using the Hölder-like seminorms as mentioned by Thomás, one using the Fourier coefficients (see Fractional Sobolev Spaces on Wikipedia) and one using interpolation between $L^2(Ω')$ and $H^1(Ω')$ (see [1] again).

For understanding the actual behavior of functions in $H^{1/2}(Ω')$ the definition using the Hölder-like norm is probably the best: $$H^{1/2}(Ω') = \left\{ u ∈ L^2(Ω') \;|\; \| u \|_{L^2(Ω')} + \int_{Ω'}\int_{Ω'}\frac{|u(x)-u(y)|^2}{|x-y|^{n+1}} dx \, dy < \infty \right\}$$ Note that the additional integral term is somewhat like a Hölder condition. I like to think of $H^1(Ω) \subset H^{1/2}(Ω) \subset L^2(Ω)$ as something analogous to $C^1(Ω) \subset C^{1/2}(Ω) \subset C^0(Ω)$ in terms of regularity. That this is really analogous can be made precise using interpolation theory.

[1] Lions, J. L., & Magenes, E. (1972). Non-Homogeneous Boundary Value Problems and Applications.

• Thank you for your answer! It makes things at least a little bit clearer. – Löwe Simon Jan 12 '15 at 16:26
• I'm sorry, but do you know any type of reference for a proof of the equivalence between the quotient norm and the Hölder-like norm? – mattia.penati Nov 4 '15 at 17:34
• Should the inclusion $L^2(\Omega)\subset H^{1/2}(\Omega)\subset H^1(\Omega)$ be the other way around? – Jack Sep 11 '16 at 12:39
• @mattia.penati No, sorry. I would start looking at Lions' book, though. – Three.OneFour Sep 12 '16 at 8:58
• @Jack Yes, of course. Thanks, fixed. – Three.OneFour Sep 12 '16 at 9:00