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.

  • $\begingroup$ $f\in H^r(\mathbb R^n)$ 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 $L^2(\mathbb R^n)$. $\endgroup$ – user8268 Jan 7 '15 at 14:17
  • $\begingroup$ @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? $\endgroup$ – Löwe Simon Jan 7 '15 at 14:21
  • $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ Thank you for your answer! It makes things at least a little bit clearer. $\endgroup$ – Löwe Simon Jan 12 '15 at 16:26
  • 3
    $\begingroup$ 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? $\endgroup$ – mattia.penati Nov 4 '15 at 17:34
  • 1
    $\begingroup$ Should the inclusion $L^2(\Omega)\subset H^{1/2}(\Omega)\subset H^1(\Omega)$ be the other way around? $\endgroup$ – Jack Sep 11 '16 at 12:39
  • $\begingroup$ @mattia.penati No, sorry. I would start looking at Lions' book, though. $\endgroup$ – Three.OneFour Sep 12 '16 at 8:58
  • $\begingroup$ @Jack Yes, of course. Thanks, fixed. $\endgroup$ – Three.OneFour Sep 12 '16 at 9:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.