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.

  • 1
    $\begingroup$ $f\in H^r(\mathbb R^n)$ 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, 2015 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? $\endgroup$ Jan 7, 2015 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, 2015 at 1:34
  • 1
    $\begingroup$ @LöweSimon: That is not a stupid question. It is a extremely good question. $\endgroup$
    – Axion004
    Oct 15, 2019 at 20:59
  • $\begingroup$ The answer to this question really depends on the context. One thing is $H^{1/2}(\mathbb R^n)$, one thing is $H^{1/2}(M)$ when $M$ is a Riemannian manifold, yet another thing is $H^{1/2}(\partial \Omega)$ when $\Omega$ is a domain. I recommend the hitchhiker's guide to fractional Sobolev spaces. $\endgroup$ Dec 4, 2020 at 16:49

1 Answer 1


$\newcommand{\tr}{\operatorname{tr}}$ There are multiple definitions of $H^{1/2}(\partial Ω)$ which are equivalent if the boundary is regular enough (Lipschitz continuous). The technically simplest, and how it usually appears in lectures on weak solutions for partial differential equations, is as the range of the trace operator $\tr \colon H^1(Ω) \to L^2(\partial Ω)$:

$$ \begin{align}H^{1/2}(\partial Ω) &:= \tr(H^1(\Omega)) := \{ v \in L^2(\partial Ω) \;|\; \exists u \in H^1(Ω) : \tr(u) =v \},\\ \| v \|_{H^{1/2}(\partial Ω)} &:= \inf \{ \| \tilde u \|_{H^1(Ω)} \;|\; \tilde u \in H^1(Ω) \land \tr(\tilde u) = v \}.\end{align}$$

The definition of the norm arises as follows. By the First isomorphism theorem for Banach Spaces, the trace operator induces an isomorphism $$ \begin{align}\widehat{\tr}\colon H^1(Ω) / \operatorname{ker} \tr &\to \tr(H^1(\Omega)), \\ [u] &\mapsto \tr(u)\end{align} $$ where $\operatorname{ker} \tr$ is the kernel of the trace operator, $[u] \in H^1(Ω) / \operatorname{ker} \tr$ denotes an equivalence class with representative $u \in H^1(Ω)$ and the norm on the quotient space is given by $$ \| [u] \|_{H^1(Ω) / \operatorname{ker} \tr} := \inf \{ \| \tilde u \|_{H^1(Ω)} \;|\; \tilde u \in [u] \}. $$ This is a general construct for the quotient norm on Banach spaces. As a side remark, there holds $\operatorname{ker} \tr = H^1_0(Ω)$ (the latter space is defined as completion of $C^\infty_0(\Omega)$ in $H^1(\Omega)$). One can then define a norm on $H^{1/2}(\partial Ω)$ using $\widehat{\tr}$:

$$ \| v \|_{H^{1/2}(\partial Ω)} := \| \widehat{\tr}^{-1}(v) \|_{H^1(Ω) / \operatorname{ker} \tr} = \inf \{ \| \tilde u \|_{H^1(Ω)} \;|\; \tilde u \in \widehat{\tr}^{-1}(v) \}$$

using that $\tilde u \in \widehat{\tr}^{-1}(v)$ if and only if $\tilde u \in H^1(\Omega)$ and $\tr(\tilde u) = v$ one arrives at the expression of the norm given in the beginning.

This definition of $H^{1/2}(\partial Ω)$ is not very useful if one wishes to check whether a specific function $v \in L^2(\partial Ω)$ is in $H^{1/2}(\partial Ω)$ and it does not explain the name $H^{1/2}(\partial Ω)$ (which came later historically).

The other definition of $H^{1/2}(\partial Ω)$ I present here is 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 (Sobolev-Slobodeckij norm) is probably the best: $$H^{1/2}(Ω') = \left\{ v ∈ L^2(Ω') \;\middle|\; \| v \|_{L^2(Ω')} + \int_{Ω'}\int_{Ω'}\frac{|v(x)-v(y)|^2}{|x-y|^{n+1}} dx \, dy < \infty \right\}$$ 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, which allows one to define spaces $H^s(Ω)$ for any $0 < s < 1$ "in-between" $L^2(\Omega)$ and $H^1(Ω)$, where the trace space appears as special case for $s = 1/2$.

The only source claiming the equivalence of the norms I know of is [2], but my Italian is not sufficient to follow the argument.

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

[2] Gagliardo, E. (1957). Caratterizzazioni delle tracce sulla frontiera relative ad alcune classi di funzioni in n variabili. Rendiconti Del Seminario Matematico Della Università Di Padova, 27, 284–305.

  • $\begingroup$ Thank you for your answer! It makes things at least a little bit clearer. $\endgroup$ Jan 12, 2015 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? $\endgroup$ Nov 4, 2015 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? $\endgroup$
    – user9464
    Sep 11, 2016 at 12:39
  • $\begingroup$ @mattia.penati No, sorry. I would start looking at Lions' book, though. $\endgroup$ Sep 12, 2016 at 8:58
  • $\begingroup$ @Jack Yes, of course. Thanks, fixed. $\endgroup$ Sep 12, 2016 at 9:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .