Let $\Omega\subset\mathbb R^n$ be a bounded Lipschitz domain.

How does one prove that the inclusion $H^1(\partial\Omega) \subset H^{\frac 12}(\partial\Omega)$ is continuous?

I define $H^{\frac 1 2}(\partial\Omega)$ to be space of functions $u \in L^2(\partial\Omega)$ such that $$|u|_{H^{\frac 12}(\partial\Omega)} = \int_{\partial\Omega}\int_{\partial\Omega} \frac{|u(x)-u(y)|^2}{|x-y|^n} < \infty$$ and give it the norm $$\lVert \cdot \rVert_{H^{\frac 12}(\partial\Omega)}^2 = \lVert \cdot \rVert_{L^2(\partial\Omega)}^2 + |\cdot|_{H^{\frac 12}(\partial\Omega)}^2$$

There is the following result: (1) If $D \subset \mathbb{R}^n$ is an open Lipschitz domain with bounded boundary then $H^1(D) \subset H^{\frac 12}(D)$ is continuous.

We can use an alternative norm on $H^{\frac 12}(\partial\Omega)$ by using chart maps (which are Lipschitz) to map neighbourhoods of the boundary onto subsets $D_i$ of Euclidean space. But AFAIK we do not know that these subsets $D_i$ are Lipschitz, so we cannot apply result (1).

Edit: A reference to this result would be great too.

(I ask here because my question on M.SE did not receive attention, hope that is OK..)

  • $\begingroup$ Isn't the map in question simply the identity map? $\endgroup$ – username Jan 15 '14 at 15:11
  • $\begingroup$ @AthanagorWurlitzer Yes it is identity map $i:H^1(\partial\Omega) \to H^{\frac 12}(\partial\Omega)$. Note the norms are different (I added some details). $\endgroup$ – soup Jan 15 '14 at 15:13
  • $\begingroup$ And what norm to you use for $H^1(\partial\Omega)$? $\endgroup$ – username Jan 16 '14 at 12:39
  • $\begingroup$ @AthanagorWurlitzer Hmm, I think: around each point of the boundary, there is a neighbourhood $U_i$ and a Lipschitz function $f_i:U_i \to D_i \subset \mathbb{R}^n$. Then the norm of a function $u$ can be defined as the $H^1$ norm of the sum of the functions $u \circ f_i^{-1}$ (probably need a partition of unity somewhere) $\endgroup$ – soup Jan 16 '14 at 12:49
  • 1
    $\begingroup$ The question seems to be flawed. Nobody is interesting in an embedding $H^1\subset H^{1/2}$. People are instead interested into trace theorems stating that the restriction $u\mapsto u|_{\partial\Omega}$ extends continuously as an operator $\gamma_0:H^1(\Omega)\rightarrow H^{1/2}(\partial\Omega)$. Would you reformulate your question ? $\endgroup$ – Denis Serre Dec 17 '18 at 15:20

Lions and Magenes define $H^\frac12$ by complex interpolation, hence your desired property holds by assumption; but they always assume $\Omega$ to have smooth boundary.

In the books of Adams and Grisvard you will find some related results, but - as far as I have (quickly) seen - not exactly what you look like.

  • $\begingroup$ Thanks. I found the details in the book "S.A. Sauter and C. Schwab. Boundary Element Methods." $\endgroup$ – soup Jan 23 '14 at 14:36

Here is a direct argument based on the definition by the Gagliardo norm $$ \Vert u \Vert_{H^{1/2}}^2 = \int_{\mathbb{R}^N}\int_{\mathbb{R}^N} \frac{\vert u (x) - u (y)\vert^2}{ \vert x - y \vert^{N + 1}}\,dx\,dy + \int_{\mathbb{R}^N} \vert u \vert^2. $$

In a first step you apply a weighted Hardy inequality on the ball $B_1 (y)$ to write $$ \begin{split} \int_{\mathbb{R}^N} \frac{\vert u (x) - u (y)\vert^2}{ \vert x - y \vert^{N + 1}}\,dx\,dy &= \int_{B_1 (y)} \frac{\vert u (x) - u (y)\vert^2}{ \vert x - y \vert^{N + 1}}\,dx\,dy + \int_{\mathbb{R}^N \setminus B_1 (y)} \frac{\vert u (x) - u (y)\vert^2}{ \vert x - y \vert^{N + 1}}\,dx\,dy\\ &\le \int_{B_1 (y)} \frac{\vert \nabla u (x)\vert^2}{\vert x - y \vert^{N - 1}} + \vert{u(x)}\vert^2\,dx + 2 \int_{\mathbb{R}^{N }\setminus B_1 (y)} \frac{\vert u (x) \vert^2}{\vert x - y\vert^{N + 1}} \,dx + 2 \vert u (y)\vert^2 \int_{\mathbb{R}^{N} \setminus B_1} \frac{1}{\vert z\vert^{N + 1}} \,dx. \end{split} $$ The conclusion comes by integrating with respect to $y \in \mathbb{R}^N$ and then applying Fubini's theorem and using the facts that $$ \int_{B_1} \frac{1}{\lvert z \rvert^{N - 1}} \,dz < \infty \text{ and }\int_{\mathbb{R}^N \setminus B_1} \frac{1}{\lvert z \rvert^{N + 1}} \,dz < \infty . $$

To cover a smooth domain, the function can be extended to the whole space $\mathbb{R}^N$.

This argument also works for the spaces $W^{s, p}$ and $W^{1, p}$, where $$ \Vert u \Vert_{W^{s, p}}^p = \int_{\mathbb{R}^N}\int_{\mathbb{R}^N} \frac{\vert u (x) - u (y)\vert^p}{ \vert x - y \vert^{N + sp}}\,dx\,dy + \int_{\mathbb{R}^N} \vert u \vert^p. $$


In fact a substantially stronger embedding is true:

Theorem. If $n\geq 2$, $p>1$, and $\Omega\subset\mathbb{R}^n$ is a bounded domain with Lipschitz boundary, then $W^{1,p}(\partial\Omega)\subset W^{1-\frac{1}{q},q}(\partial\Omega)$, where $q=\frac{np}{n-1}$. That is, there is a bounded linear extension operator $$ E:W^{1,p}(\partial\Omega)\to W^{1,q}(\Omega)\cap C^\infty(\Omega). $$

When $p=2$ i.e., $W^{1,2}(\partial\Omega)=H^1(\partial\Omega)$, we have $q=\frac{2n}{n-1}>2$. Hence the extension is $$ E:H^1(\partial\Omega)\to W^{1,q}(\Omega)\subset H^1(\Omega), $$ proving (indirectly) that $H^1(\partial\Omega)\subset H^{\frac{1}{2}}(\partial\Omega)$ (because traces of $H^1(\Omega)$ are in $H^{\frac{1}{2}}(\partial\Omega)$.

The above theorem is known. For a reference to a proof, see: https://mathoverflow.net/a/322635/121665.


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