Showing $H^1(\partial\Omega) \subset H^{\frac 12}(\partial\Omega)$ is continuous? 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..)
 A: 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.
A: 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 
$$
  \Vert u \Vert_{W^{s, p}}^p = \int_{\mathbb{R}^N}\int_{\mathbb{R}^N} \frac{\vert u (x) - u (y)\vert^p}{ \vert x - y \vert^{N + sp}}\,dx\,dy + \int_{\mathbb{R}^N} \vert u \vert^p.
$$
A: 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.
