I have a question about the traces of functions in $W^{1,2}$.


Let $D$ be a connected open subset of $\mathbb{R}^d$.We denote $W^{1,2}(D)$ by 
\begin{align\*}
W^{1,2}(D)=\{f \in L^{2}(D,dx) \mid \partial f/\partial x\_i \in L^{2}(D,dx)\},
\end{align\*}
where $\partial f/\partial x\_i$ is the distributional derivative of $f$.


According to Giovanni Leoni's book [A First Course in Sobolev Spaces](https://rads.stackoverflow.com/amzn/click/0821847686), if $\partial D$ is uniformly Lipschitz, there is a bounded linear operator $T:W^{1,2}(D) \to L^{2}(\partial D, \sigma)$ such that 
$$Tf=f \text{ on }\partial D \text{ for all } f \in W^{1,2}(D) \cap C(\bar{D}).$$
Here, $\sigma$ is the $d-1$-dimensional Hausdorff measure on $\partial D$. 
 $T$ is often called trace operator and $Tf$ means "the value of $f$ on $\partial D$".
It is also known that $T$ is injective and $T(W^{1,2}(D))$ is identified with the Besov space $B^{1/2,2}(\partial D, \sigma)$:


\begin{equation\*}
B^{1/2,2}(\partial D)=\{f \in L^{2}(\partial D, \sigma)\mid \|f\|\_{B^{1/2,2}(\partial D,\sigma)}<\infty \},
\end{equation\*}
where 
\begin{equation\*}
\|f\|^2\_{B^{1/2,2}}(\partial D,\sigma):=\int\_{\partial D}f^{2}\,d\sigma+\int\_{\partial D}\int\_{\partial D}\frac{|f(x)-f(y)|^2}{|x-y|^d}\,\sigma(dx)\sigma(dy).
\end{equation\*}


**My question**


As mentioned above, if $\partial D$ is uniformly Lipschitz, $$T(W^{1,2}(D))=B^{1/2,2}(\partial D, \sigma).$$
Can we show the above identity for more general domains? For general Lipschitz domains, can we construct a trace operator and obtain the above identity?



A Characterization of the traces of functions in $W^{1,2}$
I think the most natural assumption is that $D$ is a **bilipschitzian image** of a smooth domain $D\_0$, since a change of variables $A:D\_0\to D$ with $a|x-y|\le|A(x)-A(y)|\le b|x-y|$ ($a>0$) preserves $H^1(D\_0)$ and the space of traces $H^{1/2}(\partial D\_0)$ as defined with $\int\int\frac{|f(x)-f(y)|^2}{|x-y|^d}\ d\sigma(x)\ d\sigma(y)<\infty$.


