# A Characterization of the traces of functions in $W^{1,2}$

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, 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?

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 $\int\int\frac{|f(x)-f(y)|^2}{|x-y|^d}\ d\sigma(x)\ d\sigma(y)<\infty$.
You may take a look at the book ¨Function spaces on subsets of $\mathbb{R}^n$¨ by Jonsson and Wallin. The trace results there are more general than Lipschitz.