Trace theorem for $L^2$ spaces. In chapter 5 of Evans Partial differtial Equations (Section 5.5) he defines the trace operator for a bounded domain $\Omega$ with given smoothness as,
$T:H^1(\Omega)\rightarrow L^2(\partial\Omega)$
Where this operator is continuous i.e., $\|\gamma u\|_{L^2(\partial\Omega)}\leq C\|u\|_{H^1(\Omega)}$. 
My question is does there exist a continuous trace operator for function in $L^2(\Omega)$. That is to say $\gamma: L^2(\Omega)\rightarrow L^2(\partial\Omega)$ s.t, $\|\gamma u\|_{L^2(\partial\Omega)}\leq C\|u\|_{L^2(\Omega)}$.
The trace operator is defined for continuous functions and then extended to functions in $H^1$. Since continuous functions are dense in $L^2$ I believe the result should hold. 
Second question that I have is whether the trace operator has a continuous inverse i.e, is there a $T^{-1}:L^2(\partial\Omega)\rightarrow H^1(\Omega)$ that is also continuous. I can find a result for $\Omega=\mathbb{R}^d$ but not for bounded $\Omega$.
Thank you for any help.
 A: 1) No. The boundary of the domain in general has zero measure. Evaluating a $L^2$-function on a zero measure set is not well-defined. The continuation argument does not work, as the mapping $\gamma : H^1 \subset L^2(\Omega) \to L^2(\partial\Omega)$ is not bounded in the sense that there is no $C>0$ such that $\|\tau u\|_{L^2(\partial\Omega)}\le C \|u\|_{L^2(\Omega)}$ for all $u\in H^1$.
2) This is not true. The trace operator from $H^1$ to $L^2(\partial\Gamma)$ is compact in general. You have to use Sobolev-Slobodecki spaces to get invertibility.
A: Here is another argument which shows that there cannot be a trace operator $\gamma$ for $L^2(\Omega)$ with 


*

*$\gamma(f) = f_{|\partial\Omega}$ for all $f \in C(\bar\Omega)$,

*$\gamma \colon L^2(\Omega) \to L^2(\partial\Omega)$ is continuous.


First, we have $\gamma(\varphi) = 0$ for all $\varphi \in C_c^\infty(\Omega)$, since these functions are continuous and vanish on $\partial\Omega$.
Since $C_c^\infty(\Omega)$ is dense in $L^2(\Omega)$, we get $\gamma(f) = 0$ for all $f \in L^2(\Omega)$. This, however, contradicts $\gamma(1) = 1$ (here we used that constant functions are continuous).
