# 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.

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.
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).