A question about a trace operator (is this right?)

Question

Suppose I have proven that for $u \in H^1(\Omega) \cap C^1(\bar \Omega)$ that $$|u|_{L^2(\partial\Omega)} \leq f|u|_{H^1(\Omega)}$$ for some constant $f$.

Let $T:H^1(\Omega) \to L^2(\partial\Omega)$ be the trace operator. Then can I say that by the above $$|Tu|_{L^2(\partial\Omega)} \leq f|u|_{H^1(\Omega)}$$ holds for all $u \in H^1(\Omega)\cap C^1(\bar \Omega)$. Then by density
$$|Tu|_{L^2(\partial\Omega)} \leq f|u|_{H^1(\Omega)}$$ holds for all $u \in H^1(\Omega).$

Is this valid? I am not sure about my second displayed equation.

Answer 1

That seems ok. For $u\in H^1(\Omega)\cap\mathcal{C}^1(\bar \Omega)$ you just define $Tu$ to be the the restriction of $u$ to the boundary, $u\big|_{\partial\Omega}$. Then your second displayed equation is basically the same as the first one.

For general $u\in H^1(\Omega)$ you can take aproximations by continuous functions. In other words, you define $Tu=\lim\limits_{n\rightarrow \infty}Tu_n$ where $u_n\rightarrow u$ in $H^1(\Omega)$ and $u_n\in H^1(\Omega)\cap\mathcal{C}^1(\bar \Omega)$. There is some work to do, though. You have to show that $Tu_n$ converges in $L^2(\partial\Omega)$ and that the definition is independent of the sequence chosen. This can be done with your second inequality.

Answer 2

You have to be careful: $H^1(\Omega) \cap C^1(\bar\Omega) = C^1(\bar\Omega)$ is not always dense in $H^1(\Omega)$! E.g. take $\Omega = (-1,0) \cup (0,1)$. Then, the function $$u(x) = \text{sign}(x)$$ belongs to $H^1(\Omega)$, but cannot be approximated by functions in $C^1(\bar\Omega)$. The same is true for $C(\bar\Omega)$.

Hence, you need some regularity of the boundary.

If you would have the density, then you can invoke the following general argument: Let $X,Y$ be normed linear spaces and let $D \subset X$ be a dense subspace. For every bounded linear operator $T : D \to Y$, there exists a unique, bounded linear extension $T : X \to Y$. The proof of this statement is rather trivial.