Why $\|v\|_{H^1}\leq \|g\|_{H^{1/2}}$ for $g\in H^{1/2}(\partial \Omega )$?

Question

Let $\Omega \subset \mathbb R^n$ open, smooth and bounded. I know that the trace $v\longmapsto v|_{\partial \Omega }$ is continuous and surjective $H^{1}(\Omega )\longrightarrow H^{1/2}(\partial \Omega )$. Let $g\in H^{1/2}(\partial \Omega ) $. Then, there is $v\in H^1(\Omega )$ s.t. $v|_{\partial \Omega }=g$. Since the trace is continuous, there is $C>0$ s.t. $\|g\|_{H^{1/2}(\partial \Omega )}\leq C\|v\|_{H^1(\Omega )}$. But in an exercise, it's written that there is a constant $D>0$ s.t. $$\|v\|_{H^1(\Omega )}\leq D\|g\|_{H^{1/2}(\partial \Omega )}.$$ To me it's an error, and I ask to my teacher, but he told me that it's not an error, but I don't have a theorem that say that. So

1) Is this inequality true ? And if yes, why ?

2) If it's true, can we says that $\|\cdot \|_{H^1}$ and $\|\cdot \|_{H^{1/2}}$ are equivalent ?

Answer 1

This inequality in general is evidently false. Just take any nonzero function that vanishes on the boundary.

You're probably mixing these two statements:

(1) If $v \in H^1(\Omega)$ is any function and $g \in H^{1/2}(\partial \Omega)$ is its trace, then $$ \| g \|_{H^{1/2}(\partial \Omega)} \le C \| v \|_{H^1(\Omega)}. $$

(2) If $g \in H^{1/2}(\partial \Omega)$ is any function, then there exists $v \in H^1(\Omega)$ such that $g$ is its trace and $$ \| v \|_{H^1(\Omega)} \le C \| g \|_{H^{1/2}(\partial \Omega)}. $$

Consider also:

(3) The trace map $H^1(\Omega) \to H^{1/2}(\partial \Omega)$ is surjective.

The proof of (3) that I know is by taking $g \in H^{1/2}(\partial \Omega)$ and giving an explicit formula for $v \in H^1(\Omega)$ with the right trace; this $v$ happens to satisfy (2).

Anyway, (2) follows from (1) and (3) via the open mapping theorem.

Answer 2

The inequality is indeed correct, when is related to an extension theorem, and it is not related to the trace imbedding theorem.

It says the following:

If $\partial\Omega$ is sufficiently smooth and bounded, then there exists a bounded extension operator $$ T: H^{1/2}(\partial\Omega)\to H^1(\Omega). $$ The proof of this is not totally trivial.

Sketch of proof. For each $x_0\in\partial\Omega$, there is an open neighbourhood $U$ of $x_0$, and a smooth function $g : U\to \mathbb R$, such that $$ U\cap\partial \Omega=\{x: g(x)=0\}\quad\text{and}\quad U\cap\Omega=\{x: g(x)<0\}. $$ Next, cover $\partial\Omega$ with finitely many such neighbourhoods. In each neighbourhood a smooth function defined on $\partial\Omega\cap U$ can be extended to $\overline{\Omega}\cap U$, smoothly, with the $H^1-$norm of the extension controlled by the $H^{1/2}-$norm on the boundary. Finally, introducing a partition of the identity, supported on these open neighbourhoods, one can define the extension of a smooth function on $\partial\Omega$.