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Let $\Omega\subset \mathbb{R}^n$ be a bounded domain with infinitely smooth boundary. Define the Sobolev norm on $C^\infty(\bar \Omega)$ $$||u||_{W^{1,2}}:=\sqrt{\int_\Omega (|\nabla u|^2+u^2)dx}.$$ Let us denote by $W_0^{1,2}$ the completion of the space of smooth compactly supported functions in $\Omega$ with respect to this norm.

Let $u\in W^{1,2}_0\cap C(\bar \Omega)$. Is it true that $u$ vanishes on $\partial \Omega$?

Apologies if this question is not of the research level.

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  • $\begingroup$ Sure, the trace (restriction to boundary) map $r: C^\infty(\overline \Omega) \to C^\infty(\partial \Omega)$ is continuous when you equip the domain with the $W^{1,2}$ norm and the codomain with the $L^2$ norm, or when you equip both with the $C^0$ norm, and hence extends to a continuous map $(W^{1,2} \cap C^0)(\overline \Omega) \to (L^2 \cap C^0)(\partial \Omega)$. Because the map $r$ vanishes on the set of compactly supported smooth functions in $\Omega$, it also vanishes on its closure. The continuity in $C^0$ is obvious, while the continuity in $W^{1,2} \to L^2$ is in many PDE books. $\endgroup$
    – mme
    Commented May 29, 2019 at 0:06
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    $\begingroup$ @MikeMiller: I'm not sure I follow your argument. The closure of the compactly supported smooth functions within the space $W^{1,2}(\Omega) \cap C^0(\overline{\Omega})$ might be smaller than $W^{1,2}_0(\Omega)$. $\endgroup$
    – Jochen Glueck
    Commented May 29, 2019 at 5:28

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For the sake of completeness, an expansion on the comment by Mike Miller:

In Evans/Gariepy, Thm. 4.3.1, it is proven that if the boundary of $\Omega$ is Lipschitz, then for $1 \leq p < \infty$ there is a continuous linear trace operator $T$ from $W^{1,p}(\Omega)$ to $L^p(\Omega;\mathcal{H}_{n-1})$ which satisfies $$Tf = f \quad \text{on $\partial \Omega$}$$ if $f \in W^{1,p}(\Omega) \cap C(\overline\Omega)$.

Observing that $W^{1,p}_0(\Omega)$ is a closed subspace of $W^{1,p}(\Omega)$ and taking $g \in W^{1,p}_0(\Omega)$ and a sequence $(g_k) \subset C_c^\infty(\Omega)$ such that $g_k \to g$ in the $W^{1,p}(\Omega)$ norm, we obtain $$Tg = \lim_k Tg_k = g_k = 0 \quad \text{in $L^p(\partial\Omega,\mathcal{H}_{n-1})$}.$$ Thus, if $g \in W^{1,p}_0(\Omega) \cap C(\overline\Omega)$, then $0 = Tg = g$ everywhere on $\partial\Omega$.

It is maybe worthwile to note (Thm. 5.3.2 in Evans/Gariepy) that $T$ is in fact given by $$Tf(x) := \lim_{r\searrow0}\frac1{|B_r(x) \cap \Omega|} \int_{B_r(x) \cap \Omega} f$$ which can be a useful thing to look at also in situations where $\partial\Omega$ is less regular.

Evans, Lawrence C.; Gariepy, Ronald F., Measure theory and fine properties of functions, Studies in Advanced Mathematics. Boca Raton: CRC Press. viii, 268 p. (1992). ZBL0804.28001.

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The claim is true, and there are likely many different arguments. The one I like is as follows.

  1. Suppose that $u \in W^{1,2}(\Omega)$. Then there is $h \in W^{1,2}(\Omega)$ which is harmonic in $\Omega$ (in the sense that $\int \langle\nabla h, \nabla v\rangle = 0$ for every $v \in W^{1,2}_0(\Omega)$) and $h - u \in W^{1,2}_0(\Omega)$.

  2. If additionally $u \in C(\overline{\Omega})$, then $h$ is given as a Poisson integral of the boundary values of $u$. In particular, if $h = 0$ in $\Omega$, then $u = h = 0$ on $\partial \Omega$.

  3. However, if $u \in W^{1,2}_0(\Omega)$, then $h = u + (h - u) \in W^{1,2}_0(\Omega)$, and so $\int \langle \nabla h, \nabla h\rangle = 0$, that is, $h$ is identically zero.

It follows that if $u \in C(\overline{\Omega}) \cap W^{1,2}_0(\Omega)$, then $h = 0$ in $\Omega$ (by item 3) and consequently $u = h = 0$ on $\partial \Omega$ (by item 2).

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