# Question on Sobolev spaces in domains with boundary

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.

• 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.
– mme
May 29 '19 at 0:06
• @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 $W^{1,2}_0(\Omega)$. May 29 '19 at 5:28

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.

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).