# Sobolev space boundary value in PDE

Let $\Omega$ be a open bounded set. There is a unique $u \in H^1(\Omega)$ such that $$-\Delta u = f \text{ on \Omega}$$ $$u|_{\partial \Omega} = g$$

But how can we write $u|_{\delta \Omega}$ when $\Omega$ is an open set so doesn't contain its boundary??? Note that I am not asking about the trace operator. I am not asking how we can specify something on a null set. I am asking why we can say that $$u(\text{some boundary point})$$ makes sense when the boundary is not in $\Omega$ and we only know that $u \in H^1(\Omega)$, not $u \in H^1(\overline \Omega).$

• Restriction on boundary is valid for any $C^{\infty}(\overline \Omega)$. Hence we define the trace: $T:C^{\infty}(\overline \Omega)\rightarrow L^2(\partial \Omega)$.
• $C^{\infty}(\overline \Omega)$ is dense in $H^1(\Omega) = W^{1,2}(\Omega)$, hence we have a continuous linear extension $\widetilde{T}:H^1(\Omega) \rightarrow L^2(\partial \Omega)$.
• The trace of any $u \in H^1(\Omega)$ should be: consider $\{u_n\}\subset C^{\infty}(\overline \Omega)$, let $u_n\to u$ under $H^1$-norm, define $$u|_{\partial \Omega} = \lim_{n\to \infty}Tu_n$$
Lastly, we could not define $$u(\text{some boundary point})$$ the trace operator is rather a densely defined operator (well-defined on a dense subset).
It's possible that $u \in H^1(\Omega)$ isn't even continuous on $\Omega$, so it's not even clear what we'd mean by $u(\text{some interior point})$. But if $\partial\Omega$ is $C^1$, then the trace of $u$ on $\partial\Omega$ is a well-defined function in $L^2(\partial\Omega)$, and this is what we mean when we write $u|_{\partial\Omega}$. No claim is being made about any pointwise values.