A distributional normal derivative for functions in $H^1(\Omega)$

Question

Let $\Omega$ be a smooth bounded domain with $\partial\Omega = \Gamma$. I have read this.

For all $u \in H^1(\Omega)$ such that $-\Delta u = g \in L^2(\Omega)$ in distribution, we can define the normal derivative $u_\nu \in H^{-\frac 12}(\Gamma)$ such that $$\langle u_\nu, \psi \rangle_{H^{-\frac 12}(\Gamma), H^{\frac 12}(\Gamma)} = -\int_\Omega gD\psi + \int_\Omega \nabla u \nabla D\psi$$ where $D\psi \in H^1(\Omega)$ is an extension of $\psi \in H^{\frac 12}(\Gamma)$.

(eg. here) In other words, we have this notion of a weak normal derivative for $H^1$ functions (usually we need $u \in H^2$).

Question: why do we need to ask for $\Delta u \in L^2(\Omega)$? Why not just define the normal derivative like so: $$\langle u_\nu, \psi \rangle_{H^{-\frac 12}(\Gamma), H^{\frac 12}(\Gamma)} = \langle \Delta u, D\psi\rangle_{H^{1}(\Omega)^*, H^1(\Omega)} + \int_\Omega \nabla u \nabla D\psi?$$

I have not seen something like this in any text except these lecture notes. Can anyone tell me why?

Answer 1

It is not possible to define a normal derivative for all $u \in H^1(\Omega)$ which depends continuously on $u$.

The reason is that all $C_c^\infty(\Omega)$ is dense in $H^1_0(\Omega)$, but all $u \in C_c^\infty(\Omega)$ have zero normal derivative. If we would have a normal derivative depending continuously on $u \in H^1(\Omega)$, this implies that it is zero for all $u \in H^1_0(\Omega)$ and this is not correct.

Your definition fails since $\Delta u \in H^1(\Omega)^*$ is meaningless for $u \in H^1(\Omega)$ (how do you define it?).