Sobolev functions counterexample

Question

Let $A=(0,1)^{d}$.Does anyone have a simple example of a funtion in $H_0^1(A)\cap H^2(A)$ that is not in $H^2_0(A)$?

Thanks a lot.

Answer 1

As PhoemueX already said, you need a function, which has zero boundary values, but the derivative is non-zero at the boundary.

In particular, you can take $$ u(x) = \prod_{i=1}^d x_i \, (1-x_i). $$ Then, it is easy to see that $u(x) = 0$ for $x \in \partial[(0,1)^d]$, but $\nabla u(x) \ne 0$ for $x \in \partial[(0,1)^d]$. Since $u$ is smooth on $[0,1]^d$, it belongs to $H^2 \cap H_0^1$, but not to $H_0^2$.

Edit: Let me briefly describe that the above properties imply that $u$ does not belong to $H_0^2((0,1)^d)$, albeit $\Omega := (0,1)^d$ has only a Lipschitz boundary.

To the contrary, assume that $u \in H_0^2(\Omega)$. By definition, there exists a sequence $u_n \in C_c^\infty(\Omega)$, such that $u_n \to u$ in $H^2(\Omega)$. The trace operator $T$ maps $H^1(\Omega)$ continuously into $L^2(\partial\Omega)$. Hence, $0 = T(\nabla u_n) \to T(\nabla u)$ in $L^2(\partial\Omega)$. This is a contradiction to $\nabla u \ne 0$ on $\partial\Omega$.