space of robin boundary condition Let $\Omega$ be a bounded open domain in $\mathbb{R}^n$. Let $K$ be a subset defined as
$$
K=\{u\in H^1(\Omega)|\frac{\partial u}{\partial v}+u=0\ \text{on }\partial\Omega\}
$$
Is the subspace $K$ closed in $H^1(\Omega)$?
 A: This is not true. Take $\Omega=(0,1)$. Let $\epsilon\in (0,1/4)$. Define
$$
u_\epsilon(x) = \begin{cases}
1+x & \text{ if } x\in (0,\epsilon)\\
1+2\epsilon- x & \text{ if } x\in (\epsilon,2\epsilon)\\
1 &  \text{ if } x\in (2\epsilon,1-2\epsilon)\\
1+(x-(1-2\epsilon)) &  \text{ if } x\in (1-2\epsilon,1-\epsilon)\\
2-x &  \text{ if } x\in (1-\epsilon,1),
\end{cases}
$$
so $u_\epsilon$ is constant one except for two hat-shaped perturbations of height $\epsilon$ and width $2\epsilon$. Clearly, $u_\epsilon\in H^1(0,1)$. Moreover, $u_\epsilon$ is continuously differentiable near the boundary of $\Omega$. And $u_\epsilon \in K$.
In addition $u_\epsilon \to 1$ in $H^1(0,1)$: $\|u_\epsilon-1\|_{L^\infty(0,1)}\le \epsilon$, $\|\nabla u_\epsilon\|_{L^\infty(0,1)}=1$, and measure of support of $u_\epsilon-1$ is equal to $4\epsilon$. But $1\not \in K$.
A: You cannot define $\frac{\partial u}{\partial v}$ on $\partial\Omega$ for $u \in H^1(\Omega)$. You need more regularity of $u$, e.g., $\Delta u \in L^2(\Omega)$ for the definition of the normal derivative.
