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This is not true. Take a sequence of points $(x_n) \subset D$ which converge to a point on the boundary (or even worse: whose accumulation points are $\partial D$). Then, for every point $x_n$ we can …

This is not true in the case of $H^{-1}(\Omega) = H_0^1(\Omega)^*$ (real spaces), where $\Omega \subset \mathbb R^d$ is bounded: In https://math.stackexchange.com/questions/336834/decomposition-of-fu …

You have that $H_0^1(\Omega, D)$ is a subspace of $H^1(D)$ (equipped with the same norm). Due to the Lipschitz condition on $D$, $H^1(D)$ is compactly embedded in $L^2(D)$ (standard-Rellich-Kondrachov …

This is not true. Take $\Omega = (-1,1)$ and functions $u_M$ like (I hope that I got the constants right) $$ u_M(x) =\begin{cases} -M^2 (|x|-1)(|x|-1+1/M) + M & \text{for } |x| > 1-1/(2M) \\ 1/4 + M & …

I already asked this on M.SE, but get no answers. Is there a (simple) example of a finely open set (i.e. w.r.t. the fine topology in potential theory) $O$ in $\mathbb R^n$, $n \ge 2$, which is not op …

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 …

This functional is sequentially weakly lower semicontinuous under fairly mild assumptions on $f$. We need that $f$ is non-negative, continuous and bounded from above. Let $u_n \rightharpoonup u$ in $ …

This works (only?) for $p = 2$. Let us denote the solution of the PDE on $\Omega$ by $v$. Then, the variational formulations of the PDEs are $$\int_D \nabla u \cdot \nabla z - fz \,\mathrm{d}x = 0 \q …

You can manipulate the left-hand side to \begin{equation*} \lVert (Dv)^2 \rVert_{H^{3k-2}} = \lVert D v \rVert_{W^{3k-2,4}}^2 \le \lVert v \rVert_{W^{3k-1,4}}^2. \end{equation*} (Note that what w …

As mentioned by @TaQ, the embedding $W(V,H) \hookrightarrow C([0,T];V)$ is, in general, not true. However, if your estimate would be true, you can extend the embedding operator from the dense subspace …

One positive answer is that this set is $p$-quasi-open, see some resource about capacity theory, e.g., here: https://math.stackexchange.com/questions/48776/capacity-theory-beginner-resources.

I think that such a space will not exists. In particular your requirements (1) and (2) contradicts in the following sense: (1) says basically, that the norm in $V$ is weaker than the $L^\infty(\Omega) …

I will consider the case $k = 1$ and $p = 2$ (some arguments may generalize to $k \in \mathbb{N}$). Let us use the norm $\|u\|^2 = \|u\|^2 + \|\nabla u\|^2$ in $H^1(\Omega)$ (both are $L^2$-norms). Th …