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Use cutoff functions as in attempt 2, but construct $\psi$ more carefully. This is a very standard construction. Start with a function $\psi \in C^\infty_c(\mathbb{R}^d)$ having $\psi = 1$ on the ba …

Let $A = \{u = k\}$. Let $\varphi_n : \mathbb{R} \to [0,1]$ be a smooth function supported inside $[k-\frac{1}{n}, k+\frac{1}{n}]$, with $\varphi_n(k)=1$. Then $\varphi_n(u) \to 1_A$, pointwise and …

A typical counterexample is like the following. For concreteness, let's take $p=1$ and $n=2$, and $\Omega = B(0,1) \subset \mathbb{R}^2$. Let $$f_n(t) = \begin{cases} nt, & 0 \le t \le 1/n \\ 1, & t …

More of a comment than an answer, but... I don't know a reference, but perhaps you could prove it as follows. For $u \in W^{1,p}(\Omega)$ with nonnegative trace, write $u = u^+ - u^-$ which are both …

At least for $1 < p < \infty$ the answer is no. By Banach-Alaoglu, after passing to a subsequence we can assume the $u_n$ converge weakly in $W^{1,p}$ to some $v \in W^{1,p}$. Now let $w \in L^2$ be …

Let's call your new norm $\|\cdot\|_a$ and reserve $\|\cdot\|$ for the usual $L^2$ norm. Integration by parts shows us $\|u\|_a^2 = \|\nabla u\|^2 + a\|u\|^2$. If $a > 0$ then this is easy, so let $ …

(More an extended comment than a full answer) Your $L$ is the Neumann Laplacian and so functions in its domain have to satisfy Neumann boundary conditions, informally speaking. The domain certainly …

I'm not immediately sure how to formulate a strong maximum principle if you are trying to avoid "classical" techniques where you understand the solution as being regular enough to be well-defined poin …

Note that by this definition the vector space $H^{1/2}(\partial \Omega)$ is isomorphic to the quotient of $H^1(\Omega)$ by the kernel of $\operatorname{tr}$, which you can observe is a closed subspace …

For question 2, if you had $\Omega$ in place of $\bar{\Omega}$, the conditions you state would still be satisfied, but there are other interesting conditions that would not be. It would fail to be a …

(The question got modified and this may not answer it anymore.) It is possible to have a function $f$ which is $C^\infty$ and bounded on $\Omega$ and is in $W^{1,p}_0(\Omega)$, yet does not have any …

It certainly is measurable. In fact, you may find an explicit formula for it. If we take $I = (0,1)$, then $g_s$ is simply given by $g_s(t) = \operatorname{min}(s,t) - st$. How did I find this? We …

In fact, for $k > \frac{m}{2}$, any sequence of continuous functions $f_n$ which is bounded in $H^k$ norm must have a subsequence converging uniformly on compact sets. This proves your claim, since i …

Consider $d=1$, $s=1$, $\Omega=(-1,1) \subset \mathbb{R}^1$. We can find a function $g \in L^2(-1,1)$ (or even continuous) for which $g(0)$ is arbitrarily large but $\|g\|_{L^2(-1,1)}$ is arbitrarily …