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It does not remain true. If $\omega=B(0,1)$ and $\Omega=B(0,2)$ and $f(x)=1-|x|^2$, then $f\in H^1_0(\omega)\cap H^2(\omega)$ but the extension by zero is not in $H^2(\Omega)$.

Piecewise linear functions are not dense in $W^{1,\infty}(\Omega;\mathbb R^n)$ for any open set $\Omega\subset\mathbb R^m$. If it were true for $\Omega$, it would also be true for any open subset, in …

Since $f'$ is bounded, it is clear that $\nabla_gf(u)$ is in $L^2$ if $\nabla_gu$ is. It just remains to check that your formula indeed gives the weak gradient of $f(u)$. Weak derivatives on manifolds …

Let me expand my comment into an answer. Let $s>\frac12$ and pick any $r\in(\frac12,s)$. Interpolating between Sobolev spaces gives $\|u\|_{H^r_P}\lesssim\|u\|_{H^0_P}^\delta\|u\|_{H^s_P}^{1-\delta}$ …

I assume $u\in H^1(\Omega)$ for some nice domain $\Omega\subset\mathbb R^n$, and you wonder why $Lu\in H^{-1}(\Omega)$. (Correct me if I'm wrong.) For any $v\in H^1_0$, formal integration by parts giv …

If $\partial U$ has measure zero (which follows from $C^1$ regularity, for example), then $L^1(U)=L^1(\bar U)$. Since $L^1(U)\subset L^1_{\text{loc}}(U)$, $L^1(\bar U)$ is just a special case. Deriva …

Your space $W^{s,p}$ is the same as $F^s_{p2}$, not $F^s_{pp}$. See these lecture notes. The definition of $H^{s,p}$ (which you call $W^{s,p}$) is given on page 34 and the result is on page 52.

Yes. It might be more convenient to think of $H^{1/2}(\partial\Omega)$ as a quotient space: $$ H^{1/2}(\partial\Omega)=H^1(\Omega)/H^1_0(\Omega). $$ Let $T:H^1(\Omega)\to H^{1/2}(\partial\Omega)$ be t …

The trace only depends on values near the boundary. That is, if $\phi\in C^\infty([0,\infty))$ is one in a neighborhood of zero, then $\operatorname{tr}_\Omega(\phi u)=\operatorname{tr}_\Omega(u)$ for …

First, $\xi$ cannot be surjective. Consider two distinct functions in $H^1(\Omega)$ with the same boundary values; they cannot both be in the range of $\xi$. This is a general property of quotient map …

There is generally no $v\in H^1_0$ with $\nabla v=\chi_{\{|\nabla u|>1\}}\nabla u$. Consider for example $u:(-1/2,2)\to\mathbb R$, $u(x)=\min\{1-x/2,1+2x\}$. Now $u\in H^1_0((-1/2,2))$ but there is no …

The two norms are not equivalent for functions with finite $\|\cdot\|_1$ norm. Let $L>0$ be a large integer and let $f$ be the characteristic function of the cube $[0,L]^d\subset\mathbb Z^d$. In the f …

The property $F(0)=0$ follows from applying $F$ to the zero function, but the regularity is not as good as you ask for. As a partial counterexample, let $F(x)=|x|$ and $s=1$. Now $F(f)\in H^1$ for al …