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There's no reason for such a strong convergence ; in the periodic setting $\Omega=\mathbf{T}^2$ you can think of $u_n(x) := |n|^{-4}e^{in\cdot x}$ which is indeed uniformly bounded in $H^4(\mathbf{T}^ …

For $k=1$, the proof works the same on $\mathbb{R}$ ; you only need to check that compactly supported functions (no smoothness here) are dense in $H^1(\mathbb{R},\rho(x)dx)$ and this can be done using …

The unit cube has lipschitz regularity so yes, you have an extension operator. Note that your extra condition on the compactness of the support can be ensured easily once you have an extension (just m …

Yes, this is called Moser-Trudinger inequality. See for instance Theorem 1.67 of this book.

For $\varepsilon>0$ consider a continuous function $\theta_\varepsilon:\mathbf{R}_{>0}\rightarrow\mathbf{R}_{>0}$ equaling the identity map on $I_\varepsilon:=(\varepsilon,\varepsilon^{-1})$ and const …

I don't think so. If this was true this would imply boundedness in $L^2$ for your sequence and in the particular case when $f=0$, this would mean that strong convergence in $H^{-2}$ implies boundednes …

In dimension 2, if you pull off the half-segment $\{1/2\}\times(0,1/2]$ from the square $(0,1)^2$ you obtain an open set $\Omega$ which does not satisfy the lipschitz condition (nor the cone condition …

Since all your spaces are Banach, an inclusion like $L^q(\Omega)\subset W^{s,p}(\Omega)$ implies (by the closed graph theorem) a continuous embedding. Such an embedding is impossible due to compactnes …

When I introduce distributions to students for the first time I usually emphasize the difference between $f$ (locally integrable function) and $T_f$ (the corresponding distribution). For a couple of l …

This may be an overkill : you can use the closed graph theorem. If $(u_n)_n$ converges to $u$ in $W^{s,p}_0(\Omega)$ and $(E(u_n))_n$ converges to $v$ in $W^{s,p}(\mathbf{R}^d)$, then both convergence …

For $p>\max(2,d)$ you have by Sobolev embedding $$\|f-\overline{f}\|_\infty \lesssim_{\Omega,p} \|f-\overline{f}\|_p + \|\nabla f\|_p.$$ The interpolation $L^p(\Omega) = [L^2(\Omega),L^\infty(\Omega) …

Only a partial answer : (2) seems strange. In dimension $1$, if $u$ does not change sign, your setting includes the one of $u^\alpha \in H^1(0,1)$ (boundary values are irrelevant here) for some $\alph …

I think in dimension 1 you won't be able to produce a counterexample, see §4 of V. V. Zhikov, "Weighted Sobolev spaces", Mat. Sb., 189:8 (1998), 27–58; Sb. Math., 189:8 (1998), 1139–1170 For some co …

First, since you have $H^1(0,T)$ imbedds in $\mathscr{C}^0([0,T])$, $z$ can be seen as an element of $\mathscr{C}^0([0,T];L^2(\Omega))$ and you can speak without ambiguity of $z(t_1)$ and $z(t_2)$. No …

For 1., if I am not mistaking you're searching for time-periodic functions enjoying Sobolev regularity in the space variable so the Sobolev regularity is not really linked with the periodicity: you're …