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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 …

I think there's a mistake in the definition of your norm $\|\cdot\|_\kappa$ : it does not seem to be equivalent to the $H^1(\Omega)$ norm (since there's no gradients involved). I would more simply con …

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) …

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 …

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

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 …

Hi, I guess what is done in the Evan's PDE book is for $W^{1,2}_0(\Omega)$ functions, right ? If you look at all $W^{1,2}(\Omega)$, $\mathscr{D}(\Omega)$ (test functions) is not a dense subspace. He …

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 …

First, note that the embedding $H^1(\mathbf{R}^3)\hookrightarrow L^p_{\text{loc}}(\mathbf{R}^3)$ is compact only for $p<6$, and the "loc" is mandatory for this compactness to hold. I know that you did …

Thanks to Giorgio's comment I found the good reference. In fact De Vore and Lorentz give a refined estimate (Theorem 6.1, Chapter 7) in comparison with the Bramble-Hilbert Lemma I've just cited : $$ \ …

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 …

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 …

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 …

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 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 …