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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 …

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 …

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 …

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 …

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

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 …

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 : $$ \ …

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 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 …

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 …

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 …