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The assumption that $\Omega$ is bounded is in fact required. (So your attempt is the correct proof, once you fix the omission in the statement.) Counterexample: let $\Omega$ be the upper half plane. L …

As noted by Christian Remling, the Fourier transform of $|x|^{-1}$ is $|\xi|^{-2}$, so the relevant integral you wish to bound is $$\iiint |\eta - \xi|^{-2} \bar{\hat{u}}(\xi) |\eta|^{2\alpha} \hat{u} …

Since you tagged reference-request: In the PDE/harmonic analysis literature this is a consequence of the Calderon-Zygmund Inequality, it is one of the main tools for studying elliptic regularity theor …

What you want is not possible. The basic idea is that it is possible for a diffeo $\Omega_1\to \Omega_2$ that extends to a homeo $\bar{\Omega}_1 \to \bar{\Omega}_2$ to "almost fold up" $\partial\Omega …

Assuming the inequality you hoped for is $\| \phi \otimes \phi \|_{\dot{H}^1} \lesssim \|\phi\|_{\dot{H}^{1/2}}^2$, I claim that this is still impossible. Let $D = [0,2\pi]$ again. Set $$ \phi_N(x) = …

Question 1 (that higher derivatives are not used) is yes. Question 2 (getting decay without weights) is no. Without weights, let $u$ be a compactly supported smooth function. Let $f_k(x) = u(x - k v) …

The key computation is the commutator $$ \Lambda^s (xf) - x \Lambda^s f. $$ You can check this "classically" in the case $s = 4$ to find $$ (1 - \Delta)^2 (xf) = x (1-\Delta)^2 f - 4 (1-\Delta) f'$$ w …

Since the OP asked for a discussion of features, I provide one by way of an explanation of Christian Remling's counterexample: Holder's inequality states that $$ | \int fg | \leq \| f\|_p \|g\|_q $$ i …

Here's an elementary proof of a related inequality (with non-sharp constants), which may explain what Iosif said in his answer. For ease of typing, instead of $\lambda^{-k}$ I will just type $y \ll 1$ …

See Jonsson, A.; Wallin, Hans, A Whitney extension theorem in (L^p) and Besov spaces, Ann. Inst. Fourier 28, No. 1, 139-192 (1978). ZBL0369.46031. Proposition 7.1 in Section 7.3 is exactly what you ar …

The first two are equivalent, as the $H^1(\Omega)$ norm and $H^1(\mathbb{R}^d)$ norm coincide for $C^\infty_c(\Omega)$ functions. The third is in general different: If you let $d = 1$ and $\Omega = \m …

A bit of a pet peeve of mine: the negative Sobolev spaces are spaces of distributions. However, your question (phrased as asking $f = 0$ a.e.) presupposes that elements of $H^{-1}$ can be represented …

In $d \geq 3$ the answer is no from scaling argument. WLOG we can assume $0\in \Omega$ (by translation) and that $B(0,r_0)\subset\Omega$. Take $\phi\in C^\infty_0(B(0,r_0))\subset C^\infty_0(\Omega)$. …

First, you got the scaling wrong. The correct scaling for $$ |x|^\alpha u(x) \lesssim \|u\|_{L^2}^{1-\beta} \|\nabla u\|_{L^2}^\beta $$ would be $\alpha = \frac{N}{2} - \beta$ where $N$ is the spatia …

The best constant is just the multiplicative inverse of the smallest positive eigenvalue of the Laplacian on the sphere. On $\mathbb{S}^N$ this the smallest eigenvalue is $N$, so $C(N)$ in that case e …