Search Results

Start with Pietro's example in the linked question. Fix $u$. Define $$ v_{\delta,M} = M\delta^{-n/2^*} u(x/\delta) $$ We know that $\|\nabla v_{\delta,M}\|_2 = M\|\nabla u\|_2$ and $\|v_{\delta,M}\|_ …

The power of 10 from Michael Renardy's answer is in fact optimal, which follows from the fact that for $y\approx 0$, the vector fields $\partial_y$ and $x\partial_y - y\partial_x $ are parallel. We …

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 …

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 …

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) …

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 …

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 …

Taking the Fourier transform and using $L^2$ orthogonality you are equivalently trying to estimate $$ \sum_{i + j +k = 0} \hat{u}_i \hat{u}_j \hat{u}_{k} |i+j|^{n+1}|k|^{n} $$ Now from the equality …

Okay, the estimate actually holds. I had a "moment" yesterday when I first read gerw's argument, which is actually subtly flawed. Let me quickly illustrate: To estimate $\| u^2\|_{H^1}$ you need to e …

Locally if you take $$ u = r^{-2/p} $$ you see that $$ \Delta u = - (n - 3 - \frac2p) \frac2p \frac{u}{r^2} = - (n-3-\frac2p)\frac2p u^{p+1}$$ Set your $c$ to be the function on the RHS. If I did my …

By Plancherel $$ \| f - S_k f\|_{H^s}^2 \approx \int (1 + |\xi|^2)^{s} (1 - \chi(2^{-k}\xi))^2 |\hat{f}|^2 ~d\xi $$ By almost orthogonality you can write $$ (1 - \chi(2^{-k}\xi))^2 \approx \sum_{\ell …

(Making CW as this is an extended comment.) Let's generalize slightly1: let $X\hookrightarrow Y$ be an embedding of Banach spaces, and let $S\subset Y$ be a closed subspace with the induced norm. Your …

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 …

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 $\phi\in C^\infty_0(B(0,r_0))\subset C^\infty_0(\Omega)$. …

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 …