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Take $u$ in $\mathscr S'(\mathbb R^2)$ with $$ \hat u(\xi)=\frac{\mathbf 1(\vert\xi\vert\ge 2)}{\vert\xi\vert^3 \ln\vert\xi\vert},\quad \vert \xi\vert^2\hat u(\xi)=\frac{\mathbf 1(\vert\xi\vert\ge 2)} …

Continuity is a local property: functions which are in your $L^{2,s}$ are locally in $L^2$, so functions in $H^{2,s}$ are locally in the Sobolev space $H^2(\mathbb R^3)$, thus are continuous functions …

Let us start with pointing out that $f\in H^\sigma$ is equivalent to $$ (\langle n\rangle^\sigma\hat f(n))_{n\in \mathbb Z}\in \ell^2(\mathbb Z), \quad \text{with $\langle n\rangle=\sqrt{1+n^2}$.} $$ …

(1) Let me answer first to the last question: $\Delta \vert x\vert$ is homogeneous of degree $-1$ and radial. On $\mathbb R^d$ ($d\ge 2$) it is $$ (\partial_r^2+\frac{d-1}{r}\partial_r)(r)=\frac{d-1}{ …

The answer is negative: take $f$ smooth compactly supported in $(-1/4,1/4)$ equal to 1 in $(-1/8,1/8)$, take $g(x) =f(x+1)$ so that $g$ is supported where $-1/4<x+1<1/4,$ i.e. $-5/4<x<-3/4$ so tha …

From your assumptions, you have a $C^2$ function $\rho:\mathbb R^d\rightarrow \mathbb R$, such that $$ D=\{x\in \mathbb R^d, \rho(x)<0\}, \quad \partial D=\{x\in \mathbb R^d, \rho(x)=0\}, $$ and $ x\i …

Let us assume that $p=2$, and let us consider $$ \cup_{s>0} H^s(\mathbb R^d)\subset H^0(\mathbb R^d)=L^2(\mathbb R^d). $$ The above inclusion is strict. Let us consider $u\in L^2(\mathbb R^d)$ define …

The Sobolev space $H^s(\mathbb T)$ is $W^{s,2}(\mathbb T)$ and is also the Besov space $B^{s}_{2,2}(\mathbb T)$. On the other hand the set of Hölder continuous functions with index $\alpha\in (0,1)$ i …

The function $\vert x\vert^{\alpha-n}$ is radial homogeneous of degree $\alpha-n$, so its Fourier transform is radial homogeneous of degree $-(\alpha-n)-n=-\alpha$ (both locally integrable since $\alp …

So $\sigma=\sum_{1\le j\le n}\sigma_j(x)\frac{\partial}{\partial x_j}$ is a vector field with distributions coefficients $\sigma_j$ and divergence in $L^2$: $$ \sum_{1\le j\le n}\frac{\partial \sigma …

I will more comfortable with the notation $v_\epsilon=\hat{u_\epsilon}$; you have then $$ v_\epsilon(t,x)=\alpha(t,x)+\int_0^t\int e^{2πix(\xi+\epsilon\phi(s,\xi))} \hat{v_\epsilon}(s,\xi) d\xi ds=\al …

You have done half of the job. We have the absolutely convergent integral which is such that $$ f(\xi)=\int_{\mathbb R^d} \frac{\vert e^{2iπ y\cdot \xi}-1\vert^2}{\vert y\vert^{d+2s}} dy =c_{s,d}\vert …

Too long for comment. take for instance $f=0, g=0$. Then the mapping $h\mapsto u$ is a pseudo-differential operator which will have some Sobolev continuity properties for spaces $W^{s,p}$ with $p\in …

There are two different parts in elliptic regularity theory. The first and easier is interior regularity, which can be proven for $p\in (1,+\infty)$ essentially by the same method as for $p=2$, usin …

I believe that the answer is yes in dimension $d\ge 2$, from the following result. Theorem. Let $\Omega$ be an open subset of $\mathbb R^d$, let $X$ be a Lipschitz vector field on $\Omega$ and let $u …