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The main difficulty in the proof of the rule is to prove that $\nabla u=0$ a.e. on the set $u^{-1}(\Sigma)$, where $\Sigma$ is the set where $F$ is not differentiable; and where $\nabla u=0$ one defin …

In Christ-Weinstein, JFA 100 (1991) 87-109 you can find the fractional chain rule (Proposition 3.1) $$ \|F(u)\|_{\dot H^s_r}\le C \|F'(u)\|_{L^p}\|u\|_{\dot H^s_q} $$ where $s\in(0,1)$, $p,q,r\in(1,\i …

The answer is positive. If $T$ is a continuous functional on $W^{s,p}$ then $TJ^{-s}$ is a continuous functional on $L^p$, which can be represented by an element $u\in L^{p'}$: thus for all $f\in L^p$ …

Sure. You can even deduce the result for the dotted norms by the non-dotted one, via scaling. Anyway, the Fourier transform argument is sufficient in this case. Namely, if $u$ is Schwartz class then o …

I would say so. Denote your integral by $b_u(x)=\int_{|x-y|=r}u(y)dH^{n-1}$. Approximate $u$ in $H^1$ with test functions $u_j$. The property is certainly true for $u_j$ thus it is enough to prove tha …

Your question is not clear. If you want an embedding into the space of continuous bounded functions on R, then all you need is an estimate on each bounded subinterval, which you say you already unders …

It is well known that the $H^{1/2}( R)$ norm does not control the $L^\infty$ norm. This means that there exists a sequence of test functions whose $H^{1/2}$ norm tends to zero, and whose maximum value …

This is standard and well explained in several treatises: the space $W^{s,p}$ belongs to the scale of Besov spaces, while $H^{s,p}$ is in the scale of Triebel-Lizorkin spaces. The two scales satisfy m …

Since $Tu=T(\chi u)$ for any smooth cutoff $\chi$ equal to 1 near the boundary, the norm of the trace operator should be independent of the size of the open set (maybe it could be influenced by the do …

Take $s_2=-1/2$, $s_1=-1/2-\delta$, and apply the to-be-disproved estimate to a standard delta sequence $\rho_\epsilon=\epsilon^{-1}\rho(x/\epsilon)$. Since we can take $r=\epsilon$, on the Fourier si …

No time to give a complete answer but just a hint to a possible direction. Sobolev spaces in $R^n$ arise as the largest possible spaces on which some functional ('energy') can be defined. So they are …

To my taste, the cleanest approach to the extension problem is contained in Stein's 1970 book "Singular integrals and differentiability properties of functions". For a bounded open set with Lipschitz …

Of course yes, basically you achieve compactness with $H^1_r$ because you have local regularity plus decay at infinity (pointwise decay like $|x|^{(1-n)/2}$ to be precise, by Strauss-type inequalities …