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A Sobolev space is a vector space of functions equipped with a norm that is a combination of Lp-norms of the function itself and its derivatives up to a given order.
1
vote
The dependence of constant in a trace theorem on the diameter of domain
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 …
2
votes
Embeddings of Sobolev spaces
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 …
6
votes
Accepted
Sobolev imbedding
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 …
2
votes
Smallness of cut-off functions at critical Sobolev regularity
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 …
5
votes
Riesz potential and homogeneous Sobolev spaces
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 …
11
votes
Accepted
Chain rule in Sobolev space
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 …
6
votes
Smooth Sobolev extension from $W^{1,p}(U)$ to $W^{1,p} (\mathbb{R}^n) $
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 …
5
votes
Sobolev embedding in the space of continuous functions
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 …
3
votes
Accepted
$H^s$ norm of non-integer power of functions
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 …
2
votes
References on duality of fractional order Sobolev spaces
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 $f\in L^p$ …
10
votes
Comparison of two versions of fractional Sobolev spaces: do we have $W^{s,p}(\mathbb{R}^{n})...
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 …
11
votes
Sobolev spaces and geometry
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 …
1
vote
Accepted
About the continuity of the integral on the boundary of a ball
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 …