Weak convergence of weak derivative implies strong convergence
$begingroup$
Note that here both $w_n$ and $phi_n$ are vector functions.
The following is given :
$ w_n rightarrow w$ weakly in $ L^q(B; {mathbb{R}}^M)$
& ${curl[w]}_{n=1}^{infty}$ is precompact in $ W^{-1,s}(B; R^{M times M})$ for a certain $ s>1$.
Also, it is assumed that $ w_n = nabla _xphi_n$ where $int_{mathbb{R}^M} phi_n dy =0$.
Then using the Sobolev embedding theorem, that by compactness, it is concluded that $phi_n rightarrow phi$ strongly in $ L^q(mathbb{R}^M)$.
My question is how can the above conclusion be made? How can we conclude that $ phi_n in L^q(mathbb{R}^M)$ ?
(Source - Mathematical theory of compressible viscous fluids by Feireisl. The above occurs in the proof of Div Curl lemma for a special case)
real-analysis functional-analysis derivatives sobolev-spaces
asked Dec 10 '18 at 18:35
Erm20Erm20
355311
$endgroup$
add a comment |
$begingroup$
Note that here both $w_n$ and $phi_n$ are vector functions.
The following is given :
$ w_n rightarrow w$ weakly in $ L^q(B; {mathbb{R}}^M)$
& ${curl[w]}_{n=1}^{infty}$ is precompact in $ W^{-1,s}(B; R^{M times M})$ for a certain $ s>1$.
Also, it is assumed that $ w_n = nabla _xphi_n$ where $int_{mathbb{R}^M} phi_n dy =0$.
Then using the Sobolev embedding theorem, that by compactness, it is concluded that $phi_n rightarrow phi$ strongly in $ L^q(mathbb{R}^M)$.
My question is how can the above conclusion be made? How can we conclude that $ phi_n in L^q(mathbb{R}^M)$ ?
(Source - Mathematical theory of compressible viscous fluids by Feireisl. The above occurs in the proof of Div Curl lemma for a special case)
real-analysis functional-analysis derivatives sobolev-spaces
asked Dec 10 '18 at 18:35
Erm20Erm20
355311
$endgroup$
add a comment |
$begingroup$
Note that here both $w_n$ and $phi_n$ are vector functions.
The following is given :
$ w_n rightarrow w$ weakly in $ L^q(B; {mathbb{R}}^M)$
& ${curl[w]}_{n=1}^{infty}$ is precompact in $ W^{-1,s}(B; R^{M times M})$ for a certain $ s>1$.
Also, it is assumed that $ w_n = nabla _xphi_n$ where $int_{mathbb{R}^M} phi_n dy =0$.
Then using the Sobolev embedding theorem, that by compactness, it is concluded that $phi_n rightarrow phi$ strongly in $ L^q(mathbb{R}^M)$.
My question is how can the above conclusion be made? How can we conclude that $ phi_n in L^q(mathbb{R}^M)$ ?
(Source - Mathematical theory of compressible viscous fluids by Feireisl. The above occurs in the proof of Div Curl lemma for a special case)
real-analysis functional-analysis derivatives sobolev-spaces
asked Dec 10 '18 at 18:35
Erm20Erm20
355311
$endgroup$
Note that here both $w_n$ and $phi_n$ are vector functions.
The following is given :
$ w_n rightarrow w$ weakly in $ L^q(B; {mathbb{R}}^M)$
& ${curl[w]}_{n=1}^{infty}$ is precompact in $ W^{-1,s}(B; R^{M times M})$ for a certain $ s>1$.
Also, it is assumed that $ w_n = nabla _xphi_n$ where $int_{mathbb{R}^M} phi_n dy =0$.
Then using the Sobolev embedding theorem, that by compactness, it is concluded that $phi_n rightarrow phi$ strongly in $ L^q(mathbb{R}^M)$.
My question is how can the above conclusion be made? How can we conclude that $ phi_n in L^q(mathbb{R}^M)$ ?
(Source - Mathematical theory of compressible viscous fluids by Feireisl. The above occurs in the proof of Div Curl lemma for a special case)
real-analysis functional-analysis derivatives sobolev-spaces
real-analysis functional-analysis derivatives sobolev-spaces
asked Dec 10 '18 at 18:35
Erm20Erm20
355311
asked Dec 10 '18 at 18:35
Erm20Erm20
355311
edited Dec 10 '18 at 19:09
Erm20
asked Dec 10 '18 at 18:35
Erm20Erm20
355311
asked Dec 10 '18 at 18:35
Erm20Erm20
355311
asked Dec 10 '18 at 18:35
Erm20Erm20
355311
355311
add a comment |
add a comment |
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