Does strong convergence in $H$ (or $L^2$) imply convergence in $V$ (or $W_0^{1,2}$)?
$begingroup$
Suppose:
$mathbf{V}$ and $mathbf{H}$ are Hilbert spaces.
$mathbf{V} hookrightarrow mathbf{H}$ is compact embedding.
$mathbf{V}$ is dense in $mathbf{H}$.
For example $mathbf{V} = W_0^{1,2}(Omega)$ and $mathbf{H} = L^2(Omega)$ for bounded $Omega in mathbb{R}^n$.
Let $mathbf{u}_n,mathbf{u} in mathbf{V}$. It's implied from the Rellich-Kondrachov embeddidng theorem that bounded $mathbf{u}_n$ in $mathbf{V}$ has convergent subsequence in $mathbf{H}$.
I am asking the opposite question:
Consider a bounded sequence $mathbf{u}_n in mathbf{V}$. If $mathbf{u}_n to mathbf{u}$ strongly in $mathbf{H}$, does it weakly convergent in $mathbf{V}$? Does it converge to the same $mathbf{u}$?
A more general question:
What requirements the sequence $(mathbf{u}_n)_{n in mathbb{N}}$ should have in $mathbf{H}$, so that $mathbf{u}_n$ is
- weakly convergenent in $mathbf{V}$?
- Strongly convergent in $mathbf{V}$?
Here is a similar question, but I do not know how this is shown.
functional-analysis hilbert-spaces sobolev-spaces weak-convergence
asked Dec 12 '18 at 1:29
SiaSia
1064
$endgroup$
add a comment |
$begingroup$
Suppose:
$mathbf{V}$ and $mathbf{H}$ are Hilbert spaces.
$mathbf{V} hookrightarrow mathbf{H}$ is compact embedding.
$mathbf{V}$ is dense in $mathbf{H}$.
For example $mathbf{V} = W_0^{1,2}(Omega)$ and $mathbf{H} = L^2(Omega)$ for bounded $Omega in mathbb{R}^n$.
Let $mathbf{u}_n,mathbf{u} in mathbf{V}$. It's implied from the Rellich-Kondrachov embeddidng theorem that bounded $mathbf{u}_n$ in $mathbf{V}$ has convergent subsequence in $mathbf{H}$.
I am asking the opposite question:
Consider a bounded sequence $mathbf{u}_n in mathbf{V}$. If $mathbf{u}_n to mathbf{u}$ strongly in $mathbf{H}$, does it weakly convergent in $mathbf{V}$? Does it converge to the same $mathbf{u}$?
A more general question:
What requirements the sequence $(mathbf{u}_n)_{n in mathbb{N}}$ should have in $mathbf{H}$, so that $mathbf{u}_n$ is
- weakly convergenent in $mathbf{V}$?
- Strongly convergent in $mathbf{V}$?
Here is a similar question, but I do not know how this is shown.
functional-analysis hilbert-spaces sobolev-spaces weak-convergence
asked Dec 12 '18 at 1:29
SiaSia
1064
$endgroup$
add a comment |
$begingroup$
Suppose:
$mathbf{V}$ and $mathbf{H}$ are Hilbert spaces.
$mathbf{V} hookrightarrow mathbf{H}$ is compact embedding.
$mathbf{V}$ is dense in $mathbf{H}$.
For example $mathbf{V} = W_0^{1,2}(Omega)$ and $mathbf{H} = L^2(Omega)$ for bounded $Omega in mathbb{R}^n$.
Let $mathbf{u}_n,mathbf{u} in mathbf{V}$. It's implied from the Rellich-Kondrachov embeddidng theorem that bounded $mathbf{u}_n$ in $mathbf{V}$ has convergent subsequence in $mathbf{H}$.
I am asking the opposite question:
Consider a bounded sequence $mathbf{u}_n in mathbf{V}$. If $mathbf{u}_n to mathbf{u}$ strongly in $mathbf{H}$, does it weakly convergent in $mathbf{V}$? Does it converge to the same $mathbf{u}$?
A more general question:
What requirements the sequence $(mathbf{u}_n)_{n in mathbb{N}}$ should have in $mathbf{H}$, so that $mathbf{u}_n$ is
- weakly convergenent in $mathbf{V}$?
- Strongly convergent in $mathbf{V}$?
Here is a similar question, but I do not know how this is shown.
functional-analysis hilbert-spaces sobolev-spaces weak-convergence
asked Dec 12 '18 at 1:29
SiaSia
1064
$endgroup$
Suppose:
$mathbf{V}$ and $mathbf{H}$ are Hilbert spaces.
$mathbf{V} hookrightarrow mathbf{H}$ is compact embedding.
$mathbf{V}$ is dense in $mathbf{H}$.
For example $mathbf{V} = W_0^{1,2}(Omega)$ and $mathbf{H} = L^2(Omega)$ for bounded $Omega in mathbb{R}^n$.
Let $mathbf{u}_n,mathbf{u} in mathbf{V}$. It's implied from the Rellich-Kondrachov embeddidng theorem that bounded $mathbf{u}_n$ in $mathbf{V}$ has convergent subsequence in $mathbf{H}$.
I am asking the opposite question:
Consider a bounded sequence $mathbf{u}_n in mathbf{V}$. If $mathbf{u}_n to mathbf{u}$ strongly in $mathbf{H}$, does it weakly convergent in $mathbf{V}$? Does it converge to the same $mathbf{u}$?
A more general question:
What requirements the sequence $(mathbf{u}_n)_{n in mathbb{N}}$ should have in $mathbf{H}$, so that $mathbf{u}_n$ is
- weakly convergenent in $mathbf{V}$?
- Strongly convergent in $mathbf{V}$?
Here is a similar question, but I do not know how this is shown.
functional-analysis hilbert-spaces sobolev-spaces weak-convergence
functional-analysis hilbert-spaces sobolev-spaces weak-convergence
asked Dec 12 '18 at 1:29
SiaSia
1064
asked Dec 12 '18 at 1:29
SiaSia
1064
edited Dec 12 '18 at 6:28
Sia
asked Dec 12 '18 at 1:29
SiaSia
1064
asked Dec 12 '18 at 1:29
SiaSia
1064
asked Dec 12 '18 at 1:29
SiaSia
1064
1064
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $(u_n)$ is bounded in $V$ and $u_nto u$ in $H$, then $u_nrightharpoonup u$ in $V$:
As $(u_n)$ is bounded in $V$, it has a weakly converging subsequence $u_{n_k}rightharpoonup v$ in $V$. The continuity of the embedding into $H$ implies $v=u$ and $uin V$.
Using this argument, one can prove that every subsequence of $(u_n)$ has a subsequence that converges weakly to $u$ in $V$. Hence $u_nrightharpoonup u$ in $V$.
answered Dec 12 '18 at 15:18
dawdaw
25k1745
$endgroup$
$begingroup$
Thank you @daw. As I understood from your answer, we do not need $H$ then. As long as $u_n in V$ is bounded, it has a weakly convergent subsequence $u_n rightharpoonup u in V$. Furthermore, if $u_n to v$ in $H$, then $u = v$.
$endgroup$
– Sia
Dec 13 '18 at 1:33
$begingroup$
A furthure question @daw: is it a consequence of Banach-Alaoglu theorem?
$endgroup$
– Sia
Dec 13 '18 at 1:42
$begingroup$
yes to both......
$endgroup$
– daw
Dec 13 '18 at 7:14
add a comment |
$begingroup$
From ${u_n},u in H$ we can't imply ${u_n},u in V$.
answered Dec 12 '18 at 5:08
Trần Quang MinhTrần Quang Minh
3347
$endgroup$
$begingroup$
I assumed $u_n in V$, and $u_n$ strongly converge to $u$ in $H$. I edited the question.
$endgroup$
– Sia
Dec 12 '18 at 5:10
add a comment |
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2 Answers
2
active
oldest
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2 Answers
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$begingroup$
If $(u_n)$ is bounded in $V$ and $u_nto u$ in $H$, then $u_nrightharpoonup u$ in $V$:
As $(u_n)$ is bounded in $V$, it has a weakly converging subsequence $u_{n_k}rightharpoonup v$ in $V$. The continuity of the embedding into $H$ implies $v=u$ and $uin V$.
Using this argument, one can prove that every subsequence of $(u_n)$ has a subsequence that converges weakly to $u$ in $V$. Hence $u_nrightharpoonup u$ in $V$.
answered Dec 12 '18 at 15:18
dawdaw
25k1745
$endgroup$
$begingroup$
Thank you @daw. As I understood from your answer, we do not need $H$ then. As long as $u_n in V$ is bounded, it has a weakly convergent subsequence $u_n rightharpoonup u in V$. Furthermore, if $u_n to v$ in $H$, then $u = v$.
$endgroup$
– Sia
Dec 13 '18 at 1:33
$begingroup$
A furthure question @daw: is it a consequence of Banach-Alaoglu theorem?
$endgroup$
– Sia
Dec 13 '18 at 1:42
$begingroup$
yes to both......
$endgroup$
– daw
Dec 13 '18 at 7:14
add a comment |
$begingroup$
If $(u_n)$ is bounded in $V$ and $u_nto u$ in $H$, then $u_nrightharpoonup u$ in $V$:
As $(u_n)$ is bounded in $V$, it has a weakly converging subsequence $u_{n_k}rightharpoonup v$ in $V$. The continuity of the embedding into $H$ implies $v=u$ and $uin V$.
Using this argument, one can prove that every subsequence of $(u_n)$ has a subsequence that converges weakly to $u$ in $V$. Hence $u_nrightharpoonup u$ in $V$.
answered Dec 12 '18 at 15:18
dawdaw
25k1745
$endgroup$
$begingroup$
Thank you @daw. As I understood from your answer, we do not need $H$ then. As long as $u_n in V$ is bounded, it has a weakly convergent subsequence $u_n rightharpoonup u in V$. Furthermore, if $u_n to v$ in $H$, then $u = v$.
$endgroup$
– Sia
Dec 13 '18 at 1:33
$begingroup$
A furthure question @daw: is it a consequence of Banach-Alaoglu theorem?
$endgroup$
– Sia
Dec 13 '18 at 1:42
$begingroup$
yes to both......
$endgroup$
– daw
Dec 13 '18 at 7:14
add a comment |
$begingroup$
If $(u_n)$ is bounded in $V$ and $u_nto u$ in $H$, then $u_nrightharpoonup u$ in $V$:
As $(u_n)$ is bounded in $V$, it has a weakly converging subsequence $u_{n_k}rightharpoonup v$ in $V$. The continuity of the embedding into $H$ implies $v=u$ and $uin V$.
Using this argument, one can prove that every subsequence of $(u_n)$ has a subsequence that converges weakly to $u$ in $V$. Hence $u_nrightharpoonup u$ in $V$.
answered Dec 12 '18 at 15:18
dawdaw
25k1745
$endgroup$
If $(u_n)$ is bounded in $V$ and $u_nto u$ in $H$, then $u_nrightharpoonup u$ in $V$:
As $(u_n)$ is bounded in $V$, it has a weakly converging subsequence $u_{n_k}rightharpoonup v$ in $V$. The continuity of the embedding into $H$ implies $v=u$ and $uin V$.
Using this argument, one can prove that every subsequence of $(u_n)$ has a subsequence that converges weakly to $u$ in $V$. Hence $u_nrightharpoonup u$ in $V$.
answered Dec 12 '18 at 15:18
dawdaw
25k1745
answered Dec 12 '18 at 15:18
dawdaw
25k1745
answered Dec 12 '18 at 15:18
dawdaw
25k1745
answered Dec 12 '18 at 15:18
dawdaw
25k1745
25k1745
$begingroup$
Thank you @daw. As I understood from your answer, we do not need $H$ then. As long as $u_n in V$ is bounded, it has a weakly convergent subsequence $u_n rightharpoonup u in V$. Furthermore, if $u_n to v$ in $H$, then $u = v$.
$endgroup$
– Sia
Dec 13 '18 at 1:33
$begingroup$
A furthure question @daw: is it a consequence of Banach-Alaoglu theorem?
$endgroup$
– Sia
Dec 13 '18 at 1:42
$begingroup$
yes to both......
$endgroup$
– daw
Dec 13 '18 at 7:14
add a comment |
$begingroup$
Thank you @daw. As I understood from your answer, we do not need $H$ then. As long as $u_n in V$ is bounded, it has a weakly convergent subsequence $u_n rightharpoonup u in V$. Furthermore, if $u_n to v$ in $H$, then $u = v$.
$endgroup$
– Sia
Dec 13 '18 at 1:33
$begingroup$
A furthure question @daw: is it a consequence of Banach-Alaoglu theorem?
$endgroup$
– Sia
Dec 13 '18 at 1:42
$begingroup$
yes to both......
$endgroup$
– daw
Dec 13 '18 at 7:14
$begingroup$
Thank you @daw. As I understood from your answer, we do not need $H$ then. As long as $u_n in V$ is bounded, it has a weakly convergent subsequence $u_n rightharpoonup u in V$. Furthermore, if $u_n to v$ in $H$, then $u = v$.
$endgroup$
– Sia
Dec 13 '18 at 1:33
$begingroup$
Thank you @daw. As I understood from your answer, we do not need $H$ then. As long as $u_n in V$ is bounded, it has a weakly convergent subsequence $u_n rightharpoonup u in V$. Furthermore, if $u_n to v$ in $H$, then $u = v$.
$endgroup$
– Sia
Dec 13 '18 at 1:33
$begingroup$
A furthure question @daw: is it a consequence of Banach-Alaoglu theorem?
$endgroup$
– Sia
Dec 13 '18 at 1:42
$begingroup$
A furthure question @daw: is it a consequence of Banach-Alaoglu theorem?
$endgroup$
– Sia
Dec 13 '18 at 1:42
$begingroup$
yes to both......
$endgroup$
– daw
Dec 13 '18 at 7:14
$begingroup$
yes to both......
$endgroup$
– daw
Dec 13 '18 at 7:14
add a comment |
$begingroup$
From ${u_n},u in H$ we can't imply ${u_n},u in V$.
answered Dec 12 '18 at 5:08
Trần Quang MinhTrần Quang Minh
3347
$endgroup$
$begingroup$
I assumed $u_n in V$, and $u_n$ strongly converge to $u$ in $H$. I edited the question.
$endgroup$
– Sia
Dec 12 '18 at 5:10
add a comment |
$begingroup$
From ${u_n},u in H$ we can't imply ${u_n},u in V$.
answered Dec 12 '18 at 5:08
Trần Quang MinhTrần Quang Minh
3347
$endgroup$
$begingroup$
I assumed $u_n in V$, and $u_n$ strongly converge to $u$ in $H$. I edited the question.
$endgroup$
– Sia
Dec 12 '18 at 5:10
add a comment |
$begingroup$
From ${u_n},u in H$ we can't imply ${u_n},u in V$.
answered Dec 12 '18 at 5:08
Trần Quang MinhTrần Quang Minh
3347
$endgroup$
From ${u_n},u in H$ we can't imply ${u_n},u in V$.
answered Dec 12 '18 at 5:08
Trần Quang MinhTrần Quang Minh
3347
answered Dec 12 '18 at 5:08
Trần Quang MinhTrần Quang Minh
3347
answered Dec 12 '18 at 5:08
Trần Quang MinhTrần Quang Minh
3347
answered Dec 12 '18 at 5:08
Trần Quang MinhTrần Quang Minh
3347
3347
$begingroup$
I assumed $u_n in V$, and $u_n$ strongly converge to $u$ in $H$. I edited the question.
$endgroup$
– Sia
Dec 12 '18 at 5:10
add a comment |
$begingroup$
I assumed $u_n in V$, and $u_n$ strongly converge to $u$ in $H$. I edited the question.
$endgroup$
– Sia
Dec 12 '18 at 5:10
$begingroup$
I assumed $u_n in V$, and $u_n$ strongly converge to $u$ in $H$. I edited the question.
$endgroup$
– Sia
Dec 12 '18 at 5:10
$begingroup$
I assumed $u_n in V$, and $u_n$ strongly converge to $u$ in $H$. I edited the question.
$endgroup$
– Sia
Dec 12 '18 at 5:10
add a comment |
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