Existence of convergent sequence implies convergence of function as $ttoinfty$ The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Convergent sequence on a step functionConvergence of products of sequences implies convergence of sequence itselfa.e. convergence to a constant implies the sup of the sequence is almost surely finiteExistence of a subsequence with weak-* convergent tracesUniform convergence of sequence of dense subspacesIntegral Convergence Estimate with cut-off functionNorm Inequality for 1 Dimensional Sobolev SpaceConfirmation of convergence in subdomainWeak Convergence from Strong ConvergenceProperties of convergence in $L^infty$
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Existence of convergent sequence implies convergence of function as $ttoinfty$
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Convergent sequence on a step functionConvergence of products of sequences implies convergence of sequence itselfa.e. convergence to a constant implies the sup of the sequence is almost surely finiteExistence of a subsequence with weak-* convergent tracesUniform convergence of sequence of dense subspacesIntegral Convergence Estimate with cut-off functionNorm Inequality for 1 Dimensional Sobolev SpaceConfirmation of convergence in subdomainWeak Convergence from Strong ConvergenceProperties of convergence in $L^infty$
$begingroup$
Let $Omega subset mathbbR$ be an unbounded domain and $u in C([0,infty);H_0^1(Omega))$. Assume there exists a sequence $t_n_ninmathbbN$ such that $t_ntoinfty$ and $||u(t_n)||_H_0^1(Omega)to 0$ as $ntoinfty$. Is it possible to show that $||u(t)||_H_0^1(Omega)to 0$ as $ttoinfty$? I am not sure how to show this if this is possible. If this is not possible, then is there any counter example? Moreover, can I still obtain similar result if I replace $H_0^1(Omega)$ with $X$ a Banach space?
Any hint is much appreciated! Thank you!
sequences-and-series functional-analysis convergence sobolev-spaces
asked Mar 25 at 7:28
Evan William ChandraEvan William Chandra
676313
$endgroup$
add a comment |
$begingroup$
Let $Omega subset mathbbR$ be an unbounded domain and $u in C([0,infty);H_0^1(Omega))$. Assume there exists a sequence $t_n_ninmathbbN$ such that $t_ntoinfty$ and $||u(t_n)||_H_0^1(Omega)to 0$ as $ntoinfty$. Is it possible to show that $||u(t)||_H_0^1(Omega)to 0$ as $ttoinfty$? I am not sure how to show this if this is possible. If this is not possible, then is there any counter example? Moreover, can I still obtain similar result if I replace $H_0^1(Omega)$ with $X$ a Banach space?
Any hint is much appreciated! Thank you!
sequences-and-series functional-analysis convergence sobolev-spaces
asked Mar 25 at 7:28
Evan William ChandraEvan William Chandra
676313
$endgroup$
add a comment |
$begingroup$
Let $Omega subset mathbbR$ be an unbounded domain and $u in C([0,infty);H_0^1(Omega))$. Assume there exists a sequence $t_n_ninmathbbN$ such that $t_ntoinfty$ and $||u(t_n)||_H_0^1(Omega)to 0$ as $ntoinfty$. Is it possible to show that $||u(t)||_H_0^1(Omega)to 0$ as $ttoinfty$? I am not sure how to show this if this is possible. If this is not possible, then is there any counter example? Moreover, can I still obtain similar result if I replace $H_0^1(Omega)$ with $X$ a Banach space?
Any hint is much appreciated! Thank you!
sequences-and-series functional-analysis convergence sobolev-spaces
asked Mar 25 at 7:28
Evan William ChandraEvan William Chandra
676313
$endgroup$
Let $Omega subset mathbbR$ be an unbounded domain and $u in C([0,infty);H_0^1(Omega))$. Assume there exists a sequence $t_n_ninmathbbN$ such that $t_ntoinfty$ and $||u(t_n)||_H_0^1(Omega)to 0$ as $ntoinfty$. Is it possible to show that $||u(t)||_H_0^1(Omega)to 0$ as $ttoinfty$? I am not sure how to show this if this is possible. If this is not possible, then is there any counter example? Moreover, can I still obtain similar result if I replace $H_0^1(Omega)$ with $X$ a Banach space?
Any hint is much appreciated! Thank you!
sequences-and-series functional-analysis convergence sobolev-spaces
sequences-and-series functional-analysis convergence sobolev-spaces
asked Mar 25 at 7:28
Evan William ChandraEvan William Chandra
676313
asked Mar 25 at 7:28
Evan William ChandraEvan William Chandra
676313
asked Mar 25 at 7:28
Evan William ChandraEvan William Chandra
676313
asked Mar 25 at 7:28
Evan William ChandraEvan William Chandra
676313
asked Mar 25 at 7:28
Evan William ChandraEvan William Chandra
676313
676313
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $u(t)=sin(tpi)f$ where $f$ is a fixed non-zero element of $H_0^1(Omega)$. Take $t_n=n$.
answered Mar 25 at 7:34
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
$endgroup$
$begingroup$
what if I add one more condition such that $forall n in mathbbN, u(t_n)neq 0$?
$endgroup$
– Evan William Chandra
Mar 25 at 8:02
1
$begingroup$
Just add $frac 1 t$ to $sin(tpi)$.
$endgroup$
– Kavi Rama Murthy
Mar 25 at 8:09
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $u(t)=sin(tpi)f$ where $f$ is a fixed non-zero element of $H_0^1(Omega)$. Take $t_n=n$.
answered Mar 25 at 7:34
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
$endgroup$
$begingroup$
what if I add one more condition such that $forall n in mathbbN, u(t_n)neq 0$?
$endgroup$
– Evan William Chandra
Mar 25 at 8:02
1
$begingroup$
Just add $frac 1 t$ to $sin(tpi)$.
$endgroup$
– Kavi Rama Murthy
Mar 25 at 8:09
add a comment |
$begingroup$
Let $u(t)=sin(tpi)f$ where $f$ is a fixed non-zero element of $H_0^1(Omega)$. Take $t_n=n$.
answered Mar 25 at 7:34
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
$endgroup$
$begingroup$
what if I add one more condition such that $forall n in mathbbN, u(t_n)neq 0$?
$endgroup$
– Evan William Chandra
Mar 25 at 8:02
1
$begingroup$
Just add $frac 1 t$ to $sin(tpi)$.
$endgroup$
– Kavi Rama Murthy
Mar 25 at 8:09
add a comment |
$begingroup$
Let $u(t)=sin(tpi)f$ where $f$ is a fixed non-zero element of $H_0^1(Omega)$. Take $t_n=n$.
answered Mar 25 at 7:34
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
$endgroup$
Let $u(t)=sin(tpi)f$ where $f$ is a fixed non-zero element of $H_0^1(Omega)$. Take $t_n=n$.
answered Mar 25 at 7:34
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
answered Mar 25 at 7:34
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
answered Mar 25 at 7:34
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
answered Mar 25 at 7:34
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
74.6k53270
$begingroup$
what if I add one more condition such that $forall n in mathbbN, u(t_n)neq 0$?
$endgroup$
– Evan William Chandra
Mar 25 at 8:02
1
$begingroup$
Just add $frac 1 t$ to $sin(tpi)$.
$endgroup$
– Kavi Rama Murthy
Mar 25 at 8:09
add a comment |
$begingroup$
what if I add one more condition such that $forall n in mathbbN, u(t_n)neq 0$?
$endgroup$
– Evan William Chandra
Mar 25 at 8:02
1
$begingroup$
Just add $frac 1 t$ to $sin(tpi)$.
$endgroup$
– Kavi Rama Murthy
Mar 25 at 8:09
$begingroup$
what if I add one more condition such that $forall n in mathbbN, u(t_n)neq 0$?
$endgroup$
– Evan William Chandra
Mar 25 at 8:02
$begingroup$
what if I add one more condition such that $forall n in mathbbN, u(t_n)neq 0$?
$endgroup$
– Evan William Chandra
Mar 25 at 8:02
1
1
$begingroup$
Just add $frac 1 t$ to $sin(tpi)$.
$endgroup$
– Kavi Rama Murthy
Mar 25 at 8:09
$begingroup$
Just add $frac 1 t$ to $sin(tpi)$.
$endgroup$
– Kavi Rama Murthy
Mar 25 at 8:09
add a comment |
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