Bounded sequence in $L^infty$ which converges in $L^1$
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I have sequence in $L^infty(mathbb{R}^n)cap L^1(mathbb{R}^n)$ such that
1. $(u_n)_n$ is bounded in $L^infty$ : there exists $a>0$ such that $|u_n|_{L^infty}leq a$.
2.$(u_n)_n$ converges (strongly) in $L^1$ : there exists $uin L^1(mathbb{R}^n)$ such that $|u_n-u|_{L^1}to0$.
Is it true that the limit $u$ actually belongs to $L^infty(mathbb{R}^n)$ and that $|u|_{L^infty}leq a$ ? I guess that a strating point could be to take a weak-* converging subsequence in $L^infty$ and then prove that this limit is $u$...
functional-analysis lp-spaces weak-convergence strong-convergence
edited Dec 30 '18 at 16:56
Bernard
125k743119
asked Dec 30 '18 at 16:40
user630504user630504
132
$endgroup$
add a comment |
$begingroup$
I have sequence in $L^infty(mathbb{R}^n)cap L^1(mathbb{R}^n)$ such that
1. $(u_n)_n$ is bounded in $L^infty$ : there exists $a>0$ such that $|u_n|_{L^infty}leq a$.
2.$(u_n)_n$ converges (strongly) in $L^1$ : there exists $uin L^1(mathbb{R}^n)$ such that $|u_n-u|_{L^1}to0$.
Is it true that the limit $u$ actually belongs to $L^infty(mathbb{R}^n)$ and that $|u|_{L^infty}leq a$ ? I guess that a strating point could be to take a weak-* converging subsequence in $L^infty$ and then prove that this limit is $u$...
functional-analysis lp-spaces weak-convergence strong-convergence
edited Dec 30 '18 at 16:56
Bernard
125k743119
asked Dec 30 '18 at 16:40
user630504user630504
132
$endgroup$
3
$begingroup$
Just use that $L^1$-convergence implies convergence a.e. along a subsequence. This subsequence has $L^infty$-norm bounded by $a$ and you are done.
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– Severin Schraven
Dec 30 '18 at 17:25
2
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Your approach is also valid. Since $L^1$ is seperable, Banach-Alaoglu implies that there is a weak-star convergent subsequence converging to $u$. Then $|u|_inftyle a$ follows.
$endgroup$
– Song
Dec 30 '18 at 17:32
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Thank you both !
$endgroup$
– user630504
Dec 30 '18 at 17:48
add a comment |
$begingroup$
I have sequence in $L^infty(mathbb{R}^n)cap L^1(mathbb{R}^n)$ such that
1. $(u_n)_n$ is bounded in $L^infty$ : there exists $a>0$ such that $|u_n|_{L^infty}leq a$.
2.$(u_n)_n$ converges (strongly) in $L^1$ : there exists $uin L^1(mathbb{R}^n)$ such that $|u_n-u|_{L^1}to0$.
Is it true that the limit $u$ actually belongs to $L^infty(mathbb{R}^n)$ and that $|u|_{L^infty}leq a$ ? I guess that a strating point could be to take a weak-* converging subsequence in $L^infty$ and then prove that this limit is $u$...
functional-analysis lp-spaces weak-convergence strong-convergence
edited Dec 30 '18 at 16:56
Bernard
125k743119
asked Dec 30 '18 at 16:40
user630504user630504
132
$endgroup$
I have sequence in $L^infty(mathbb{R}^n)cap L^1(mathbb{R}^n)$ such that
1. $(u_n)_n$ is bounded in $L^infty$ : there exists $a>0$ such that $|u_n|_{L^infty}leq a$.
2.$(u_n)_n$ converges (strongly) in $L^1$ : there exists $uin L^1(mathbb{R}^n)$ such that $|u_n-u|_{L^1}to0$.
Is it true that the limit $u$ actually belongs to $L^infty(mathbb{R}^n)$ and that $|u|_{L^infty}leq a$ ? I guess that a strating point could be to take a weak-* converging subsequence in $L^infty$ and then prove that this limit is $u$...
functional-analysis lp-spaces weak-convergence strong-convergence
functional-analysis lp-spaces weak-convergence strong-convergence
edited Dec 30 '18 at 16:56
Bernard
125k743119
asked Dec 30 '18 at 16:40
user630504user630504
132
edited Dec 30 '18 at 16:56
Bernard
125k743119
asked Dec 30 '18 at 16:40
user630504user630504
132
edited Dec 30 '18 at 16:56
Bernard
125k743119
edited Dec 30 '18 at 16:56
Bernard
125k743119
edited Dec 30 '18 at 16:56
Bernard
125k743119
125k743119
asked Dec 30 '18 at 16:40
user630504user630504
132
asked Dec 30 '18 at 16:40
user630504user630504
132
asked Dec 30 '18 at 16:40
user630504user630504
132
132
3
$begingroup$
Just use that $L^1$-convergence implies convergence a.e. along a subsequence. This subsequence has $L^infty$-norm bounded by $a$ and you are done.
$endgroup$
– Severin Schraven
Dec 30 '18 at 17:25
2
$begingroup$
Your approach is also valid. Since $L^1$ is seperable, Banach-Alaoglu implies that there is a weak-star convergent subsequence converging to $u$. Then $|u|_inftyle a$ follows.
$endgroup$
– Song
Dec 30 '18 at 17:32
$begingroup$
Thank you both !
$endgroup$
– user630504
Dec 30 '18 at 17:48
add a comment |
3
$begingroup$
Just use that $L^1$-convergence implies convergence a.e. along a subsequence. This subsequence has $L^infty$-norm bounded by $a$ and you are done.
$endgroup$
– Severin Schraven
Dec 30 '18 at 17:25
2
$begingroup$
Your approach is also valid. Since $L^1$ is seperable, Banach-Alaoglu implies that there is a weak-star convergent subsequence converging to $u$. Then $|u|_inftyle a$ follows.
$endgroup$
– Song
Dec 30 '18 at 17:32
$begingroup$
Thank you both !
$endgroup$
– user630504
Dec 30 '18 at 17:48
3
3
$begingroup$
Just use that $L^1$-convergence implies convergence a.e. along a subsequence. This subsequence has $L^infty$-norm bounded by $a$ and you are done.
$endgroup$
– Severin Schraven
Dec 30 '18 at 17:25
$begingroup$
Just use that $L^1$-convergence implies convergence a.e. along a subsequence. This subsequence has $L^infty$-norm bounded by $a$ and you are done.
$endgroup$
– Severin Schraven
Dec 30 '18 at 17:25
2
2
$begingroup$
Your approach is also valid. Since $L^1$ is seperable, Banach-Alaoglu implies that there is a weak-star convergent subsequence converging to $u$. Then $|u|_inftyle a$ follows.
$endgroup$
– Song
Dec 30 '18 at 17:32
$begingroup$
Your approach is also valid. Since $L^1$ is seperable, Banach-Alaoglu implies that there is a weak-star convergent subsequence converging to $u$. Then $|u|_inftyle a$ follows.
$endgroup$
– Song
Dec 30 '18 at 17:32
$begingroup$
Thank you both !
$endgroup$
– user630504
Dec 30 '18 at 17:48
$begingroup$
Thank you both !
$endgroup$
– user630504
Dec 30 '18 at 17:48
add a comment |
2 Answers
2
active
oldest
votes
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For a more direct proof, let $lambda$ denote the Lebesgue measure and $delta>0$. Using Chebychev's inequality we get $$lambdaleft(left{x: lvert u(x) rvert geq a+delta right}right) leq lambdaleft(left{x: lvert u(x) -u_{n}(x) rvert geq delta/2 right} right) + underbrace{lambda left(x: lvert u_{n}(x)rvert geq a + delta/2 right)}_{=0} leq\ frac{2}{delta} , lvertlvert u-u_{n}rvertrvert_{L^{1}}rightarrow 0$$
as $nrightarrow infty$.
Since $delta>0$ was arbitrary, we arrive at the conclusion that $u$ is essentially bounded in $mathbb{R}^{n}$ with norm less than $a$.
answered Dec 30 '18 at 17:48
TheOscillatorTheOscillator
2,1891817
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add a comment |
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Since $u_nto u$ in $L^1,$ there is a subsequence $u_{n_k}$ that converges to $u$ a.e. We have each $|u_{n_k}|le a$ a.e., so $|u|le a$ a.e., giving the result.
answered Dec 30 '18 at 19:28
zhw.zhw.
75.4k43275
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
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votes
$begingroup$
For a more direct proof, let $lambda$ denote the Lebesgue measure and $delta>0$. Using Chebychev's inequality we get $$lambdaleft(left{x: lvert u(x) rvert geq a+delta right}right) leq lambdaleft(left{x: lvert u(x) -u_{n}(x) rvert geq delta/2 right} right) + underbrace{lambda left(x: lvert u_{n}(x)rvert geq a + delta/2 right)}_{=0} leq\ frac{2}{delta} , lvertlvert u-u_{n}rvertrvert_{L^{1}}rightarrow 0$$
as $nrightarrow infty$.
Since $delta>0$ was arbitrary, we arrive at the conclusion that $u$ is essentially bounded in $mathbb{R}^{n}$ with norm less than $a$.
answered Dec 30 '18 at 17:48
TheOscillatorTheOscillator
2,1891817
$endgroup$
add a comment |
$begingroup$
For a more direct proof, let $lambda$ denote the Lebesgue measure and $delta>0$. Using Chebychev's inequality we get $$lambdaleft(left{x: lvert u(x) rvert geq a+delta right}right) leq lambdaleft(left{x: lvert u(x) -u_{n}(x) rvert geq delta/2 right} right) + underbrace{lambda left(x: lvert u_{n}(x)rvert geq a + delta/2 right)}_{=0} leq\ frac{2}{delta} , lvertlvert u-u_{n}rvertrvert_{L^{1}}rightarrow 0$$
as $nrightarrow infty$.
Since $delta>0$ was arbitrary, we arrive at the conclusion that $u$ is essentially bounded in $mathbb{R}^{n}$ with norm less than $a$.
answered Dec 30 '18 at 17:48
TheOscillatorTheOscillator
2,1891817
$endgroup$
add a comment |
$begingroup$
For a more direct proof, let $lambda$ denote the Lebesgue measure and $delta>0$. Using Chebychev's inequality we get $$lambdaleft(left{x: lvert u(x) rvert geq a+delta right}right) leq lambdaleft(left{x: lvert u(x) -u_{n}(x) rvert geq delta/2 right} right) + underbrace{lambda left(x: lvert u_{n}(x)rvert geq a + delta/2 right)}_{=0} leq\ frac{2}{delta} , lvertlvert u-u_{n}rvertrvert_{L^{1}}rightarrow 0$$
as $nrightarrow infty$.
Since $delta>0$ was arbitrary, we arrive at the conclusion that $u$ is essentially bounded in $mathbb{R}^{n}$ with norm less than $a$.
answered Dec 30 '18 at 17:48
TheOscillatorTheOscillator
2,1891817
$endgroup$
For a more direct proof, let $lambda$ denote the Lebesgue measure and $delta>0$. Using Chebychev's inequality we get $$lambdaleft(left{x: lvert u(x) rvert geq a+delta right}right) leq lambdaleft(left{x: lvert u(x) -u_{n}(x) rvert geq delta/2 right} right) + underbrace{lambda left(x: lvert u_{n}(x)rvert geq a + delta/2 right)}_{=0} leq\ frac{2}{delta} , lvertlvert u-u_{n}rvertrvert_{L^{1}}rightarrow 0$$
as $nrightarrow infty$.
Since $delta>0$ was arbitrary, we arrive at the conclusion that $u$ is essentially bounded in $mathbb{R}^{n}$ with norm less than $a$.
answered Dec 30 '18 at 17:48
TheOscillatorTheOscillator
2,1891817
edited Dec 31 '18 at 14:04
answered Dec 30 '18 at 17:48
TheOscillatorTheOscillator
2,1891817
answered Dec 30 '18 at 17:48
TheOscillatorTheOscillator
2,1891817
answered Dec 30 '18 at 17:48
TheOscillatorTheOscillator
2,1891817
2,1891817
add a comment |
add a comment |
$begingroup$
Since $u_nto u$ in $L^1,$ there is a subsequence $u_{n_k}$ that converges to $u$ a.e. We have each $|u_{n_k}|le a$ a.e., so $|u|le a$ a.e., giving the result.
answered Dec 30 '18 at 19:28
zhw.zhw.
75.4k43275
$endgroup$
add a comment |
$begingroup$
Since $u_nto u$ in $L^1,$ there is a subsequence $u_{n_k}$ that converges to $u$ a.e. We have each $|u_{n_k}|le a$ a.e., so $|u|le a$ a.e., giving the result.
answered Dec 30 '18 at 19:28
zhw.zhw.
75.4k43275
$endgroup$
add a comment |
$begingroup$
Since $u_nto u$ in $L^1,$ there is a subsequence $u_{n_k}$ that converges to $u$ a.e. We have each $|u_{n_k}|le a$ a.e., so $|u|le a$ a.e., giving the result.
answered Dec 30 '18 at 19:28
zhw.zhw.
75.4k43275
$endgroup$
Since $u_nto u$ in $L^1,$ there is a subsequence $u_{n_k}$ that converges to $u$ a.e. We have each $|u_{n_k}|le a$ a.e., so $|u|le a$ a.e., giving the result.
answered Dec 30 '18 at 19:28
zhw.zhw.
75.4k43275
answered Dec 30 '18 at 19:28
zhw.zhw.
75.4k43275
answered Dec 30 '18 at 19:28
zhw.zhw.
75.4k43275
answered Dec 30 '18 at 19:28
zhw.zhw.
75.4k43275
75.4k43275
add a comment |
add a comment |
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Just use that $L^1$-convergence implies convergence a.e. along a subsequence. This subsequence has $L^infty$-norm bounded by $a$ and you are done.
$endgroup$
– Severin Schraven
Dec 30 '18 at 17:25
2
$begingroup$
Your approach is also valid. Since $L^1$ is seperable, Banach-Alaoglu implies that there is a weak-star convergent subsequence converging to $u$. Then $|u|_inftyle a$ follows.
$endgroup$
– Song
Dec 30 '18 at 17:32
$begingroup$
Thank you both !
$endgroup$
– user630504
Dec 30 '18 at 17:48