Mixing
Can a sequence of functions converge to $0$ in only one $L_p$ space?
Let $1<p<infty$, and suppose that $f_n$ converges to $0$ in the $L_q$ norm for all $qneq p$ where $1leq q<infty$? My question is, is it possible for $f_n$ to not converge to $0$ in the $L_p$ norm?
If there doesn't exist such a sequence of functions, does there at least a sequence of functions which does not converges to $0$ in some $L_p$ for $1leq p<infty$ but converges to $0$ in every $L_q$ for $1<q<p$, and another sequence of functions which does not converge to $0$ in $L_p$ but converges to $0$ in every $L_q$ for $q>p$?
measure-theory lebesgue-integral lp-spaces
asked Nov 20 '18 at 5:09
Keshav Srinivasan
2,09111441
add a comment |
Let $1<p<infty$, and suppose that $f_n$ converges to $0$ in the $L_q$ norm for all $qneq p$ where $1leq q<infty$? My question is, is it possible for $f_n$ to not converge to $0$ in the $L_p$ norm?
If there doesn't exist such a sequence of functions, does there at least a sequence of functions which does not converges to $0$ in some $L_p$ for $1leq p<infty$ but converges to $0$ in every $L_q$ for $1<q<p$, and another sequence of functions which does not converge to $0$ in $L_p$ but converges to $0$ in every $L_q$ for $q>p$?
measure-theory lebesgue-integral lp-spaces
asked Nov 20 '18 at 5:09
Keshav Srinivasan
2,09111441
add a comment |
Let $1<p<infty$, and suppose that $f_n$ converges to $0$ in the $L_q$ norm for all $qneq p$ where $1leq q<infty$? My question is, is it possible for $f_n$ to not converge to $0$ in the $L_p$ norm?
If there doesn't exist such a sequence of functions, does there at least a sequence of functions which does not converges to $0$ in some $L_p$ for $1leq p<infty$ but converges to $0$ in every $L_q$ for $1<q<p$, and another sequence of functions which does not converge to $0$ in $L_p$ but converges to $0$ in every $L_q$ for $q>p$?
measure-theory lebesgue-integral lp-spaces
asked Nov 20 '18 at 5:09
Keshav Srinivasan
2,09111441
Let $1<p<infty$, and suppose that $f_n$ converges to $0$ in the $L_q$ norm for all $qneq p$ where $1leq q<infty$? My question is, is it possible for $f_n$ to not converge to $0$ in the $L_p$ norm?
If there doesn't exist such a sequence of functions, does there at least a sequence of functions which does not converges to $0$ in some $L_p$ for $1leq p<infty$ but converges to $0$ in every $L_q$ for $1<q<p$, and another sequence of functions which does not converge to $0$ in $L_p$ but converges to $0$ in every $L_q$ for $q>p$?
measure-theory lebesgue-integral lp-spaces
measure-theory lebesgue-integral lp-spaces
asked Nov 20 '18 at 5:09
Keshav Srinivasan
2,09111441
asked Nov 20 '18 at 5:09
Keshav Srinivasan
2,09111441
edited Nov 20 '18 at 5:22
asked Nov 20 '18 at 5:09
Keshav Srinivasan
2,09111441
asked Nov 20 '18 at 5:09
Keshav Srinivasan
2,09111441
asked Nov 20 '18 at 5:09
Keshav Srinivasan
2,09111441
2,09111441
add a comment |
add a comment |
1 Answer
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No such sequence exists by Hölder's inequality. This is in Folland's book.
For the second question, you can do something analogous to what I did in my previous answer.
answered Nov 20 '18 at 5:15
mathworker21
8,6371928
Thanks for your answer, but I was hoping for something with $L_q$'s on both sides. That is, I want $p$ to be greater than $1$ and to have the sequence converge for all $L_q$ where $q$ can be either less than or greater than $p$ (but still gretar than or equal to 1). I'll modify my question to change greater than or equal to greater than.
– Keshav Srinivasan
Nov 20 '18 at 5:20
@KeshavSrinivasan ok I changed my answer
– mathworker21
Nov 20 '18 at 5:25
OK, so there's no $L_p$ with $L_q$'s on both sides with the property I want. But your previous answer at least addressed the case where it converges to $0$ in all $L_q$'s for $q>1$ but not for $L_1$. But now what about the opposite case, a sequence of functions which converges to $0$ in all $L_q$'s $q$ between $1$ and $p$ (where $p$>1), but doesn't converge to $0$ in $L_p$?
– Keshav Srinivasan
Nov 20 '18 at 5:32
$n1_{[0,frac{1}{n}]}$ I think does the job
– mathworker21
Nov 20 '18 at 5:51
What is the $p$ it does the job for?
– Keshav Srinivasan
Nov 20 '18 at 5:57
|
show 1 more comment
Your Answer
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1 Answer
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1 Answer
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oldest
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votes
No such sequence exists by Hölder's inequality. This is in Folland's book.
For the second question, you can do something analogous to what I did in my previous answer.
answered Nov 20 '18 at 5:15
mathworker21
8,6371928
Thanks for your answer, but I was hoping for something with $L_q$'s on both sides. That is, I want $p$ to be greater than $1$ and to have the sequence converge for all $L_q$ where $q$ can be either less than or greater than $p$ (but still gretar than or equal to 1). I'll modify my question to change greater than or equal to greater than.
– Keshav Srinivasan
Nov 20 '18 at 5:20
@KeshavSrinivasan ok I changed my answer
– mathworker21
Nov 20 '18 at 5:25
OK, so there's no $L_p$ with $L_q$'s on both sides with the property I want. But your previous answer at least addressed the case where it converges to $0$ in all $L_q$'s for $q>1$ but not for $L_1$. But now what about the opposite case, a sequence of functions which converges to $0$ in all $L_q$'s $q$ between $1$ and $p$ (where $p$>1), but doesn't converge to $0$ in $L_p$?
– Keshav Srinivasan
Nov 20 '18 at 5:32
$n1_{[0,frac{1}{n}]}$ I think does the job
– mathworker21
Nov 20 '18 at 5:51
What is the $p$ it does the job for?
– Keshav Srinivasan
Nov 20 '18 at 5:57
|
show 1 more comment
No such sequence exists by Hölder's inequality. This is in Folland's book.
For the second question, you can do something analogous to what I did in my previous answer.
answered Nov 20 '18 at 5:15
mathworker21
8,6371928
Thanks for your answer, but I was hoping for something with $L_q$'s on both sides. That is, I want $p$ to be greater than $1$ and to have the sequence converge for all $L_q$ where $q$ can be either less than or greater than $p$ (but still gretar than or equal to 1). I'll modify my question to change greater than or equal to greater than.
– Keshav Srinivasan
Nov 20 '18 at 5:20
@KeshavSrinivasan ok I changed my answer
– mathworker21
Nov 20 '18 at 5:25
OK, so there's no $L_p$ with $L_q$'s on both sides with the property I want. But your previous answer at least addressed the case where it converges to $0$ in all $L_q$'s for $q>1$ but not for $L_1$. But now what about the opposite case, a sequence of functions which converges to $0$ in all $L_q$'s $q$ between $1$ and $p$ (where $p$>1), but doesn't converge to $0$ in $L_p$?
– Keshav Srinivasan
Nov 20 '18 at 5:32
$n1_{[0,frac{1}{n}]}$ I think does the job
– mathworker21
Nov 20 '18 at 5:51
What is the $p$ it does the job for?
– Keshav Srinivasan
Nov 20 '18 at 5:57
|
show 1 more comment
No such sequence exists by Hölder's inequality. This is in Folland's book.
For the second question, you can do something analogous to what I did in my previous answer.
answered Nov 20 '18 at 5:15
mathworker21
8,6371928
No such sequence exists by Hölder's inequality. This is in Folland's book.
For the second question, you can do something analogous to what I did in my previous answer.
answered Nov 20 '18 at 5:15
mathworker21
8,6371928
edited Nov 20 '18 at 5:25
answered Nov 20 '18 at 5:15
mathworker21
8,6371928
answered Nov 20 '18 at 5:15
mathworker21
8,6371928
answered Nov 20 '18 at 5:15
mathworker21
8,6371928
8,6371928
Thanks for your answer, but I was hoping for something with $L_q$'s on both sides. That is, I want $p$ to be greater than $1$ and to have the sequence converge for all $L_q$ where $q$ can be either less than or greater than $p$ (but still gretar than or equal to 1). I'll modify my question to change greater than or equal to greater than.
– Keshav Srinivasan
Nov 20 '18 at 5:20
@KeshavSrinivasan ok I changed my answer
– mathworker21
Nov 20 '18 at 5:25
OK, so there's no $L_p$ with $L_q$'s on both sides with the property I want. But your previous answer at least addressed the case where it converges to $0$ in all $L_q$'s for $q>1$ but not for $L_1$. But now what about the opposite case, a sequence of functions which converges to $0$ in all $L_q$'s $q$ between $1$ and $p$ (where $p$>1), but doesn't converge to $0$ in $L_p$?
– Keshav Srinivasan
Nov 20 '18 at 5:32
$n1_{[0,frac{1}{n}]}$ I think does the job
– mathworker21
Nov 20 '18 at 5:51
What is the $p$ it does the job for?
– Keshav Srinivasan
Nov 20 '18 at 5:57
|
show 1 more comment
Thanks for your answer, but I was hoping for something with $L_q$'s on both sides. That is, I want $p$ to be greater than $1$ and to have the sequence converge for all $L_q$ where $q$ can be either less than or greater than $p$ (but still gretar than or equal to 1). I'll modify my question to change greater than or equal to greater than.
– Keshav Srinivasan
Nov 20 '18 at 5:20
@KeshavSrinivasan ok I changed my answer
– mathworker21
Nov 20 '18 at 5:25
OK, so there's no $L_p$ with $L_q$'s on both sides with the property I want. But your previous answer at least addressed the case where it converges to $0$ in all $L_q$'s for $q>1$ but not for $L_1$. But now what about the opposite case, a sequence of functions which converges to $0$ in all $L_q$'s $q$ between $1$ and $p$ (where $p$>1), but doesn't converge to $0$ in $L_p$?
– Keshav Srinivasan
Nov 20 '18 at 5:32
$n1_{[0,frac{1}{n}]}$ I think does the job
– mathworker21
Nov 20 '18 at 5:51
What is the $p$ it does the job for?
– Keshav Srinivasan
Nov 20 '18 at 5:57
Thanks for your answer, but I was hoping for something with $L_q$'s on both sides. That is, I want $p$ to be greater than $1$ and to have the sequence converge for all $L_q$ where $q$ can be either less than or greater than $p$ (but still gretar than or equal to 1). I'll modify my question to change greater than or equal to greater than.
– Keshav Srinivasan
Nov 20 '18 at 5:20
Thanks for your answer, but I was hoping for something with $L_q$'s on both sides. That is, I want $p$ to be greater than $1$ and to have the sequence converge for all $L_q$ where $q$ can be either less than or greater than $p$ (but still gretar than or equal to 1). I'll modify my question to change greater than or equal to greater than.
– Keshav Srinivasan
Nov 20 '18 at 5:20
@KeshavSrinivasan ok I changed my answer
– mathworker21
Nov 20 '18 at 5:25
@KeshavSrinivasan ok I changed my answer
– mathworker21
Nov 20 '18 at 5:25
OK, so there's no $L_p$ with $L_q$'s on both sides with the property I want. But your previous answer at least addressed the case where it converges to $0$ in all $L_q$'s for $q>1$ but not for $L_1$. But now what about the opposite case, a sequence of functions which converges to $0$ in all $L_q$'s $q$ between $1$ and $p$ (where $p$>1), but doesn't converge to $0$ in $L_p$?
– Keshav Srinivasan
Nov 20 '18 at 5:32
OK, so there's no $L_p$ with $L_q$'s on both sides with the property I want. But your previous answer at least addressed the case where it converges to $0$ in all $L_q$'s for $q>1$ but not for $L_1$. But now what about the opposite case, a sequence of functions which converges to $0$ in all $L_q$'s $q$ between $1$ and $p$ (where $p$>1), but doesn't converge to $0$ in $L_p$?
– Keshav Srinivasan
Nov 20 '18 at 5:32
$n1_{[0,frac{1}{n}]}$ I think does the job
– mathworker21
Nov 20 '18 at 5:51
$n1_{[0,frac{1}{n}]}$ I think does the job
– mathworker21
Nov 20 '18 at 5:51
What is the $p$ it does the job for?
– Keshav Srinivasan
Nov 20 '18 at 5:57
What is the $p$ it does the job for?
– Keshav Srinivasan
Nov 20 '18 at 5:57
|
show 1 more comment
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