Find the value of p for which the following series converges? [closed]












-3












$begingroup$


Let $x = (x_1, x_2, ldots )in l^4$, $xneq 0$. For which one of the following values of $p$, the series
$$
sum_{i=1}^{infty} x_i y_i
$$
converges for every $y = (y_1, y_2, ldots )in l^p$?




  • A) 1

  • B) 2

  • C) 3

  • D) 4


What I tried: I tried to use Hölder's inequality, I am almost done, I think I am just missing a little.



$|sum_{i=1}^{infty} x_i y_i|le
sum_{i=1}^{infty} |x_i y_i|le ||x||_4 ||y||_p $



where $p$ is the Hölder conjugate of $4$ which gives $p=4/3$.






convergence lp-spaces







share|cite|improve this question











$endgroup$



closed as unclear what you're asking by Did, Leucippus, metamorphy, mrtaurho, Cesareo Jan 30 at 8:58


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.














  • 1




    $begingroup$
    "I have tried to use the Holder's inequality to get something" Excellent! Please show how you tried.
    $endgroup$
    – Did
    Jan 30 at 7:23










  • $begingroup$
    Right -- and this points clearly at a unique correct option... But maybe what you are missing is that the family of spaces $(ell^p)_{p>0}$ is increasing... Thus, you might want to show this fact and then to apply it to your setting.
    $endgroup$
    – Did
    Jan 30 at 7:37












  • $begingroup$
    Is it $||y||_{4/3}le ||y||_1$ impling that p= 1 will work??
    $endgroup$
    – Devendra Singh Rana
    Jan 30 at 7:45












  • $begingroup$
    Well, $ell^1subsetell^{4/3}subsetell^2subsetell^3subsetell^4$ and every $y$ in $ell^{4/3}$ works, so...
    $endgroup$
    – Did
    Jan 30 at 7:45


















-3












$begingroup$


Let $x = (x_1, x_2, ldots )in l^4$, $xneq 0$. For which one of the following values of $p$, the series
$$
sum_{i=1}^{infty} x_i y_i
$$
converges for every $y = (y_1, y_2, ldots )in l^p$?




  • A) 1

  • B) 2

  • C) 3

  • D) 4


What I tried: I tried to use Hölder's inequality, I am almost done, I think I am just missing a little.



$|sum_{i=1}^{infty} x_i y_i|le
sum_{i=1}^{infty} |x_i y_i|le ||x||_4 ||y||_p $



where $p$ is the Hölder conjugate of $4$ which gives $p=4/3$.






convergence lp-spaces







share|cite|improve this question











$endgroup$



closed as unclear what you're asking by Did, Leucippus, metamorphy, mrtaurho, Cesareo Jan 30 at 8:58


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.














  • 1




    $begingroup$
    "I have tried to use the Holder's inequality to get something" Excellent! Please show how you tried.
    $endgroup$
    – Did
    Jan 30 at 7:23










  • $begingroup$
    Right -- and this points clearly at a unique correct option... But maybe what you are missing is that the family of spaces $(ell^p)_{p>0}$ is increasing... Thus, you might want to show this fact and then to apply it to your setting.
    $endgroup$
    – Did
    Jan 30 at 7:37












  • $begingroup$
    Is it $||y||_{4/3}le ||y||_1$ impling that p= 1 will work??
    $endgroup$
    – Devendra Singh Rana
    Jan 30 at 7:45












  • $begingroup$
    Well, $ell^1subsetell^{4/3}subsetell^2subsetell^3subsetell^4$ and every $y$ in $ell^{4/3}$ works, so...
    $endgroup$
    – Did
    Jan 30 at 7:45
















-3












-3








-3





$begingroup$


Let $x = (x_1, x_2, ldots )in l^4$, $xneq 0$. For which one of the following values of $p$, the series
$$
sum_{i=1}^{infty} x_i y_i
$$
converges for every $y = (y_1, y_2, ldots )in l^p$?




  • A) 1

  • B) 2

  • C) 3

  • D) 4


What I tried: I tried to use Hölder's inequality, I am almost done, I think I am just missing a little.



$|sum_{i=1}^{infty} x_i y_i|le
sum_{i=1}^{infty} |x_i y_i|le ||x||_4 ||y||_p $



where $p$ is the Hölder conjugate of $4$ which gives $p=4/3$.






convergence lp-spaces







share|cite|improve this question











$endgroup$




Let $x = (x_1, x_2, ldots )in l^4$, $xneq 0$. For which one of the following values of $p$, the series
$$
sum_{i=1}^{infty} x_i y_i
$$
converges for every $y = (y_1, y_2, ldots )in l^p$?




  • A) 1

  • B) 2

  • C) 3

  • D) 4


What I tried: I tried to use Hölder's inequality, I am almost done, I think I am just missing a little.



$|sum_{i=1}^{infty} x_i y_i|le
sum_{i=1}^{infty} |x_i y_i|le ||x||_4 ||y||_p $



where $p$ is the Hölder conjugate of $4$ which gives $p=4/3$.






convergence lp-spaces




convergence lp-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 30 at 7:48









Did

249k23227466




249k23227466










asked Jan 30 at 7:15









Devendra Singh RanaDevendra Singh Rana

7871416




7871416




closed as unclear what you're asking by Did, Leucippus, metamorphy, mrtaurho, Cesareo Jan 30 at 8:58


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.









closed as unclear what you're asking by Did, Leucippus, metamorphy, mrtaurho, Cesareo Jan 30 at 8:58


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.










  • 1




    $begingroup$
    "I have tried to use the Holder's inequality to get something" Excellent! Please show how you tried.
    $endgroup$
    – Did
    Jan 30 at 7:23










  • $begingroup$
    Right -- and this points clearly at a unique correct option... But maybe what you are missing is that the family of spaces $(ell^p)_{p>0}$ is increasing... Thus, you might want to show this fact and then to apply it to your setting.
    $endgroup$
    – Did
    Jan 30 at 7:37












  • $begingroup$
    Is it $||y||_{4/3}le ||y||_1$ impling that p= 1 will work??
    $endgroup$
    – Devendra Singh Rana
    Jan 30 at 7:45












  • $begingroup$
    Well, $ell^1subsetell^{4/3}subsetell^2subsetell^3subsetell^4$ and every $y$ in $ell^{4/3}$ works, so...
    $endgroup$
    – Did
    Jan 30 at 7:45
















  • 1




    $begingroup$
    "I have tried to use the Holder's inequality to get something" Excellent! Please show how you tried.
    $endgroup$
    – Did
    Jan 30 at 7:23










  • $begingroup$
    Right -- and this points clearly at a unique correct option... But maybe what you are missing is that the family of spaces $(ell^p)_{p>0}$ is increasing... Thus, you might want to show this fact and then to apply it to your setting.
    $endgroup$
    – Did
    Jan 30 at 7:37












  • $begingroup$
    Is it $||y||_{4/3}le ||y||_1$ impling that p= 1 will work??
    $endgroup$
    – Devendra Singh Rana
    Jan 30 at 7:45












  • $begingroup$
    Well, $ell^1subsetell^{4/3}subsetell^2subsetell^3subsetell^4$ and every $y$ in $ell^{4/3}$ works, so...
    $endgroup$
    – Did
    Jan 30 at 7:45










1




1




$begingroup$
"I have tried to use the Holder's inequality to get something" Excellent! Please show how you tried.
$endgroup$
– Did
Jan 30 at 7:23




$begingroup$
"I have tried to use the Holder's inequality to get something" Excellent! Please show how you tried.
$endgroup$
– Did
Jan 30 at 7:23












$begingroup$
Right -- and this points clearly at a unique correct option... But maybe what you are missing is that the family of spaces $(ell^p)_{p>0}$ is increasing... Thus, you might want to show this fact and then to apply it to your setting.
$endgroup$
– Did
Jan 30 at 7:37






$begingroup$
Right -- and this points clearly at a unique correct option... But maybe what you are missing is that the family of spaces $(ell^p)_{p>0}$ is increasing... Thus, you might want to show this fact and then to apply it to your setting.
$endgroup$
– Did
Jan 30 at 7:37














$begingroup$
Is it $||y||_{4/3}le ||y||_1$ impling that p= 1 will work??
$endgroup$
– Devendra Singh Rana
Jan 30 at 7:45






$begingroup$
Is it $||y||_{4/3}le ||y||_1$ impling that p= 1 will work??
$endgroup$
– Devendra Singh Rana
Jan 30 at 7:45














$begingroup$
Well, $ell^1subsetell^{4/3}subsetell^2subsetell^3subsetell^4$ and every $y$ in $ell^{4/3}$ works, so...
$endgroup$
– Did
Jan 30 at 7:45






$begingroup$
Well, $ell^1subsetell^{4/3}subsetell^2subsetell^3subsetell^4$ and every $y$ in $ell^{4/3}$ works, so...
$endgroup$
– Did
Jan 30 at 7:45












1 Answer
1






active

oldest

votes


















1












$begingroup$

If $x in ell^{q}$ for every $q in [1,infty]$ then the series converges in all 4 cases. Assume that $x in ell^{4}setminus ell^{q}$ for all $q <4$. In this case the answer is A). You already know that the series converges if $y in ell ^{4/3}$. If $y in ell ^{p}$ with $p leq frac 4 3$ then $y in ell ^{4/3}$ so the series converges. To show that the series may not converge in cases B), C) and D) take $y_n=sign(x_n)|x_n|^{alpha}$ for a suitable $alpha$. I will leave the details to you.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That pretty much helps the deal
    $endgroup$
    – Devendra Singh Rana
    Jan 30 at 7:50


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

If $x in ell^{q}$ for every $q in [1,infty]$ then the series converges in all 4 cases. Assume that $x in ell^{4}setminus ell^{q}$ for all $q <4$. In this case the answer is A). You already know that the series converges if $y in ell ^{4/3}$. If $y in ell ^{p}$ with $p leq frac 4 3$ then $y in ell ^{4/3}$ so the series converges. To show that the series may not converge in cases B), C) and D) take $y_n=sign(x_n)|x_n|^{alpha}$ for a suitable $alpha$. I will leave the details to you.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That pretty much helps the deal
    $endgroup$
    – Devendra Singh Rana
    Jan 30 at 7:50
















1












$begingroup$

If $x in ell^{q}$ for every $q in [1,infty]$ then the series converges in all 4 cases. Assume that $x in ell^{4}setminus ell^{q}$ for all $q <4$. In this case the answer is A). You already know that the series converges if $y in ell ^{4/3}$. If $y in ell ^{p}$ with $p leq frac 4 3$ then $y in ell ^{4/3}$ so the series converges. To show that the series may not converge in cases B), C) and D) take $y_n=sign(x_n)|x_n|^{alpha}$ for a suitable $alpha$. I will leave the details to you.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That pretty much helps the deal
    $endgroup$
    – Devendra Singh Rana
    Jan 30 at 7:50














1












1








1





$begingroup$

If $x in ell^{q}$ for every $q in [1,infty]$ then the series converges in all 4 cases. Assume that $x in ell^{4}setminus ell^{q}$ for all $q <4$. In this case the answer is A). You already know that the series converges if $y in ell ^{4/3}$. If $y in ell ^{p}$ with $p leq frac 4 3$ then $y in ell ^{4/3}$ so the series converges. To show that the series may not converge in cases B), C) and D) take $y_n=sign(x_n)|x_n|^{alpha}$ for a suitable $alpha$. I will leave the details to you.






share|cite|improve this answer









$endgroup$



If $x in ell^{q}$ for every $q in [1,infty]$ then the series converges in all 4 cases. Assume that $x in ell^{4}setminus ell^{q}$ for all $q <4$. In this case the answer is A). You already know that the series converges if $y in ell ^{4/3}$. If $y in ell ^{p}$ with $p leq frac 4 3$ then $y in ell ^{4/3}$ so the series converges. To show that the series may not converge in cases B), C) and D) take $y_n=sign(x_n)|x_n|^{alpha}$ for a suitable $alpha$. I will leave the details to you.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 30 at 7:46









Kavi Rama MurthyKavi Rama Murthy

71.9k53170




71.9k53170












  • $begingroup$
    That pretty much helps the deal
    $endgroup$
    – Devendra Singh Rana
    Jan 30 at 7:50


















  • $begingroup$
    That pretty much helps the deal
    $endgroup$
    – Devendra Singh Rana
    Jan 30 at 7:50
















$begingroup$
That pretty much helps the deal
$endgroup$
– Devendra Singh Rana
Jan 30 at 7:50




$begingroup$
That pretty much helps the deal
$endgroup$
– Devendra Singh Rana
Jan 30 at 7:50



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