$y in mathbb{K^N}$ a scalar sequence, if $sum_{n geq 1} x(n)y(n)$ is bounded,does $x in l_{p^*}$?












1














Given $y in mathbb{K^N}$ a scalar sequence, and $1<p<infty$.



Suppose that, $forall x in l_p$, the series $sum_{n geq 1} x(n)y(n)$ is bounded. Show that $x in l_{p^*}$.



With $l_{p^*}$ I mean the dual of $l_p$, if I am not wrong, we have $(l_p)^*=l_{p^*}$.
Any suggestion? I'm not really keen in dual... what should I demonstrate?






sequences-and-series functional-analysis lp-spaces dual-spaces







share|cite|improve this question
























  • so you mean $p^*=frac{p}{1-p}$? In that case, you are right, $l_p^*=l_{p^*}$ and you want to show that $yin l_{p^*}$ right?
    – Pink Panther
    2 days ago








  • 1




    @PinkPanther Yes!
    – Maggie94
    2 days ago
















1














Given $y in mathbb{K^N}$ a scalar sequence, and $1<p<infty$.



Suppose that, $forall x in l_p$, the series $sum_{n geq 1} x(n)y(n)$ is bounded. Show that $x in l_{p^*}$.



With $l_{p^*}$ I mean the dual of $l_p$, if I am not wrong, we have $(l_p)^*=l_{p^*}$.
Any suggestion? I'm not really keen in dual... what should I demonstrate?






sequences-and-series functional-analysis lp-spaces dual-spaces







share|cite|improve this question
























  • so you mean $p^*=frac{p}{1-p}$? In that case, you are right, $l_p^*=l_{p^*}$ and you want to show that $yin l_{p^*}$ right?
    – Pink Panther
    2 days ago








  • 1




    @PinkPanther Yes!
    – Maggie94
    2 days ago














1












1








1







Given $y in mathbb{K^N}$ a scalar sequence, and $1<p<infty$.



Suppose that, $forall x in l_p$, the series $sum_{n geq 1} x(n)y(n)$ is bounded. Show that $x in l_{p^*}$.



With $l_{p^*}$ I mean the dual of $l_p$, if I am not wrong, we have $(l_p)^*=l_{p^*}$.
Any suggestion? I'm not really keen in dual... what should I demonstrate?






sequences-and-series functional-analysis lp-spaces dual-spaces







share|cite|improve this question















Given $y in mathbb{K^N}$ a scalar sequence, and $1<p<infty$.



Suppose that, $forall x in l_p$, the series $sum_{n geq 1} x(n)y(n)$ is bounded. Show that $x in l_{p^*}$.



With $l_{p^*}$ I mean the dual of $l_p$, if I am not wrong, we have $(l_p)^*=l_{p^*}$.
Any suggestion? I'm not really keen in dual... what should I demonstrate?






sequences-and-series functional-analysis lp-spaces dual-spaces




sequences-and-series functional-analysis lp-spaces dual-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Bernard

118k639112




118k639112










asked 2 days ago









Maggie94Maggie94

966




966












  • so you mean $p^*=frac{p}{1-p}$? In that case, you are right, $l_p^*=l_{p^*}$ and you want to show that $yin l_{p^*}$ right?
    – Pink Panther
    2 days ago








  • 1




    @PinkPanther Yes!
    – Maggie94
    2 days ago


















  • so you mean $p^*=frac{p}{1-p}$? In that case, you are right, $l_p^*=l_{p^*}$ and you want to show that $yin l_{p^*}$ right?
    – Pink Panther
    2 days ago








  • 1




    @PinkPanther Yes!
    – Maggie94
    2 days ago
















so you mean $p^*=frac{p}{1-p}$? In that case, you are right, $l_p^*=l_{p^*}$ and you want to show that $yin l_{p^*}$ right?
– Pink Panther
2 days ago






so you mean $p^*=frac{p}{1-p}$? In that case, you are right, $l_p^*=l_{p^*}$ and you want to show that $yin l_{p^*}$ right?
– Pink Panther
2 days ago






1




1




@PinkPanther Yes!
– Maggie94
2 days ago




@PinkPanther Yes!
– Maggie94
2 days ago










1 Answer
1






active

oldest

votes


















2














Note that can strenghthen the assumption to
$$
sum_{n =1}^infty |x(n)||y(n)| <infty, quadforall xin l^p.
$$ by considering $widehat{x}(n) = |x(n)|e^{-itheta_n}$ where $y(n) = |y(n)|e^{itheta_n}$. Define $$
F_k ={xin l^p;|; sum_n |x(n)||y(n)|le k}.$$ We can show by Fatou's lemma(or directly) that $F_k$ is closed in $l^p$ for all $k$. Now, since $l^p = cup_k F_k$, by Baire category theorem, there exists $k$, $zin l^p$ and $delta>0$ such that
$$
z'in l^p,;|z-z'|_pleq delta Rightarrow z'in F_k.
$$ Let $xin l^p$ be $|x|_ple 1$. Then from
$$
sum_n |z(n)+delta x(n)||y(n)|le k,
$$ we have
$$
deltasum_n |x(n)||y(n)| leq sum_n |z(n)||y(n)|+k le 2k.
$$ Now, if we take $x(n) = frac{|y(n)|^{p^*-1}}{(sum_{n=1}^N |y(n)|^{p^*})^{1/p}}1_{nleq N}$ , then $|x|_ple 1$ and it follows that
$$
left(sum_{nle N}|y(n)|^{p^*}right)^{1-1/p} = sum_{nle N}|x(n)||y(n)| le sum_{n}|x(n)||y(n)| lefrac{2k}{delta}.
$$ Take $Ntoinfty$ to get
$$
|y|_{p^*}le frac{2k}{delta}<infty.
$$






share|cite|improve this answer























  • What do you mean by $z’$?
    – Maggie94
    2 days ago










  • I used $z'$ as another generic element of $l^p$.
    – Song
    2 days ago










  • Ok thank you... and I don’t unterstand the final part... we get that $Vert yVert_{p^*} leq infty$, why this implies that $x in l_{p^*}$?
    – Maggie94
    2 days ago












  • Your question is stated misleadingly. Please check that. Either $forall xin l^p$ or $xin l^{p^*}$ should be fixed.
    – Song
    2 days ago












  • What don’t you understand? $x in l_p$ is fixed, so $forall x in l_p$ I want to show that $x in l_{p^*}$, i think the question is well written. What I don’t understand in the final part of your answer is how can I say that $x in l_{p*}$ because you get $Vert y Vert _{p^*} < infty$
    – Maggie94
    yesterday













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1 Answer
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1 Answer
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2














Note that can strenghthen the assumption to
$$
sum_{n =1}^infty |x(n)||y(n)| <infty, quadforall xin l^p.
$$ by considering $widehat{x}(n) = |x(n)|e^{-itheta_n}$ where $y(n) = |y(n)|e^{itheta_n}$. Define $$
F_k ={xin l^p;|; sum_n |x(n)||y(n)|le k}.$$ We can show by Fatou's lemma(or directly) that $F_k$ is closed in $l^p$ for all $k$. Now, since $l^p = cup_k F_k$, by Baire category theorem, there exists $k$, $zin l^p$ and $delta>0$ such that
$$
z'in l^p,;|z-z'|_pleq delta Rightarrow z'in F_k.
$$ Let $xin l^p$ be $|x|_ple 1$. Then from
$$
sum_n |z(n)+delta x(n)||y(n)|le k,
$$ we have
$$
deltasum_n |x(n)||y(n)| leq sum_n |z(n)||y(n)|+k le 2k.
$$ Now, if we take $x(n) = frac{|y(n)|^{p^*-1}}{(sum_{n=1}^N |y(n)|^{p^*})^{1/p}}1_{nleq N}$ , then $|x|_ple 1$ and it follows that
$$
left(sum_{nle N}|y(n)|^{p^*}right)^{1-1/p} = sum_{nle N}|x(n)||y(n)| le sum_{n}|x(n)||y(n)| lefrac{2k}{delta}.
$$ Take $Ntoinfty$ to get
$$
|y|_{p^*}le frac{2k}{delta}<infty.
$$






share|cite|improve this answer























  • What do you mean by $z’$?
    – Maggie94
    2 days ago










  • I used $z'$ as another generic element of $l^p$.
    – Song
    2 days ago










  • Ok thank you... and I don’t unterstand the final part... we get that $Vert yVert_{p^*} leq infty$, why this implies that $x in l_{p^*}$?
    – Maggie94
    2 days ago












  • Your question is stated misleadingly. Please check that. Either $forall xin l^p$ or $xin l^{p^*}$ should be fixed.
    – Song
    2 days ago












  • What don’t you understand? $x in l_p$ is fixed, so $forall x in l_p$ I want to show that $x in l_{p^*}$, i think the question is well written. What I don’t understand in the final part of your answer is how can I say that $x in l_{p*}$ because you get $Vert y Vert _{p^*} < infty$
    – Maggie94
    yesterday


















2














Note that can strenghthen the assumption to
$$
sum_{n =1}^infty |x(n)||y(n)| <infty, quadforall xin l^p.
$$ by considering $widehat{x}(n) = |x(n)|e^{-itheta_n}$ where $y(n) = |y(n)|e^{itheta_n}$. Define $$
F_k ={xin l^p;|; sum_n |x(n)||y(n)|le k}.$$ We can show by Fatou's lemma(or directly) that $F_k$ is closed in $l^p$ for all $k$. Now, since $l^p = cup_k F_k$, by Baire category theorem, there exists $k$, $zin l^p$ and $delta>0$ such that
$$
z'in l^p,;|z-z'|_pleq delta Rightarrow z'in F_k.
$$ Let $xin l^p$ be $|x|_ple 1$. Then from
$$
sum_n |z(n)+delta x(n)||y(n)|le k,
$$ we have
$$
deltasum_n |x(n)||y(n)| leq sum_n |z(n)||y(n)|+k le 2k.
$$ Now, if we take $x(n) = frac{|y(n)|^{p^*-1}}{(sum_{n=1}^N |y(n)|^{p^*})^{1/p}}1_{nleq N}$ , then $|x|_ple 1$ and it follows that
$$
left(sum_{nle N}|y(n)|^{p^*}right)^{1-1/p} = sum_{nle N}|x(n)||y(n)| le sum_{n}|x(n)||y(n)| lefrac{2k}{delta}.
$$ Take $Ntoinfty$ to get
$$
|y|_{p^*}le frac{2k}{delta}<infty.
$$






share|cite|improve this answer























  • What do you mean by $z’$?
    – Maggie94
    2 days ago










  • I used $z'$ as another generic element of $l^p$.
    – Song
    2 days ago










  • Ok thank you... and I don’t unterstand the final part... we get that $Vert yVert_{p^*} leq infty$, why this implies that $x in l_{p^*}$?
    – Maggie94
    2 days ago












  • Your question is stated misleadingly. Please check that. Either $forall xin l^p$ or $xin l^{p^*}$ should be fixed.
    – Song
    2 days ago












  • What don’t you understand? $x in l_p$ is fixed, so $forall x in l_p$ I want to show that $x in l_{p^*}$, i think the question is well written. What I don’t understand in the final part of your answer is how can I say that $x in l_{p*}$ because you get $Vert y Vert _{p^*} < infty$
    – Maggie94
    yesterday
















2












2








2






Note that can strenghthen the assumption to
$$
sum_{n =1}^infty |x(n)||y(n)| <infty, quadforall xin l^p.
$$ by considering $widehat{x}(n) = |x(n)|e^{-itheta_n}$ where $y(n) = |y(n)|e^{itheta_n}$. Define $$
F_k ={xin l^p;|; sum_n |x(n)||y(n)|le k}.$$ We can show by Fatou's lemma(or directly) that $F_k$ is closed in $l^p$ for all $k$. Now, since $l^p = cup_k F_k$, by Baire category theorem, there exists $k$, $zin l^p$ and $delta>0$ such that
$$
z'in l^p,;|z-z'|_pleq delta Rightarrow z'in F_k.
$$ Let $xin l^p$ be $|x|_ple 1$. Then from
$$
sum_n |z(n)+delta x(n)||y(n)|le k,
$$ we have
$$
deltasum_n |x(n)||y(n)| leq sum_n |z(n)||y(n)|+k le 2k.
$$ Now, if we take $x(n) = frac{|y(n)|^{p^*-1}}{(sum_{n=1}^N |y(n)|^{p^*})^{1/p}}1_{nleq N}$ , then $|x|_ple 1$ and it follows that
$$
left(sum_{nle N}|y(n)|^{p^*}right)^{1-1/p} = sum_{nle N}|x(n)||y(n)| le sum_{n}|x(n)||y(n)| lefrac{2k}{delta}.
$$ Take $Ntoinfty$ to get
$$
|y|_{p^*}le frac{2k}{delta}<infty.
$$






share|cite|improve this answer














Note that can strenghthen the assumption to
$$
sum_{n =1}^infty |x(n)||y(n)| <infty, quadforall xin l^p.
$$ by considering $widehat{x}(n) = |x(n)|e^{-itheta_n}$ where $y(n) = |y(n)|e^{itheta_n}$. Define $$
F_k ={xin l^p;|; sum_n |x(n)||y(n)|le k}.$$ We can show by Fatou's lemma(or directly) that $F_k$ is closed in $l^p$ for all $k$. Now, since $l^p = cup_k F_k$, by Baire category theorem, there exists $k$, $zin l^p$ and $delta>0$ such that
$$
z'in l^p,;|z-z'|_pleq delta Rightarrow z'in F_k.
$$ Let $xin l^p$ be $|x|_ple 1$. Then from
$$
sum_n |z(n)+delta x(n)||y(n)|le k,
$$ we have
$$
deltasum_n |x(n)||y(n)| leq sum_n |z(n)||y(n)|+k le 2k.
$$ Now, if we take $x(n) = frac{|y(n)|^{p^*-1}}{(sum_{n=1}^N |y(n)|^{p^*})^{1/p}}1_{nleq N}$ , then $|x|_ple 1$ and it follows that
$$
left(sum_{nle N}|y(n)|^{p^*}right)^{1-1/p} = sum_{nle N}|x(n)||y(n)| le sum_{n}|x(n)||y(n)| lefrac{2k}{delta}.
$$ Take $Ntoinfty$ to get
$$
|y|_{p^*}le frac{2k}{delta}<infty.
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago

























answered 2 days ago









SongSong

6,025318




6,025318












  • What do you mean by $z’$?
    – Maggie94
    2 days ago










  • I used $z'$ as another generic element of $l^p$.
    – Song
    2 days ago










  • Ok thank you... and I don’t unterstand the final part... we get that $Vert yVert_{p^*} leq infty$, why this implies that $x in l_{p^*}$?
    – Maggie94
    2 days ago












  • Your question is stated misleadingly. Please check that. Either $forall xin l^p$ or $xin l^{p^*}$ should be fixed.
    – Song
    2 days ago












  • What don’t you understand? $x in l_p$ is fixed, so $forall x in l_p$ I want to show that $x in l_{p^*}$, i think the question is well written. What I don’t understand in the final part of your answer is how can I say that $x in l_{p*}$ because you get $Vert y Vert _{p^*} < infty$
    – Maggie94
    yesterday




















  • What do you mean by $z’$?
    – Maggie94
    2 days ago










  • I used $z'$ as another generic element of $l^p$.
    – Song
    2 days ago










  • Ok thank you... and I don’t unterstand the final part... we get that $Vert yVert_{p^*} leq infty$, why this implies that $x in l_{p^*}$?
    – Maggie94
    2 days ago












  • Your question is stated misleadingly. Please check that. Either $forall xin l^p$ or $xin l^{p^*}$ should be fixed.
    – Song
    2 days ago












  • What don’t you understand? $x in l_p$ is fixed, so $forall x in l_p$ I want to show that $x in l_{p^*}$, i think the question is well written. What I don’t understand in the final part of your answer is how can I say that $x in l_{p*}$ because you get $Vert y Vert _{p^*} < infty$
    – Maggie94
    yesterday


















What do you mean by $z’$?
– Maggie94
2 days ago




What do you mean by $z’$?
– Maggie94
2 days ago












I used $z'$ as another generic element of $l^p$.
– Song
2 days ago




I used $z'$ as another generic element of $l^p$.
– Song
2 days ago












Ok thank you... and I don’t unterstand the final part... we get that $Vert yVert_{p^*} leq infty$, why this implies that $x in l_{p^*}$?
– Maggie94
2 days ago






Ok thank you... and I don’t unterstand the final part... we get that $Vert yVert_{p^*} leq infty$, why this implies that $x in l_{p^*}$?
– Maggie94
2 days ago














Your question is stated misleadingly. Please check that. Either $forall xin l^p$ or $xin l^{p^*}$ should be fixed.
– Song
2 days ago






Your question is stated misleadingly. Please check that. Either $forall xin l^p$ or $xin l^{p^*}$ should be fixed.
– Song
2 days ago














What don’t you understand? $x in l_p$ is fixed, so $forall x in l_p$ I want to show that $x in l_{p^*}$, i think the question is well written. What I don’t understand in the final part of your answer is how can I say that $x in l_{p*}$ because you get $Vert y Vert _{p^*} < infty$
– Maggie94
yesterday






What don’t you understand? $x in l_p$ is fixed, so $forall x in l_p$ I want to show that $x in l_{p^*}$, i think the question is well written. What I don’t understand in the final part of your answer is how can I say that $x in l_{p*}$ because you get $Vert y Vert _{p^*} < infty$
– Maggie94
yesterday




















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