$y in mathbb{K^N}$ 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













Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062693%2fy-in-mathbbkn-a-scalar-sequence-if-sum-n-geq-1-xnyn-is-bounded%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






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


















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




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062693%2fy-in-mathbbkn-a-scalar-sequence-if-sum-n-geq-1-xnyn-is-bounded%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Mario Kart Wii

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
Read more

What does “Dominus providebit” mean?

Image
5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
Read more

File:Tiny Toon Adventures Wacky Sports JP Title.png

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! File:Tiny Toon Adventures Wacky Sports JP Title.png From The Cutting Room Floor Jump to: navigation, search File File history File usage No higher resolution available. Tiny_Toon_Adventures_Wacky_Sports_JP_Title.png ‎ (160 × 144 pixels, file size: 3 KB, MIME type: image/png ) This image is a screenshot of a(n) Game Boy title screen. File history Click on a date/time to view the file as it appeared at that time. Date/Time Thumbnail Dimensions User Comment current 21:31, 20 August 2016 160 × 144 (3 KB) TheBluegeek (Talk | contribs) 17:25, 29 August 2013 160 × 144 (3 KB) Luigi-San (Talk | contribs) {{titlescreen|Game Boy}} You cannot overwrite this file. File usage The following page links to this file: Tiny Toon Advent
Read more