Show that the sum is the norm of $l_1$












2












$begingroup$


I have the space $mathbb E = l_1$ and $Tx = (x_1, 0, x_3, 0, ...).$
I need to show that $T$ is continuous.



The norm of $l_1$ is $$Vert x Vert = sum_{k=1}^∞ vert x_k vert.$$



My solution:



$Vert Tx Vert = sum_{k=1}^∞ vert (Tx)_k vert = vert 0 vert + sum_{k=0}^∞ vert x_{2k+1} vert $.



Question:



How to continue this equality to get $sum_{k=1}^infty vert x_k vert$ (the norm)?






functional-analysis continuity operator-theory norm lp-spaces







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You do not need to continue the equality. An inequality will do just as well (provided the norm is on the greater side)
    $endgroup$
    – Mindlack
    Dec 16 '18 at 11:18
















2












$begingroup$


I have the space $mathbb E = l_1$ and $Tx = (x_1, 0, x_3, 0, ...).$
I need to show that $T$ is continuous.



The norm of $l_1$ is $$Vert x Vert = sum_{k=1}^∞ vert x_k vert.$$



My solution:



$Vert Tx Vert = sum_{k=1}^∞ vert (Tx)_k vert = vert 0 vert + sum_{k=0}^∞ vert x_{2k+1} vert $.



Question:



How to continue this equality to get $sum_{k=1}^infty vert x_k vert$ (the norm)?






functional-analysis continuity operator-theory norm lp-spaces







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You do not need to continue the equality. An inequality will do just as well (provided the norm is on the greater side)
    $endgroup$
    – Mindlack
    Dec 16 '18 at 11:18














2












2








2


1



$begingroup$


I have the space $mathbb E = l_1$ and $Tx = (x_1, 0, x_3, 0, ...).$
I need to show that $T$ is continuous.



The norm of $l_1$ is $$Vert x Vert = sum_{k=1}^∞ vert x_k vert.$$



My solution:



$Vert Tx Vert = sum_{k=1}^∞ vert (Tx)_k vert = vert 0 vert + sum_{k=0}^∞ vert x_{2k+1} vert $.



Question:



How to continue this equality to get $sum_{k=1}^infty vert x_k vert$ (the norm)?






functional-analysis continuity operator-theory norm lp-spaces







share|cite|improve this question











$endgroup$




I have the space $mathbb E = l_1$ and $Tx = (x_1, 0, x_3, 0, ...).$
I need to show that $T$ is continuous.



The norm of $l_1$ is $$Vert x Vert = sum_{k=1}^∞ vert x_k vert.$$



My solution:



$Vert Tx Vert = sum_{k=1}^∞ vert (Tx)_k vert = vert 0 vert + sum_{k=0}^∞ vert x_{2k+1} vert $.



Question:



How to continue this equality to get $sum_{k=1}^infty vert x_k vert$ (the norm)?






functional-analysis continuity operator-theory norm lp-spaces




functional-analysis continuity operator-theory norm lp-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 12:36









Davide Giraudo

126k16150261




126k16150261










asked Dec 16 '18 at 11:06









PhilipPhilip

826




826








  • 2




    $begingroup$
    You do not need to continue the equality. An inequality will do just as well (provided the norm is on the greater side)
    $endgroup$
    – Mindlack
    Dec 16 '18 at 11:18














  • 2




    $begingroup$
    You do not need to continue the equality. An inequality will do just as well (provided the norm is on the greater side)
    $endgroup$
    – Mindlack
    Dec 16 '18 at 11:18








2




2




$begingroup$
You do not need to continue the equality. An inequality will do just as well (provided the norm is on the greater side)
$endgroup$
– Mindlack
Dec 16 '18 at 11:18




$begingroup$
You do not need to continue the equality. An inequality will do just as well (provided the norm is on the greater side)
$endgroup$
– Mindlack
Dec 16 '18 at 11:18










1 Answer
1






active

oldest

votes


















2












$begingroup$

It is :



$$|Tx| = |(x_1,0,x_3,0,dots)| = sum_{k=1}^infty|x_{2k+1}| leq sum_{k=1}^infty|x_k| = |x|.$$



since the absolute sum of the odd entries of the sequence is definitely smaller or equal to the total sum of absolute values of all entries.



Thus :



$$|Tx| leq |x| implies |T| leq 1$$



Now, if you take the sequence $x = (x_n) in ell^1$ with $x = (1,0,dots,0,dots)$, then :



$$|Tx| = |(1,0,dots,0,dots)| = 1 implies |T| = 1$$



Thus the operator $T$ is continuous.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I would like to have an input from the person that downvoted.
    $endgroup$
    – Rebellos
    Dec 16 '18 at 14:10











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%2f3042477%2fshow-that-the-sum-is-the-norm-of-l-1%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












$begingroup$

It is :



$$|Tx| = |(x_1,0,x_3,0,dots)| = sum_{k=1}^infty|x_{2k+1}| leq sum_{k=1}^infty|x_k| = |x|.$$



since the absolute sum of the odd entries of the sequence is definitely smaller or equal to the total sum of absolute values of all entries.



Thus :



$$|Tx| leq |x| implies |T| leq 1$$



Now, if you take the sequence $x = (x_n) in ell^1$ with $x = (1,0,dots,0,dots)$, then :



$$|Tx| = |(1,0,dots,0,dots)| = 1 implies |T| = 1$$



Thus the operator $T$ is continuous.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I would like to have an input from the person that downvoted.
    $endgroup$
    – Rebellos
    Dec 16 '18 at 14:10
















2












$begingroup$

It is :



$$|Tx| = |(x_1,0,x_3,0,dots)| = sum_{k=1}^infty|x_{2k+1}| leq sum_{k=1}^infty|x_k| = |x|.$$



since the absolute sum of the odd entries of the sequence is definitely smaller or equal to the total sum of absolute values of all entries.



Thus :



$$|Tx| leq |x| implies |T| leq 1$$



Now, if you take the sequence $x = (x_n) in ell^1$ with $x = (1,0,dots,0,dots)$, then :



$$|Tx| = |(1,0,dots,0,dots)| = 1 implies |T| = 1$$



Thus the operator $T$ is continuous.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I would like to have an input from the person that downvoted.
    $endgroup$
    – Rebellos
    Dec 16 '18 at 14:10














2












2








2





$begingroup$

It is :



$$|Tx| = |(x_1,0,x_3,0,dots)| = sum_{k=1}^infty|x_{2k+1}| leq sum_{k=1}^infty|x_k| = |x|.$$



since the absolute sum of the odd entries of the sequence is definitely smaller or equal to the total sum of absolute values of all entries.



Thus :



$$|Tx| leq |x| implies |T| leq 1$$



Now, if you take the sequence $x = (x_n) in ell^1$ with $x = (1,0,dots,0,dots)$, then :



$$|Tx| = |(1,0,dots,0,dots)| = 1 implies |T| = 1$$



Thus the operator $T$ is continuous.






share|cite|improve this answer











$endgroup$



It is :



$$|Tx| = |(x_1,0,x_3,0,dots)| = sum_{k=1}^infty|x_{2k+1}| leq sum_{k=1}^infty|x_k| = |x|.$$



since the absolute sum of the odd entries of the sequence is definitely smaller or equal to the total sum of absolute values of all entries.



Thus :



$$|Tx| leq |x| implies |T| leq 1$$



Now, if you take the sequence $x = (x_n) in ell^1$ with $x = (1,0,dots,0,dots)$, then :



$$|Tx| = |(1,0,dots,0,dots)| = 1 implies |T| = 1$$



Thus the operator $T$ is continuous.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 16 '18 at 11:37

























answered Dec 16 '18 at 11:18









RebellosRebellos

14.6k31247




14.6k31247












  • $begingroup$
    I would like to have an input from the person that downvoted.
    $endgroup$
    – Rebellos
    Dec 16 '18 at 14:10


















  • $begingroup$
    I would like to have an input from the person that downvoted.
    $endgroup$
    – Rebellos
    Dec 16 '18 at 14:10
















$begingroup$
I would like to have an input from the person that downvoted.
$endgroup$
– Rebellos
Dec 16 '18 at 14:10




$begingroup$
I would like to have an input from the person that downvoted.
$endgroup$
– Rebellos
Dec 16 '18 at 14:10


















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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3042477%2fshow-that-the-sum-is-the-norm-of-l-1%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

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
Read more

build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

Image
0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
Read more