Given an averaging operator $A: l_p rightarrow l_p$, why $A$ is not compact












2












$begingroup$


Let $A$ be an operator $A: l_p rightarrow l_p , 1 <p<infty$
$$A(x_1, ..., x_n, ...)=left(x_1, frac{x_1+x_2}{2}, ..., frac{x_1+...+x_n}{n}, ...right)$$
I want to show that operator $A$ is not compact. I want to prove it using one of the equivalent definitions of operator compactness: I want to show that the image of unit ball $A(B_1)=left{ left(x_1, frac{x_1+x_2}{2}, ..., frac{x_1+...+x_n}{n}right) : ||x_n||_{l_p}le 1 right} $ is not relatively compact.



I have a criterion for that: I know that the subset $K subset l_p $ is relatively compact iff $K$ is bounded and $lim_{Nrightarrowinfty}sup_ {x in K}sum_{n=N}^{infty}|x_n|^p=0$ so I want to show that my $A(B_1)$ doesn't satisfy this criterion.



I would be grateful for any help!






functional-analysis operator-theory compactness lp-spaces compact-operators







share|cite|improve this question











$endgroup$












  • $begingroup$
    What have you tried?
    $endgroup$
    – MisterRiemann
    Dec 9 '18 at 17:11










  • $begingroup$
    I tried to take such elements from unit ball $(1, 1, ..., 1, 0, ..., 0, ...)$-only finite number of nonzero coordinates and to look at the image of such elements after action of operator $A$
    $endgroup$
    – Anton Zagrivin
    Dec 9 '18 at 17:30










  • $begingroup$
    I though I could have $lim sup...ne 0$(from my oppost criterion), but I couldn’t find this limit
    $endgroup$
    – Anton Zagrivin
    Dec 9 '18 at 17:32










  • $begingroup$
    Maybe my entire idea is wrong and it will not work here
    $endgroup$
    – Anton Zagrivin
    Dec 9 '18 at 17:32










  • $begingroup$
    A useful keyword for request is "Cesaro" or "Cesaro mean". Related : math.stackexchange.com/q/1313738
    $endgroup$
    – Jean Marie
    Dec 9 '18 at 17:56
















2












$begingroup$


Let $A$ be an operator $A: l_p rightarrow l_p , 1 <p<infty$
$$A(x_1, ..., x_n, ...)=left(x_1, frac{x_1+x_2}{2}, ..., frac{x_1+...+x_n}{n}, ...right)$$
I want to show that operator $A$ is not compact. I want to prove it using one of the equivalent definitions of operator compactness: I want to show that the image of unit ball $A(B_1)=left{ left(x_1, frac{x_1+x_2}{2}, ..., frac{x_1+...+x_n}{n}right) : ||x_n||_{l_p}le 1 right} $ is not relatively compact.



I have a criterion for that: I know that the subset $K subset l_p $ is relatively compact iff $K$ is bounded and $lim_{Nrightarrowinfty}sup_ {x in K}sum_{n=N}^{infty}|x_n|^p=0$ so I want to show that my $A(B_1)$ doesn't satisfy this criterion.



I would be grateful for any help!






functional-analysis operator-theory compactness lp-spaces compact-operators







share|cite|improve this question











$endgroup$












  • $begingroup$
    What have you tried?
    $endgroup$
    – MisterRiemann
    Dec 9 '18 at 17:11










  • $begingroup$
    I tried to take such elements from unit ball $(1, 1, ..., 1, 0, ..., 0, ...)$-only finite number of nonzero coordinates and to look at the image of such elements after action of operator $A$
    $endgroup$
    – Anton Zagrivin
    Dec 9 '18 at 17:30










  • $begingroup$
    I though I could have $lim sup...ne 0$(from my oppost criterion), but I couldn’t find this limit
    $endgroup$
    – Anton Zagrivin
    Dec 9 '18 at 17:32










  • $begingroup$
    Maybe my entire idea is wrong and it will not work here
    $endgroup$
    – Anton Zagrivin
    Dec 9 '18 at 17:32










  • $begingroup$
    A useful keyword for request is "Cesaro" or "Cesaro mean". Related : math.stackexchange.com/q/1313738
    $endgroup$
    – Jean Marie
    Dec 9 '18 at 17:56














2












2








2


1



$begingroup$


Let $A$ be an operator $A: l_p rightarrow l_p , 1 <p<infty$
$$A(x_1, ..., x_n, ...)=left(x_1, frac{x_1+x_2}{2}, ..., frac{x_1+...+x_n}{n}, ...right)$$
I want to show that operator $A$ is not compact. I want to prove it using one of the equivalent definitions of operator compactness: I want to show that the image of unit ball $A(B_1)=left{ left(x_1, frac{x_1+x_2}{2}, ..., frac{x_1+...+x_n}{n}right) : ||x_n||_{l_p}le 1 right} $ is not relatively compact.



I have a criterion for that: I know that the subset $K subset l_p $ is relatively compact iff $K$ is bounded and $lim_{Nrightarrowinfty}sup_ {x in K}sum_{n=N}^{infty}|x_n|^p=0$ so I want to show that my $A(B_1)$ doesn't satisfy this criterion.



I would be grateful for any help!






functional-analysis operator-theory compactness lp-spaces compact-operators







share|cite|improve this question











$endgroup$




Let $A$ be an operator $A: l_p rightarrow l_p , 1 <p<infty$
$$A(x_1, ..., x_n, ...)=left(x_1, frac{x_1+x_2}{2}, ..., frac{x_1+...+x_n}{n}, ...right)$$
I want to show that operator $A$ is not compact. I want to prove it using one of the equivalent definitions of operator compactness: I want to show that the image of unit ball $A(B_1)=left{ left(x_1, frac{x_1+x_2}{2}, ..., frac{x_1+...+x_n}{n}right) : ||x_n||_{l_p}le 1 right} $ is not relatively compact.



I have a criterion for that: I know that the subset $K subset l_p $ is relatively compact iff $K$ is bounded and $lim_{Nrightarrowinfty}sup_ {x in K}sum_{n=N}^{infty}|x_n|^p=0$ so I want to show that my $A(B_1)$ doesn't satisfy this criterion.



I would be grateful for any help!






functional-analysis operator-theory compactness lp-spaces compact-operators




functional-analysis operator-theory compactness lp-spaces compact-operators






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 9:57









Davide Giraudo

126k16150261




126k16150261










asked Dec 9 '18 at 17:10









Anton ZagrivinAnton Zagrivin

1648




1648












  • $begingroup$
    What have you tried?
    $endgroup$
    – MisterRiemann
    Dec 9 '18 at 17:11










  • $begingroup$
    I tried to take such elements from unit ball $(1, 1, ..., 1, 0, ..., 0, ...)$-only finite number of nonzero coordinates and to look at the image of such elements after action of operator $A$
    $endgroup$
    – Anton Zagrivin
    Dec 9 '18 at 17:30










  • $begingroup$
    I though I could have $lim sup...ne 0$(from my oppost criterion), but I couldn’t find this limit
    $endgroup$
    – Anton Zagrivin
    Dec 9 '18 at 17:32










  • $begingroup$
    Maybe my entire idea is wrong and it will not work here
    $endgroup$
    – Anton Zagrivin
    Dec 9 '18 at 17:32










  • $begingroup$
    A useful keyword for request is "Cesaro" or "Cesaro mean". Related : math.stackexchange.com/q/1313738
    $endgroup$
    – Jean Marie
    Dec 9 '18 at 17:56


















  • $begingroup$
    What have you tried?
    $endgroup$
    – MisterRiemann
    Dec 9 '18 at 17:11










  • $begingroup$
    I tried to take such elements from unit ball $(1, 1, ..., 1, 0, ..., 0, ...)$-only finite number of nonzero coordinates and to look at the image of such elements after action of operator $A$
    $endgroup$
    – Anton Zagrivin
    Dec 9 '18 at 17:30










  • $begingroup$
    I though I could have $lim sup...ne 0$(from my oppost criterion), but I couldn’t find this limit
    $endgroup$
    – Anton Zagrivin
    Dec 9 '18 at 17:32










  • $begingroup$
    Maybe my entire idea is wrong and it will not work here
    $endgroup$
    – Anton Zagrivin
    Dec 9 '18 at 17:32










  • $begingroup$
    A useful keyword for request is "Cesaro" or "Cesaro mean". Related : math.stackexchange.com/q/1313738
    $endgroup$
    – Jean Marie
    Dec 9 '18 at 17:56
















$begingroup$
What have you tried?
$endgroup$
– MisterRiemann
Dec 9 '18 at 17:11




$begingroup$
What have you tried?
$endgroup$
– MisterRiemann
Dec 9 '18 at 17:11












$begingroup$
I tried to take such elements from unit ball $(1, 1, ..., 1, 0, ..., 0, ...)$-only finite number of nonzero coordinates and to look at the image of such elements after action of operator $A$
$endgroup$
– Anton Zagrivin
Dec 9 '18 at 17:30




$begingroup$
I tried to take such elements from unit ball $(1, 1, ..., 1, 0, ..., 0, ...)$-only finite number of nonzero coordinates and to look at the image of such elements after action of operator $A$
$endgroup$
– Anton Zagrivin
Dec 9 '18 at 17:30












$begingroup$
I though I could have $lim sup...ne 0$(from my oppost criterion), but I couldn’t find this limit
$endgroup$
– Anton Zagrivin
Dec 9 '18 at 17:32




$begingroup$
I though I could have $lim sup...ne 0$(from my oppost criterion), but I couldn’t find this limit
$endgroup$
– Anton Zagrivin
Dec 9 '18 at 17:32












$begingroup$
Maybe my entire idea is wrong and it will not work here
$endgroup$
– Anton Zagrivin
Dec 9 '18 at 17:32




$begingroup$
Maybe my entire idea is wrong and it will not work here
$endgroup$
– Anton Zagrivin
Dec 9 '18 at 17:32












$begingroup$
A useful keyword for request is "Cesaro" or "Cesaro mean". Related : math.stackexchange.com/q/1313738
$endgroup$
– Jean Marie
Dec 9 '18 at 17:56




$begingroup$
A useful keyword for request is "Cesaro" or "Cesaro mean". Related : math.stackexchange.com/q/1313738
$endgroup$
– Jean Marie
Dec 9 '18 at 17:56










1 Answer
1






active

oldest

votes


















1












$begingroup$

Let $N$ be a fixed integer and let $v=v^{(N)}$ be the vector defined by $v_i=2^{-(N+1)/p}$ for $1leqslant ileqslant 2^{N+1}$ and $0$ otherwise. Then $v$ belongs to the unit ball. Moreover, for $2^{N}+1leqslant nleqslant 2^{N+1}$, the $n$-th coordinate of $Av$, denoted $(Av)(n)$, satisfies
$$
(Av)(n)=2^{-(N+1)/p}frac 1ncdot n=2^{-(N+1)/p}
$$
hence
$$
sum_{n=2^N+1}^{2^{N+1}}leftlvert (Av)(n)rightrvert^p=2^Nleft(2^{-(N+1)/p}right)^p=2^{-1}.
$$
This proves, by the mentioned compactness criterion, that the set $left{Av^{(N)},Ngeqslant 1right}$ is not relatively compact in $ell^p$ hence that $A$ is not a compact operator.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Oh, that's what I was trying to do but couldn't achieve
    $endgroup$
    – Anton Zagrivin
    Dec 11 '18 at 7:16










  • $begingroup$
    Thank you a lot
    $endgroup$
    – Anton Zagrivin
    Dec 11 '18 at 7:16










  • $begingroup$
    You are welcome
    $endgroup$
    – Davide Giraudo
    Dec 11 '18 at 9:22











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%2f3032619%2fgiven-an-averaging-operator-a-l-p-rightarrow-l-p-why-a-is-not-compact%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









1












$begingroup$

Let $N$ be a fixed integer and let $v=v^{(N)}$ be the vector defined by $v_i=2^{-(N+1)/p}$ for $1leqslant ileqslant 2^{N+1}$ and $0$ otherwise. Then $v$ belongs to the unit ball. Moreover, for $2^{N}+1leqslant nleqslant 2^{N+1}$, the $n$-th coordinate of $Av$, denoted $(Av)(n)$, satisfies
$$
(Av)(n)=2^{-(N+1)/p}frac 1ncdot n=2^{-(N+1)/p}
$$
hence
$$
sum_{n=2^N+1}^{2^{N+1}}leftlvert (Av)(n)rightrvert^p=2^Nleft(2^{-(N+1)/p}right)^p=2^{-1}.
$$
This proves, by the mentioned compactness criterion, that the set $left{Av^{(N)},Ngeqslant 1right}$ is not relatively compact in $ell^p$ hence that $A$ is not a compact operator.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Oh, that's what I was trying to do but couldn't achieve
    $endgroup$
    – Anton Zagrivin
    Dec 11 '18 at 7:16










  • $begingroup$
    Thank you a lot
    $endgroup$
    – Anton Zagrivin
    Dec 11 '18 at 7:16










  • $begingroup$
    You are welcome
    $endgroup$
    – Davide Giraudo
    Dec 11 '18 at 9:22
















1












$begingroup$

Let $N$ be a fixed integer and let $v=v^{(N)}$ be the vector defined by $v_i=2^{-(N+1)/p}$ for $1leqslant ileqslant 2^{N+1}$ and $0$ otherwise. Then $v$ belongs to the unit ball. Moreover, for $2^{N}+1leqslant nleqslant 2^{N+1}$, the $n$-th coordinate of $Av$, denoted $(Av)(n)$, satisfies
$$
(Av)(n)=2^{-(N+1)/p}frac 1ncdot n=2^{-(N+1)/p}
$$
hence
$$
sum_{n=2^N+1}^{2^{N+1}}leftlvert (Av)(n)rightrvert^p=2^Nleft(2^{-(N+1)/p}right)^p=2^{-1}.
$$
This proves, by the mentioned compactness criterion, that the set $left{Av^{(N)},Ngeqslant 1right}$ is not relatively compact in $ell^p$ hence that $A$ is not a compact operator.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Oh, that's what I was trying to do but couldn't achieve
    $endgroup$
    – Anton Zagrivin
    Dec 11 '18 at 7:16










  • $begingroup$
    Thank you a lot
    $endgroup$
    – Anton Zagrivin
    Dec 11 '18 at 7:16










  • $begingroup$
    You are welcome
    $endgroup$
    – Davide Giraudo
    Dec 11 '18 at 9:22














1












1








1





$begingroup$

Let $N$ be a fixed integer and let $v=v^{(N)}$ be the vector defined by $v_i=2^{-(N+1)/p}$ for $1leqslant ileqslant 2^{N+1}$ and $0$ otherwise. Then $v$ belongs to the unit ball. Moreover, for $2^{N}+1leqslant nleqslant 2^{N+1}$, the $n$-th coordinate of $Av$, denoted $(Av)(n)$, satisfies
$$
(Av)(n)=2^{-(N+1)/p}frac 1ncdot n=2^{-(N+1)/p}
$$
hence
$$
sum_{n=2^N+1}^{2^{N+1}}leftlvert (Av)(n)rightrvert^p=2^Nleft(2^{-(N+1)/p}right)^p=2^{-1}.
$$
This proves, by the mentioned compactness criterion, that the set $left{Av^{(N)},Ngeqslant 1right}$ is not relatively compact in $ell^p$ hence that $A$ is not a compact operator.






share|cite|improve this answer











$endgroup$



Let $N$ be a fixed integer and let $v=v^{(N)}$ be the vector defined by $v_i=2^{-(N+1)/p}$ for $1leqslant ileqslant 2^{N+1}$ and $0$ otherwise. Then $v$ belongs to the unit ball. Moreover, for $2^{N}+1leqslant nleqslant 2^{N+1}$, the $n$-th coordinate of $Av$, denoted $(Av)(n)$, satisfies
$$
(Av)(n)=2^{-(N+1)/p}frac 1ncdot n=2^{-(N+1)/p}
$$
hence
$$
sum_{n=2^N+1}^{2^{N+1}}leftlvert (Av)(n)rightrvert^p=2^Nleft(2^{-(N+1)/p}right)^p=2^{-1}.
$$
This proves, by the mentioned compactness criterion, that the set $left{Av^{(N)},Ngeqslant 1right}$ is not relatively compact in $ell^p$ hence that $A$ is not a compact operator.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 10 '18 at 9:57

























answered Dec 9 '18 at 22:28









Davide GiraudoDavide Giraudo

126k16150261




126k16150261












  • $begingroup$
    Oh, that's what I was trying to do but couldn't achieve
    $endgroup$
    – Anton Zagrivin
    Dec 11 '18 at 7:16










  • $begingroup$
    Thank you a lot
    $endgroup$
    – Anton Zagrivin
    Dec 11 '18 at 7:16










  • $begingroup$
    You are welcome
    $endgroup$
    – Davide Giraudo
    Dec 11 '18 at 9:22


















  • $begingroup$
    Oh, that's what I was trying to do but couldn't achieve
    $endgroup$
    – Anton Zagrivin
    Dec 11 '18 at 7:16










  • $begingroup$
    Thank you a lot
    $endgroup$
    – Anton Zagrivin
    Dec 11 '18 at 7:16










  • $begingroup$
    You are welcome
    $endgroup$
    – Davide Giraudo
    Dec 11 '18 at 9:22
















$begingroup$
Oh, that's what I was trying to do but couldn't achieve
$endgroup$
– Anton Zagrivin
Dec 11 '18 at 7:16




$begingroup$
Oh, that's what I was trying to do but couldn't achieve
$endgroup$
– Anton Zagrivin
Dec 11 '18 at 7:16












$begingroup$
Thank you a lot
$endgroup$
– Anton Zagrivin
Dec 11 '18 at 7:16




$begingroup$
Thank you a lot
$endgroup$
– Anton Zagrivin
Dec 11 '18 at 7:16












$begingroup$
You are welcome
$endgroup$
– Davide Giraudo
Dec 11 '18 at 9:22




$begingroup$
You are welcome
$endgroup$
– Davide Giraudo
Dec 11 '18 at 9:22


















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%2f3032619%2fgiven-an-averaging-operator-a-l-p-rightarrow-l-p-why-a-is-not-compact%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

Berounka

Image
Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc
Read more

Fiat S.p.A.

Image
Cet article court présente un sujet plus développé dans : Fiat et Fiat Chrysler Automobiles. Fiat S.p.A. Fiat S.p.A. ( Fabbrica Italiana Automobili Torino , acronyme FIAT) est un groupe, notamment connu comme société-mère du constructeur automobile italien, Fiat. Portail des entreprises Portail de Turin Portail de l’automobile This page is only for reference, If you need detailed information, please check here
Read more

Type 'String' is not a subtype of type 'int' of 'index'

Image
up vote 0 down vote favorite My App can authenticate successfully when debugging against the Android emulator, but if I try to authenticate using debugging against the physical device (with same OS version), an error message appears after more than one minute waiting: Exception has occurred. _TypeError (type 'String' is not a subtype of type 'int' of 'index') The error message points to the following code: if (responseJson[AuthUtils.authTokenKey] != null) { AuthUtils.insertDetails( userNameController.text, passwordController.text, responseJson); ... } else { ... }; And in the DEBUG CONSOLE I get the following: I/flutter ( 9531): Auth: SocketException: OS Error: Connection timed out, errno = 110, address = example.com, port = 3897
Read more