Prove that ${x^n}$ is Cauchy in $Ssubseteq ell_infty$
Multi tool use
I'm kind of new into Functional analysis. So, I have the following question bothering me. Let $S$ be the set of sequences having only a finite number of non-zero terms. Clearly, $Ssubseteq ell_infty$. Take ${x^n}$ in $S$ where $x^n=(1,frac{1}{2},frac{1}{3},cdots,frac{1}{n},0,0,cdots).$ I want to prove that ${x^n}$ is Cauchy in $S.$
MY TRIAL
Let $mgeq n.$ Then,
begin{align}Vert x^n-x^m Vert_ {ell_infty}=suplimits_{m,nin Bbb{N}}left|frac{1}{m}-frac{1}{n} right|end{align}
By convergence of $frac{1}{n}$ and $frac{1}{m}$ for each $m$ and $n$, there exists $NinBbb{N}$ such that
begin{align}Vert x^n-x^m Vert_ {ell_infty}=suplimits_{m,nin Bbb{N}}left|frac{1}{m}-frac{1}{n} right|=frac{1}{n}-frac{1}{m}<epsilon,;;forall;mgeq ngeq N. end{align}
Please check if I'm right or wrong. If I'm wrong, kindly help out. Thanks
functional-analysis analysis convergence normed-spaces cauchy-sequences
add a comment |
I'm kind of new into Functional analysis. So, I have the following question bothering me. Let $S$ be the set of sequences having only a finite number of non-zero terms. Clearly, $Ssubseteq ell_infty$. Take ${x^n}$ in $S$ where $x^n=(1,frac{1}{2},frac{1}{3},cdots,frac{1}{n},0,0,cdots).$ I want to prove that ${x^n}$ is Cauchy in $S.$
MY TRIAL
Let $mgeq n.$ Then,
begin{align}Vert x^n-x^m Vert_ {ell_infty}=suplimits_{m,nin Bbb{N}}left|frac{1}{m}-frac{1}{n} right|end{align}
By convergence of $frac{1}{n}$ and $frac{1}{m}$ for each $m$ and $n$, there exists $NinBbb{N}$ such that
begin{align}Vert x^n-x^m Vert_ {ell_infty}=suplimits_{m,nin Bbb{N}}left|frac{1}{m}-frac{1}{n} right|=frac{1}{n}-frac{1}{m}<epsilon,;;forall;mgeq ngeq N. end{align}
Please check if I'm right or wrong. If I'm wrong, kindly help out. Thanks
functional-analysis analysis convergence normed-spaces cauchy-sequences
add a comment |
I'm kind of new into Functional analysis. So, I have the following question bothering me. Let $S$ be the set of sequences having only a finite number of non-zero terms. Clearly, $Ssubseteq ell_infty$. Take ${x^n}$ in $S$ where $x^n=(1,frac{1}{2},frac{1}{3},cdots,frac{1}{n},0,0,cdots).$ I want to prove that ${x^n}$ is Cauchy in $S.$
MY TRIAL
Let $mgeq n.$ Then,
begin{align}Vert x^n-x^m Vert_ {ell_infty}=suplimits_{m,nin Bbb{N}}left|frac{1}{m}-frac{1}{n} right|end{align}
By convergence of $frac{1}{n}$ and $frac{1}{m}$ for each $m$ and $n$, there exists $NinBbb{N}$ such that
begin{align}Vert x^n-x^m Vert_ {ell_infty}=suplimits_{m,nin Bbb{N}}left|frac{1}{m}-frac{1}{n} right|=frac{1}{n}-frac{1}{m}<epsilon,;;forall;mgeq ngeq N. end{align}
Please check if I'm right or wrong. If I'm wrong, kindly help out. Thanks
functional-analysis analysis convergence normed-spaces cauchy-sequences
I'm kind of new into Functional analysis. So, I have the following question bothering me. Let $S$ be the set of sequences having only a finite number of non-zero terms. Clearly, $Ssubseteq ell_infty$. Take ${x^n}$ in $S$ where $x^n=(1,frac{1}{2},frac{1}{3},cdots,frac{1}{n},0,0,cdots).$ I want to prove that ${x^n}$ is Cauchy in $S.$
MY TRIAL
Let $mgeq n.$ Then,
begin{align}Vert x^n-x^m Vert_ {ell_infty}=suplimits_{m,nin Bbb{N}}left|frac{1}{m}-frac{1}{n} right|end{align}
By convergence of $frac{1}{n}$ and $frac{1}{m}$ for each $m$ and $n$, there exists $NinBbb{N}$ such that
begin{align}Vert x^n-x^m Vert_ {ell_infty}=suplimits_{m,nin Bbb{N}}left|frac{1}{m}-frac{1}{n} right|=frac{1}{n}-frac{1}{m}<epsilon,;;forall;mgeq ngeq N. end{align}
Please check if I'm right or wrong. If I'm wrong, kindly help out. Thanks
functional-analysis analysis convergence normed-spaces cauchy-sequences
functional-analysis analysis convergence normed-spaces cauchy-sequences
edited Dec 13 '18 at 9:54
José Carlos Santos
152k22123225
152k22123225
asked Dec 13 '18 at 9:52
Omojola MichealOmojola Micheal
1,692324
1,692324
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Actually.$$mgeqslant nimplieslVert x^n-x^mrVert_infty=begin{cases}0&text{ if }m=n\dfrac1{n+1}&text{ otherwise,}end{cases}$$since, if $m>n$,$$lVert x^n-x^mrVert_infty=supleft{0,frac1{n+1},frac1{n+2},ldots,frac1mright}=frac1{n+1}.$$So, given $varepsilon>0$, take $Ninmathbb{N}$ such that $frac1N<varepsilon$ and then$$mgeqslant ngeqslant NimplieslVert x^n-x^mrVert_infty<varepsilon.$$
That was nice! I am grateful! +1!
– Omojola Micheal
Dec 13 '18 at 10:14
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3037822%2fprove-that-xn-is-cauchy-in-s-subseteq-ell-infty%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
Actually.$$mgeqslant nimplieslVert x^n-x^mrVert_infty=begin{cases}0&text{ if }m=n\dfrac1{n+1}&text{ otherwise,}end{cases}$$since, if $m>n$,$$lVert x^n-x^mrVert_infty=supleft{0,frac1{n+1},frac1{n+2},ldots,frac1mright}=frac1{n+1}.$$So, given $varepsilon>0$, take $Ninmathbb{N}$ such that $frac1N<varepsilon$ and then$$mgeqslant ngeqslant NimplieslVert x^n-x^mrVert_infty<varepsilon.$$
That was nice! I am grateful! +1!
– Omojola Micheal
Dec 13 '18 at 10:14
add a comment |
Actually.$$mgeqslant nimplieslVert x^n-x^mrVert_infty=begin{cases}0&text{ if }m=n\dfrac1{n+1}&text{ otherwise,}end{cases}$$since, if $m>n$,$$lVert x^n-x^mrVert_infty=supleft{0,frac1{n+1},frac1{n+2},ldots,frac1mright}=frac1{n+1}.$$So, given $varepsilon>0$, take $Ninmathbb{N}$ such that $frac1N<varepsilon$ and then$$mgeqslant ngeqslant NimplieslVert x^n-x^mrVert_infty<varepsilon.$$
That was nice! I am grateful! +1!
– Omojola Micheal
Dec 13 '18 at 10:14
add a comment |
Actually.$$mgeqslant nimplieslVert x^n-x^mrVert_infty=begin{cases}0&text{ if }m=n\dfrac1{n+1}&text{ otherwise,}end{cases}$$since, if $m>n$,$$lVert x^n-x^mrVert_infty=supleft{0,frac1{n+1},frac1{n+2},ldots,frac1mright}=frac1{n+1}.$$So, given $varepsilon>0$, take $Ninmathbb{N}$ such that $frac1N<varepsilon$ and then$$mgeqslant ngeqslant NimplieslVert x^n-x^mrVert_infty<varepsilon.$$
Actually.$$mgeqslant nimplieslVert x^n-x^mrVert_infty=begin{cases}0&text{ if }m=n\dfrac1{n+1}&text{ otherwise,}end{cases}$$since, if $m>n$,$$lVert x^n-x^mrVert_infty=supleft{0,frac1{n+1},frac1{n+2},ldots,frac1mright}=frac1{n+1}.$$So, given $varepsilon>0$, take $Ninmathbb{N}$ such that $frac1N<varepsilon$ and then$$mgeqslant ngeqslant NimplieslVert x^n-x^mrVert_infty<varepsilon.$$
answered Dec 13 '18 at 9:58
José Carlos SantosJosé Carlos Santos
152k22123225
152k22123225
That was nice! I am grateful! +1!
– Omojola Micheal
Dec 13 '18 at 10:14
add a comment |
That was nice! I am grateful! +1!
– Omojola Micheal
Dec 13 '18 at 10:14
That was nice! I am grateful! +1!
– Omojola Micheal
Dec 13 '18 at 10:14
That was nice! I am grateful! +1!
– Omojola Micheal
Dec 13 '18 at 10:14
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3037822%2fprove-that-xn-is-cauchy-in-s-subseteq-ell-infty%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
v68Wx,sCRJ,U9vcRiYMGLJjdHHf Zw2Lq2txt6f10tVcOZb2wosasRlD,oxIuNCn9LhPeR1kVIgwF Y5NG1t7 V9Juo5obB7,F Fi,j9t9e