Lemma 2.4.1 Introductory Functional Analysis - Kreyszig :Confirmation of my understandingLemma 2.4-1 in Erwin Kreyszig's “Introductory Functional Analysis with Applications”: Is there an easier proof?I need help understanding the proof of Lemma 2.4-1 from Kreyszig's Functional Analysis.Space Of All Compact Operators (Functional Analysis)Let $A , BsubseteqmathbbR$. If $A$ is closed and $B$ is compact, is $Acdot B$ closed?Erwine Kryszeg's INTRODUCTORY FUNCTIONAL ANALYSIS WITH APPLICATIONS, Section 1.5-8My proof of: Given an adherent point, some sequence converges to it.Understanding a proposition in Zeidler's book on functional analysisLinearly independent set of vectors in a normed spaceProve IVT from clopen sets in $mathbbR$.Rudin - Functional analysis excercise 12.7Question from conway's functional analysis bookHelp understanding terms related to functional analysis
At which point does a character regain all their Hit Dice?
apt-get update is failing in debian
Go Pregnant or Go Home
How does residential electricity work?
Is exact Kanji stroke length important?
How does it work when somebody invests in my business?
Was the picture area of a CRT a parallelogram (instead of a true rectangle)?
The baby cries all morning
Student evaluations of teaching assistants
How do I define a right arrow with bar in LaTeX?
Short story about space worker geeks who zone out by 'listening' to radiation from stars
Bash method for viewing beginning and end of file
Is expanding the research of a group into machine learning as a PhD student risky?
What is the intuitive meaning of having a linear relationship between the logs of two variables?
Why "be dealt cards" rather than "be dealing cards"?
Efficiently merge handle parallel feature branches in SFDX
How can I replace every global instance of "x[2]" with "x_2"
Is it okay / does it make sense for another player to join a running game of Munchkin?
Hide Select Output from T-SQL
Applicability of Single Responsibility Principle
What is the oldest known work of fiction?
Is there a good way to store credentials outside of a password manager?
How to be diplomatic in refusing to write code that breaches the privacy of our users
Can somebody explain Brexit in a few child-proof sentences?
Lemma 2.4.1 Introductory Functional Analysis - Kreyszig :Confirmation of my understanding
Lemma 2.4-1 in Erwin Kreyszig's “Introductory Functional Analysis with Applications”: Is there an easier proof?I need help understanding the proof of Lemma 2.4-1 from Kreyszig's Functional Analysis.Space Of All Compact Operators (Functional Analysis)Let $A , BsubseteqmathbbR$. If $A$ is closed and $B$ is compact, is $Acdot B$ closed?Erwine Kryszeg's INTRODUCTORY FUNCTIONAL ANALYSIS WITH APPLICATIONS, Section 1.5-8My proof of: Given an adherent point, some sequence converges to it.Understanding a proposition in Zeidler's book on functional analysisLinearly independent set of vectors in a normed spaceProve IVT from clopen sets in $mathbbR$.Rudin - Functional analysis excercise 12.7Question from conway's functional analysis bookHelp understanding terms related to functional analysis
$begingroup$
I see that questions relating to this lemma have been asked here and here .
But my question is quite different in the sense , that i believe that i have understood what the author intends to say but i need a confirmation if my understanding is correct .
The lemma is : 
So this is my understanding :
To prove the lemma , we would prove that
$$exists c> 0 ,text for which forall alpha_i in R , text we have ||alpha_1x_1 +alpha_2x_2 +alpha_3x_3 +.....+ alpha_nx_n|| geq c(|alpha_1|+|alpha_2|+|alpha_+.....+|alpha_n|) tag1 $$
where the set of vectors $$x_1,x_2,.....,x_n$$ are independent and $x_i in X$ where $X$ is a normed space with some norm.
Now the first idea that comes to me is that because the set of vectors are independent .We know , from linear algebra , that $$alpha_1x_1 +alpha_2x_2 +alpha_3x_3 +.....+ alpha_nx_n = 0 $$ only when all $alpha_i$ are zero , in which case the inequality of the lemma will hold true for all $c$ . But let $alpha_i neq 0$ . In this case the quantity $$alpha_1x_1 +alpha_2x_2 +alpha_3x_3 +.....+ alpha_nx_n tag1$$ will have some value but we don't know what it is but the norm over it would be some positive value .
$textbf(1) $So intuitively , we can think that there exists and infinitesimally small $c neq 0 $ and this proves the lemma. Is this reasoning correct ?
$textbf(2) $Now , separately i had a look at author's explanation . And this is what i think he is saying . After some initial algebraic manipulation , we arrive to prove that : $$ exists c gt 0 text such that ||beta_1x_1+beta_2x_2+beta_3x_3+....+beta_nx_n|| geq c tag2$$ where $$sum_i=1^i=n|beta_i| = 1$$ .
Now to prove $textit equation 2 $this , what the author is trying to do is that he creates a sequence which contains all the possible values for
$$beta_1x_1+beta_2x_2+beta_3x_3+....+beta_nx_n
tag 3 $$ with a norm over it .And then he proves that this sequence converges to a a point , and the value of the point is not $0$ , and hence all elements of the of the sequences are not $0$ and hence the same intution , that followed in $textitPoint (1.) $follows here .
So is this deduction by me correct ? If that is the case , why did not the author just used $textitPoint (1.) $ ?
$textbf(3) $Lastly i want to confirm my understanding of how he created the sequence : Let the sequence be denoted by $(y_m)$ with the general element given by $$y_m= beta_1^(m)x_1+beta_2^(m)x_2+beta_3^(m)x_3+....beta_n^(m)x_n$$ . So that for m =1 ,2 , we have :
$$y_1 =beta_1^(1)x_1+beta_2^(1)x_2+beta_3^(1)x_3+....beta_n^(1)x_n \
y_2 =beta_1^(2)x_1+beta_2^(2)x_2+beta_3^(2)x_3+....beta_n^(2)x_n \
.\
.\
.\
y_m =beta_1^(m)x_1+beta_2^(m)x_2+beta_3^(m)x_3+....beta_n^(m)x_n tag4
$$
Now we know that $sum_i=1^i=n|beta_i^m| =1 $ , this implies that $|beta_i^m| leq 1$ . So we can say that the sequence $$(beta_1^1,beta_1^2,beta_1^3...)$$is bounded and by the BW theorem , there exists its sub sequence which is convergent and let that point of convergence be called $beta_1$ .The author then writes $textitLet (y_1,m) textit be the corresponding sub sequence of y_m$. I want to confirm if the general element of the $(y_1,m)$ looks like : $$y_1,m = beta_1x_1+ beta_2^mx_2+beta_3^mx_3 +.....+ beta_n^mx_n$$
Is that correct ?
$textbf(4) $ And we repeat the process and reach the $textit expression 3 $.
Edit 1 : I wrongly ordered the quantifiers as pointed out by @Matthew Towers. It has been fixed now. Also the language around c has been changed to make "c" look more like a constant.
Edit 2 : Typo around $|.|$ in equation 1 fixed.
real-analysis functional-analysis normed-spaces
asked Mar 17 at 13:51
warrior_monkwarrior_monk
436
$endgroup$
add a comment |
$begingroup$
I see that questions relating to this lemma have been asked here and here .
But my question is quite different in the sense , that i believe that i have understood what the author intends to say but i need a confirmation if my understanding is correct .
The lemma is :
So this is my understanding :
To prove the lemma , we would prove that
$$exists c> 0 ,text for which forall alpha_i in R , text we have ||alpha_1x_1 +alpha_2x_2 +alpha_3x_3 +.....+ alpha_nx_n|| geq c(|alpha_1|+|alpha_2|+|alpha_+.....+|alpha_n|) tag1 $$
where the set of vectors $$x_1,x_2,.....,x_n$$ are independent and $x_i in X$ where $X$ is a normed space with some norm.
Now the first idea that comes to me is that because the set of vectors are independent .We know , from linear algebra , that $$alpha_1x_1 +alpha_2x_2 +alpha_3x_3 +.....+ alpha_nx_n = 0 $$ only when all $alpha_i$ are zero , in which case the inequality of the lemma will hold true for all $c$ . But let $alpha_i neq 0$ . In this case the quantity $$alpha_1x_1 +alpha_2x_2 +alpha_3x_3 +.....+ alpha_nx_n tag1$$ will have some value but we don't know what it is but the norm over it would be some positive value .
$textbf(1) $So intuitively , we can think that there exists and infinitesimally small $c neq 0 $ and this proves the lemma. Is this reasoning correct ?
$textbf(2) $Now , separately i had a look at author's explanation . And this is what i think he is saying . After some initial algebraic manipulation , we arrive to prove that : $$ exists c gt 0 text such that ||beta_1x_1+beta_2x_2+beta_3x_3+....+beta_nx_n|| geq c tag2$$ where $$sum_i=1^i=n|beta_i| = 1$$ .
Now to prove $textit equation 2 $this , what the author is trying to do is that he creates a sequence which contains all the possible values for
$$beta_1x_1+beta_2x_2+beta_3x_3+....+beta_nx_n
tag 3 $$ with a norm over it .And then he proves that this sequence converges to a a point , and the value of the point is not $0$ , and hence all elements of the of the sequences are not $0$ and hence the same intution , that followed in $textitPoint (1.) $follows here .
So is this deduction by me correct ? If that is the case , why did not the author just used $textitPoint (1.) $ ?
$textbf(3) $Lastly i want to confirm my understanding of how he created the sequence : Let the sequence be denoted by $(y_m)$ with the general element given by $$y_m= beta_1^(m)x_1+beta_2^(m)x_2+beta_3^(m)x_3+....beta_n^(m)x_n$$ . So that for m =1 ,2 , we have :
$$y_1 =beta_1^(1)x_1+beta_2^(1)x_2+beta_3^(1)x_3+....beta_n^(1)x_n \
y_2 =beta_1^(2)x_1+beta_2^(2)x_2+beta_3^(2)x_3+....beta_n^(2)x_n \
.\
.\
.\
y_m =beta_1^(m)x_1+beta_2^(m)x_2+beta_3^(m)x_3+....beta_n^(m)x_n tag4
$$
Now we know that $sum_i=1^i=n|beta_i^m| =1 $ , this implies that $|beta_i^m| leq 1$ . So we can say that the sequence $$(beta_1^1,beta_1^2,beta_1^3...)$$is bounded and by the BW theorem , there exists its sub sequence which is convergent and let that point of convergence be called $beta_1$ .The author then writes $textitLet (y_1,m) textit be the corresponding sub sequence of y_m$. I want to confirm if the general element of the $(y_1,m)$ looks like : $$y_1,m = beta_1x_1+ beta_2^mx_2+beta_3^mx_3 +.....+ beta_n^mx_n$$
Is that correct ?
$textbf(4) $ And we repeat the process and reach the $textit expression 3 $.
Edit 1 : I wrongly ordered the quantifiers as pointed out by @Matthew Towers. It has been fixed now. Also the language around c has been changed to make "c" look more like a constant.
Edit 2 : Typo around $|.|$ in equation 1 fixed.
real-analysis functional-analysis normed-spaces
asked Mar 17 at 13:51
warrior_monkwarrior_monk
436
$endgroup$
$begingroup$
I fixed my mistake regarding ordering of quantifiers and the confusing statement i wrote along the constant c . What about my points 1.) ,2.) , 3.) and 4.) . ? Can you explain why they are wrong ?
$endgroup$
– warrior_monk
Mar 17 at 22:55
$begingroup$
Equation (1) still isn't right: it should be $|alpha_1| + |alpha_2|+cdots$ not $|alpha_1 + alpha_2+cdots$. Point (1): no, that's not what's happening. (2) isn't the right idea. Kreyszig proceeds by contradiction: if no such $c$ existed then (for example) for each $epsilon >0$ there is a $y$ in that form with $||y || < epsilon$, so you can find a sequence of $y_m$ with $||y_m|| to 0$.
$endgroup$
– Matthew Towers
Mar 18 at 7:44
add a comment |
$begingroup$
I see that questions relating to this lemma have been asked here and here .
But my question is quite different in the sense , that i believe that i have understood what the author intends to say but i need a confirmation if my understanding is correct .
The lemma is :
So this is my understanding :
To prove the lemma , we would prove that
$$exists c> 0 ,text for which forall alpha_i in R , text we have ||alpha_1x_1 +alpha_2x_2 +alpha_3x_3 +.....+ alpha_nx_n|| geq c(|alpha_1|+|alpha_2|+|alpha_+.....+|alpha_n|) tag1 $$
where the set of vectors $$x_1,x_2,.....,x_n$$ are independent and $x_i in X$ where $X$ is a normed space with some norm.
Now the first idea that comes to me is that because the set of vectors are independent .We know , from linear algebra , that $$alpha_1x_1 +alpha_2x_2 +alpha_3x_3 +.....+ alpha_nx_n = 0 $$ only when all $alpha_i$ are zero , in which case the inequality of the lemma will hold true for all $c$ . But let $alpha_i neq 0$ . In this case the quantity $$alpha_1x_1 +alpha_2x_2 +alpha_3x_3 +.....+ alpha_nx_n tag1$$ will have some value but we don't know what it is but the norm over it would be some positive value .
$textbf(1) $So intuitively , we can think that there exists and infinitesimally small $c neq 0 $ and this proves the lemma. Is this reasoning correct ?
$textbf(2) $Now , separately i had a look at author's explanation . And this is what i think he is saying . After some initial algebraic manipulation , we arrive to prove that : $$ exists c gt 0 text such that ||beta_1x_1+beta_2x_2+beta_3x_3+....+beta_nx_n|| geq c tag2$$ where $$sum_i=1^i=n|beta_i| = 1$$ .
Now to prove $textit equation 2 $this , what the author is trying to do is that he creates a sequence which contains all the possible values for
$$beta_1x_1+beta_2x_2+beta_3x_3+....+beta_nx_n
tag 3 $$ with a norm over it .And then he proves that this sequence converges to a a point , and the value of the point is not $0$ , and hence all elements of the of the sequences are not $0$ and hence the same intution , that followed in $textitPoint (1.) $follows here .
So is this deduction by me correct ? If that is the case , why did not the author just used $textitPoint (1.) $ ?
$textbf(3) $Lastly i want to confirm my understanding of how he created the sequence : Let the sequence be denoted by $(y_m)$ with the general element given by $$y_m= beta_1^(m)x_1+beta_2^(m)x_2+beta_3^(m)x_3+....beta_n^(m)x_n$$ . So that for m =1 ,2 , we have :
$$y_1 =beta_1^(1)x_1+beta_2^(1)x_2+beta_3^(1)x_3+....beta_n^(1)x_n \
y_2 =beta_1^(2)x_1+beta_2^(2)x_2+beta_3^(2)x_3+....beta_n^(2)x_n \
.\
.\
.\
y_m =beta_1^(m)x_1+beta_2^(m)x_2+beta_3^(m)x_3+....beta_n^(m)x_n tag4
$$
Now we know that $sum_i=1^i=n|beta_i^m| =1 $ , this implies that $|beta_i^m| leq 1$ . So we can say that the sequence $$(beta_1^1,beta_1^2,beta_1^3...)$$is bounded and by the BW theorem , there exists its sub sequence which is convergent and let that point of convergence be called $beta_1$ .The author then writes $textitLet (y_1,m) textit be the corresponding sub sequence of y_m$. I want to confirm if the general element of the $(y_1,m)$ looks like : $$y_1,m = beta_1x_1+ beta_2^mx_2+beta_3^mx_3 +.....+ beta_n^mx_n$$
Is that correct ?
$textbf(4) $ And we repeat the process and reach the $textit expression 3 $.
Edit 1 : I wrongly ordered the quantifiers as pointed out by @Matthew Towers. It has been fixed now. Also the language around c has been changed to make "c" look more like a constant.
Edit 2 : Typo around $|.|$ in equation 1 fixed.
real-analysis functional-analysis normed-spaces
asked Mar 17 at 13:51
warrior_monkwarrior_monk
436
$endgroup$
I see that questions relating to this lemma have been asked here and here .
But my question is quite different in the sense , that i believe that i have understood what the author intends to say but i need a confirmation if my understanding is correct .
The lemma is :
So this is my understanding :
To prove the lemma , we would prove that
$$exists c> 0 ,text for which forall alpha_i in R , text we have ||alpha_1x_1 +alpha_2x_2 +alpha_3x_3 +.....+ alpha_nx_n|| geq c(|alpha_1|+|alpha_2|+|alpha_+.....+|alpha_n|) tag1 $$
where the set of vectors $$x_1,x_2,.....,x_n$$ are independent and $x_i in X$ where $X$ is a normed space with some norm.
Now the first idea that comes to me is that because the set of vectors are independent .We know , from linear algebra , that $$alpha_1x_1 +alpha_2x_2 +alpha_3x_3 +.....+ alpha_nx_n = 0 $$ only when all $alpha_i$ are zero , in which case the inequality of the lemma will hold true for all $c$ . But let $alpha_i neq 0$ . In this case the quantity $$alpha_1x_1 +alpha_2x_2 +alpha_3x_3 +.....+ alpha_nx_n tag1$$ will have some value but we don't know what it is but the norm over it would be some positive value .
$textbf(1) $So intuitively , we can think that there exists and infinitesimally small $c neq 0 $ and this proves the lemma. Is this reasoning correct ?
$textbf(2) $Now , separately i had a look at author's explanation . And this is what i think he is saying . After some initial algebraic manipulation , we arrive to prove that : $$ exists c gt 0 text such that ||beta_1x_1+beta_2x_2+beta_3x_3+....+beta_nx_n|| geq c tag2$$ where $$sum_i=1^i=n|beta_i| = 1$$ .
Now to prove $textit equation 2 $this , what the author is trying to do is that he creates a sequence which contains all the possible values for
$$beta_1x_1+beta_2x_2+beta_3x_3+....+beta_nx_n
tag 3 $$ with a norm over it .And then he proves that this sequence converges to a a point , and the value of the point is not $0$ , and hence all elements of the of the sequences are not $0$ and hence the same intution , that followed in $textitPoint (1.) $follows here .
So is this deduction by me correct ? If that is the case , why did not the author just used $textitPoint (1.) $ ?
$textbf(3) $Lastly i want to confirm my understanding of how he created the sequence : Let the sequence be denoted by $(y_m)$ with the general element given by $$y_m= beta_1^(m)x_1+beta_2^(m)x_2+beta_3^(m)x_3+....beta_n^(m)x_n$$ . So that for m =1 ,2 , we have :
$$y_1 =beta_1^(1)x_1+beta_2^(1)x_2+beta_3^(1)x_3+....beta_n^(1)x_n \
y_2 =beta_1^(2)x_1+beta_2^(2)x_2+beta_3^(2)x_3+....beta_n^(2)x_n \
.\
.\
.\
y_m =beta_1^(m)x_1+beta_2^(m)x_2+beta_3^(m)x_3+....beta_n^(m)x_n tag4
$$
Now we know that $sum_i=1^i=n|beta_i^m| =1 $ , this implies that $|beta_i^m| leq 1$ . So we can say that the sequence $$(beta_1^1,beta_1^2,beta_1^3...)$$is bounded and by the BW theorem , there exists its sub sequence which is convergent and let that point of convergence be called $beta_1$ .The author then writes $textitLet (y_1,m) textit be the corresponding sub sequence of y_m$. I want to confirm if the general element of the $(y_1,m)$ looks like : $$y_1,m = beta_1x_1+ beta_2^mx_2+beta_3^mx_3 +.....+ beta_n^mx_n$$
Is that correct ?
$textbf(4) $ And we repeat the process and reach the $textit expression 3 $.
Edit 1 : I wrongly ordered the quantifiers as pointed out by @Matthew Towers. It has been fixed now. Also the language around c has been changed to make "c" look more like a constant.
Edit 2 : Typo around $|.|$ in equation 1 fixed.
real-analysis functional-analysis normed-spaces
real-analysis functional-analysis normed-spaces
asked Mar 17 at 13:51
warrior_monkwarrior_monk
436
asked Mar 17 at 13:51
warrior_monkwarrior_monk
436
edited Mar 19 at 0:47
warrior_monk
asked Mar 17 at 13:51
warrior_monkwarrior_monk
436
asked Mar 17 at 13:51
warrior_monkwarrior_monk
436
asked Mar 17 at 13:51
warrior_monkwarrior_monk
436
436
$begingroup$
I fixed my mistake regarding ordering of quantifiers and the confusing statement i wrote along the constant c . What about my points 1.) ,2.) , 3.) and 4.) . ? Can you explain why they are wrong ?
$endgroup$
– warrior_monk
Mar 17 at 22:55
$begingroup$
Equation (1) still isn't right: it should be $|alpha_1| + |alpha_2|+cdots$ not $|alpha_1 + alpha_2+cdots$. Point (1): no, that's not what's happening. (2) isn't the right idea. Kreyszig proceeds by contradiction: if no such $c$ existed then (for example) for each $epsilon >0$ there is a $y$ in that form with $||y || < epsilon$, so you can find a sequence of $y_m$ with $||y_m|| to 0$.
$endgroup$
– Matthew Towers
Mar 18 at 7:44
add a comment |
$begingroup$
I fixed my mistake regarding ordering of quantifiers and the confusing statement i wrote along the constant c . What about my points 1.) ,2.) , 3.) and 4.) . ? Can you explain why they are wrong ?
$endgroup$
– warrior_monk
Mar 17 at 22:55
$begingroup$
Equation (1) still isn't right: it should be $|alpha_1| + |alpha_2|+cdots$ not $|alpha_1 + alpha_2+cdots$. Point (1): no, that's not what's happening. (2) isn't the right idea. Kreyszig proceeds by contradiction: if no such $c$ existed then (for example) for each $epsilon >0$ there is a $y$ in that form with $||y || < epsilon$, so you can find a sequence of $y_m$ with $||y_m|| to 0$.
$endgroup$
– Matthew Towers
Mar 18 at 7:44
$begingroup$
I fixed my mistake regarding ordering of quantifiers and the confusing statement i wrote along the constant c . What about my points 1.) ,2.) , 3.) and 4.) . ? Can you explain why they are wrong ?
$endgroup$
– warrior_monk
Mar 17 at 22:55
$begingroup$
I fixed my mistake regarding ordering of quantifiers and the confusing statement i wrote along the constant c . What about my points 1.) ,2.) , 3.) and 4.) . ? Can you explain why they are wrong ?
$endgroup$
– warrior_monk
Mar 17 at 22:55
$begingroup$
Equation (1) still isn't right: it should be $|alpha_1| + |alpha_2|+cdots$ not $|alpha_1 + alpha_2+cdots$. Point (1): no, that's not what's happening. (2) isn't the right idea. Kreyszig proceeds by contradiction: if no such $c$ existed then (for example) for each $epsilon >0$ there is a $y$ in that form with $||y || < epsilon$, so you can find a sequence of $y_m$ with $||y_m|| to 0$.
$endgroup$
– Matthew Towers
Mar 18 at 7:44
$begingroup$
Equation (1) still isn't right: it should be $|alpha_1| + |alpha_2|+cdots$ not $|alpha_1 + alpha_2+cdots$. Point (1): no, that's not what's happening. (2) isn't the right idea. Kreyszig proceeds by contradiction: if no such $c$ existed then (for example) for each $epsilon >0$ there is a $y$ in that form with $||y || < epsilon$, so you can find a sequence of $y_m$ with $||y_m|| to 0$.
$endgroup$
– Matthew Towers
Mar 18 at 7:44
add a comment |
0
active
oldest
votes
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%2f3151565%2flemma-2-4-1-introductory-functional-analysis-kreyszig-confirmation-of-my-unde%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3151565%2flemma-2-4-1-introductory-functional-analysis-kreyszig-confirmation-of-my-unde%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
$begingroup$
I fixed my mistake regarding ordering of quantifiers and the confusing statement i wrote along the constant c . What about my points 1.) ,2.) , 3.) and 4.) . ? Can you explain why they are wrong ?
$endgroup$
– warrior_monk
Mar 17 at 22:55
$begingroup$
Equation (1) still isn't right: it should be $|alpha_1| + |alpha_2|+cdots$ not $|alpha_1 + alpha_2+cdots$. Point (1): no, that's not what's happening. (2) isn't the right idea. Kreyszig proceeds by contradiction: if no such $c$ existed then (for example) for each $epsilon >0$ there is a $y$ in that form with $||y || < epsilon$, so you can find a sequence of $y_m$ with $||y_m|| to 0$.
$endgroup$
– Matthew Towers
Mar 18 at 7:44