A question on the completeness of $(C[a,b], sup)$ where $[a,b] subset mathbb{R}$
$begingroup$
I do this proof in 3 steps
First I pick a Cauchy sequence of $f_n(x) in C[a,b]$ and then fix an $epsilon >0$ and some $x in [a,b]$ then constitute $f(x)$ s.t $lim_{n to infty} f_n(x)=f(x)$ in absolute value.
then I try to prove that this limit $f(x)$ is continuous and $lim_{n to infty} f_n(x)=f(x)$ in sup-norm.
Which is a very long proof.
The second proof is.
Let ${f_n}$ be a Cauchy sequence in $C[a,b]$
then for all $epsilon >0$ there exists an $N(epsilon)$ s.t. for all $n,m>N(epsilon)$ we have $| f_m -f_n |_{infty} < epsilon$ which implies $| f_m(x) - f_n(x) | <epsilon$ for all $x in [a,b]$ then ${f_n(x)}$ is uniformly cauchy on $[a,b]$. Then by the following Theorem.
Theorem. A sequence ${f_n}$ of functions $f_n : A tomathbb R$ converges uniformly on
$A$ if and only if it is uniformly Cauchy on $A$.
Then we deduce that the $f_n to f$ uniformly on $[a,b]$
But given that $f_n$ are continuous functions then by uniform convergence we know that $f$ is continuous on $[a,b]$
then by definition of uniform convergence we have that for all $x in [a,b]$, for all $epsilon >0$ $exists N(epsilon )$ st $forall n>N(epsilon )$ $| f_n(x) - f(x) | < epsilon$ $implies$ $| f_n - f |_{infty} < epsilon$
Does the second proof suffice?
real-analysis functional-analysis proof-verification proof-writing banach-spaces
edited Jan 2 at 17:24
Davide Giraudo
127k16151265
asked Jan 2 at 12:42
Dreamer123Dreamer123
32729
$endgroup$
add a comment |
$begingroup$
I do this proof in 3 steps
First I pick a Cauchy sequence of $f_n(x) in C[a,b]$ and then fix an $epsilon >0$ and some $x in [a,b]$ then constitute $f(x)$ s.t $lim_{n to infty} f_n(x)=f(x)$ in absolute value.
then I try to prove that this limit $f(x)$ is continuous and $lim_{n to infty} f_n(x)=f(x)$ in sup-norm.
Which is a very long proof.
The second proof is.
Let ${f_n}$ be a Cauchy sequence in $C[a,b]$
then for all $epsilon >0$ there exists an $N(epsilon)$ s.t. for all $n,m>N(epsilon)$ we have $| f_m -f_n |_{infty} < epsilon$ which implies $| f_m(x) - f_n(x) | <epsilon$ for all $x in [a,b]$ then ${f_n(x)}$ is uniformly cauchy on $[a,b]$. Then by the following Theorem.
Theorem. A sequence ${f_n}$ of functions $f_n : A tomathbb R$ converges uniformly on
$A$ if and only if it is uniformly Cauchy on $A$.
Then we deduce that the $f_n to f$ uniformly on $[a,b]$
But given that $f_n$ are continuous functions then by uniform convergence we know that $f$ is continuous on $[a,b]$
then by definition of uniform convergence we have that for all $x in [a,b]$, for all $epsilon >0$ $exists N(epsilon )$ st $forall n>N(epsilon )$ $| f_n(x) - f(x) | < epsilon$ $implies$ $| f_n - f |_{infty} < epsilon$
Does the second proof suffice?
real-analysis functional-analysis proof-verification proof-writing banach-spaces
edited Jan 2 at 17:24
Davide Giraudo
127k16151265
asked Jan 2 at 12:42
Dreamer123Dreamer123
32729
$endgroup$
add a comment |
$begingroup$
I do this proof in 3 steps
First I pick a Cauchy sequence of $f_n(x) in C[a,b]$ and then fix an $epsilon >0$ and some $x in [a,b]$ then constitute $f(x)$ s.t $lim_{n to infty} f_n(x)=f(x)$ in absolute value.
then I try to prove that this limit $f(x)$ is continuous and $lim_{n to infty} f_n(x)=f(x)$ in sup-norm.
Which is a very long proof.
The second proof is.
Let ${f_n}$ be a Cauchy sequence in $C[a,b]$
then for all $epsilon >0$ there exists an $N(epsilon)$ s.t. for all $n,m>N(epsilon)$ we have $| f_m -f_n |_{infty} < epsilon$ which implies $| f_m(x) - f_n(x) | <epsilon$ for all $x in [a,b]$ then ${f_n(x)}$ is uniformly cauchy on $[a,b]$. Then by the following Theorem.
Theorem. A sequence ${f_n}$ of functions $f_n : A tomathbb R$ converges uniformly on
$A$ if and only if it is uniformly Cauchy on $A$.
Then we deduce that the $f_n to f$ uniformly on $[a,b]$
But given that $f_n$ are continuous functions then by uniform convergence we know that $f$ is continuous on $[a,b]$
then by definition of uniform convergence we have that for all $x in [a,b]$, for all $epsilon >0$ $exists N(epsilon )$ st $forall n>N(epsilon )$ $| f_n(x) - f(x) | < epsilon$ $implies$ $| f_n - f |_{infty} < epsilon$
Does the second proof suffice?
real-analysis functional-analysis proof-verification proof-writing banach-spaces
edited Jan 2 at 17:24
Davide Giraudo
127k16151265
asked Jan 2 at 12:42
Dreamer123Dreamer123
32729
$endgroup$
I do this proof in 3 steps
First I pick a Cauchy sequence of $f_n(x) in C[a,b]$ and then fix an $epsilon >0$ and some $x in [a,b]$ then constitute $f(x)$ s.t $lim_{n to infty} f_n(x)=f(x)$ in absolute value.
then I try to prove that this limit $f(x)$ is continuous and $lim_{n to infty} f_n(x)=f(x)$ in sup-norm.
Which is a very long proof.
The second proof is.
Let ${f_n}$ be a Cauchy sequence in $C[a,b]$
then for all $epsilon >0$ there exists an $N(epsilon)$ s.t. for all $n,m>N(epsilon)$ we have $| f_m -f_n |_{infty} < epsilon$ which implies $| f_m(x) - f_n(x) | <epsilon$ for all $x in [a,b]$ then ${f_n(x)}$ is uniformly cauchy on $[a,b]$. Then by the following Theorem.
Theorem. A sequence ${f_n}$ of functions $f_n : A tomathbb R$ converges uniformly on
$A$ if and only if it is uniformly Cauchy on $A$.
Then we deduce that the $f_n to f$ uniformly on $[a,b]$
But given that $f_n$ are continuous functions then by uniform convergence we know that $f$ is continuous on $[a,b]$
then by definition of uniform convergence we have that for all $x in [a,b]$, for all $epsilon >0$ $exists N(epsilon )$ st $forall n>N(epsilon )$ $| f_n(x) - f(x) | < epsilon$ $implies$ $| f_n - f |_{infty} < epsilon$
Does the second proof suffice?
real-analysis functional-analysis proof-verification proof-writing banach-spaces
real-analysis functional-analysis proof-verification proof-writing banach-spaces
edited Jan 2 at 17:24
Davide Giraudo
127k16151265
asked Jan 2 at 12:42
Dreamer123Dreamer123
32729
edited Jan 2 at 17:24
Davide Giraudo
127k16151265
asked Jan 2 at 12:42
Dreamer123Dreamer123
32729
edited Jan 2 at 17:24
Davide Giraudo
127k16151265
edited Jan 2 at 17:24
Davide Giraudo
127k16151265
edited Jan 2 at 17:24
Davide Giraudo
127k16151265
127k16151265
asked Jan 2 at 12:42
Dreamer123Dreamer123
32729
asked Jan 2 at 12:42
Dreamer123Dreamer123
32729
asked Jan 2 at 12:42
Dreamer123Dreamer123
32729
32729
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In your second 'proof' you are assuming what you have to prove. You have to first find a continuous function $f$ and then prove uniform convergence. The statement that ${f_n}$ is Cauchy means that given $epsilon >0$ there exists $k$ (independent of $x$) such that $|f_n(x)-f_m(x)| <epsilon$ for all $x$ for all $n,m >k$. $cdots (1)$. In particular, you can fix $x$ an conclude that the sequence ${f_n(x)}$ of real numbers is Cauchy, hence convergent. Let $f(x)$ be the limit of this sequence. In (1) if you take limit as $m to infty$ you get $|f_n(x)-f(x)| leq epsilon$ for all $x$ for all $n >k$. This means $f_n to f$ uniformly which implies that $f$ is continuous and the convergence takes place in the metric of $C[0,1]$.
answered Jan 2 at 12:54
Kavi Rama MurthyKavi Rama Murthy
63.7k42464
$endgroup$
$begingroup$
I used this theorem Theorem. A sequence ${f_n}$ of functions $f_n : A tomathbb R$ converges uniformly on $A$ if and only if it is uniformly Cauchy on $A$. which as I understood it tells me that the sequence converges uniformly to some limite I called $f$ on $[a,b]$ so the theorem tells that the limit exist but we don't know where it lives then I used another theorem which I edited in the question. Thank you
$endgroup$
– Dreamer123
Jan 8 at 17:16
1
$begingroup$
@Dreamer123 The question here is what results can be used and what cannot be used. You are using two theorems which make the question trivial, but usually a question like this is supposed to be answered without use of such theorems. I have just followed standard practice in my answer.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 23:11
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%2f3059432%2fa-question-on-the-completeness-of-ca-b-sup-where-a-b-subset-mathbb%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
$begingroup$
In your second 'proof' you are assuming what you have to prove. You have to first find a continuous function $f$ and then prove uniform convergence. The statement that ${f_n}$ is Cauchy means that given $epsilon >0$ there exists $k$ (independent of $x$) such that $|f_n(x)-f_m(x)| <epsilon$ for all $x$ for all $n,m >k$. $cdots (1)$. In particular, you can fix $x$ an conclude that the sequence ${f_n(x)}$ of real numbers is Cauchy, hence convergent. Let $f(x)$ be the limit of this sequence. In (1) if you take limit as $m to infty$ you get $|f_n(x)-f(x)| leq epsilon$ for all $x$ for all $n >k$. This means $f_n to f$ uniformly which implies that $f$ is continuous and the convergence takes place in the metric of $C[0,1]$.
answered Jan 2 at 12:54
Kavi Rama MurthyKavi Rama Murthy
63.7k42464
$endgroup$
$begingroup$
I used this theorem Theorem. A sequence ${f_n}$ of functions $f_n : A tomathbb R$ converges uniformly on $A$ if and only if it is uniformly Cauchy on $A$. which as I understood it tells me that the sequence converges uniformly to some limite I called $f$ on $[a,b]$ so the theorem tells that the limit exist but we don't know where it lives then I used another theorem which I edited in the question. Thank you
$endgroup$
– Dreamer123
Jan 8 at 17:16
1
$begingroup$
@Dreamer123 The question here is what results can be used and what cannot be used. You are using two theorems which make the question trivial, but usually a question like this is supposed to be answered without use of such theorems. I have just followed standard practice in my answer.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 23:11
add a comment |
$begingroup$
In your second 'proof' you are assuming what you have to prove. You have to first find a continuous function $f$ and then prove uniform convergence. The statement that ${f_n}$ is Cauchy means that given $epsilon >0$ there exists $k$ (independent of $x$) such that $|f_n(x)-f_m(x)| <epsilon$ for all $x$ for all $n,m >k$. $cdots (1)$. In particular, you can fix $x$ an conclude that the sequence ${f_n(x)}$ of real numbers is Cauchy, hence convergent. Let $f(x)$ be the limit of this sequence. In (1) if you take limit as $m to infty$ you get $|f_n(x)-f(x)| leq epsilon$ for all $x$ for all $n >k$. This means $f_n to f$ uniformly which implies that $f$ is continuous and the convergence takes place in the metric of $C[0,1]$.
answered Jan 2 at 12:54
Kavi Rama MurthyKavi Rama Murthy
63.7k42464
$endgroup$
$begingroup$
I used this theorem Theorem. A sequence ${f_n}$ of functions $f_n : A tomathbb R$ converges uniformly on $A$ if and only if it is uniformly Cauchy on $A$. which as I understood it tells me that the sequence converges uniformly to some limite I called $f$ on $[a,b]$ so the theorem tells that the limit exist but we don't know where it lives then I used another theorem which I edited in the question. Thank you
$endgroup$
– Dreamer123
Jan 8 at 17:16
1
$begingroup$
@Dreamer123 The question here is what results can be used and what cannot be used. You are using two theorems which make the question trivial, but usually a question like this is supposed to be answered without use of such theorems. I have just followed standard practice in my answer.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 23:11
add a comment |
$begingroup$
In your second 'proof' you are assuming what you have to prove. You have to first find a continuous function $f$ and then prove uniform convergence. The statement that ${f_n}$ is Cauchy means that given $epsilon >0$ there exists $k$ (independent of $x$) such that $|f_n(x)-f_m(x)| <epsilon$ for all $x$ for all $n,m >k$. $cdots (1)$. In particular, you can fix $x$ an conclude that the sequence ${f_n(x)}$ of real numbers is Cauchy, hence convergent. Let $f(x)$ be the limit of this sequence. In (1) if you take limit as $m to infty$ you get $|f_n(x)-f(x)| leq epsilon$ for all $x$ for all $n >k$. This means $f_n to f$ uniformly which implies that $f$ is continuous and the convergence takes place in the metric of $C[0,1]$.
answered Jan 2 at 12:54
Kavi Rama MurthyKavi Rama Murthy
63.7k42464
$endgroup$
In your second 'proof' you are assuming what you have to prove. You have to first find a continuous function $f$ and then prove uniform convergence. The statement that ${f_n}$ is Cauchy means that given $epsilon >0$ there exists $k$ (independent of $x$) such that $|f_n(x)-f_m(x)| <epsilon$ for all $x$ for all $n,m >k$. $cdots (1)$. In particular, you can fix $x$ an conclude that the sequence ${f_n(x)}$ of real numbers is Cauchy, hence convergent. Let $f(x)$ be the limit of this sequence. In (1) if you take limit as $m to infty$ you get $|f_n(x)-f(x)| leq epsilon$ for all $x$ for all $n >k$. This means $f_n to f$ uniformly which implies that $f$ is continuous and the convergence takes place in the metric of $C[0,1]$.
answered Jan 2 at 12:54
Kavi Rama MurthyKavi Rama Murthy
63.7k42464
edited Jan 3 at 6:10
answered Jan 2 at 12:54
Kavi Rama MurthyKavi Rama Murthy
63.7k42464
answered Jan 2 at 12:54
Kavi Rama MurthyKavi Rama Murthy
63.7k42464
answered Jan 2 at 12:54
Kavi Rama MurthyKavi Rama Murthy
63.7k42464
63.7k42464
$begingroup$
I used this theorem Theorem. A sequence ${f_n}$ of functions $f_n : A tomathbb R$ converges uniformly on $A$ if and only if it is uniformly Cauchy on $A$. which as I understood it tells me that the sequence converges uniformly to some limite I called $f$ on $[a,b]$ so the theorem tells that the limit exist but we don't know where it lives then I used another theorem which I edited in the question. Thank you
$endgroup$
– Dreamer123
Jan 8 at 17:16
1
$begingroup$
@Dreamer123 The question here is what results can be used and what cannot be used. You are using two theorems which make the question trivial, but usually a question like this is supposed to be answered without use of such theorems. I have just followed standard practice in my answer.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 23:11
add a comment |
$begingroup$
I used this theorem Theorem. A sequence ${f_n}$ of functions $f_n : A tomathbb R$ converges uniformly on $A$ if and only if it is uniformly Cauchy on $A$. which as I understood it tells me that the sequence converges uniformly to some limite I called $f$ on $[a,b]$ so the theorem tells that the limit exist but we don't know where it lives then I used another theorem which I edited in the question. Thank you
$endgroup$
– Dreamer123
Jan 8 at 17:16
1
$begingroup$
@Dreamer123 The question here is what results can be used and what cannot be used. You are using two theorems which make the question trivial, but usually a question like this is supposed to be answered without use of such theorems. I have just followed standard practice in my answer.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 23:11
$begingroup$
I used this theorem Theorem. A sequence ${f_n}$ of functions $f_n : A tomathbb R$ converges uniformly on $A$ if and only if it is uniformly Cauchy on $A$. which as I understood it tells me that the sequence converges uniformly to some limite I called $f$ on $[a,b]$ so the theorem tells that the limit exist but we don't know where it lives then I used another theorem which I edited in the question. Thank you
$endgroup$
– Dreamer123
Jan 8 at 17:16
$begingroup$
I used this theorem Theorem. A sequence ${f_n}$ of functions $f_n : A tomathbb R$ converges uniformly on $A$ if and only if it is uniformly Cauchy on $A$. which as I understood it tells me that the sequence converges uniformly to some limite I called $f$ on $[a,b]$ so the theorem tells that the limit exist but we don't know where it lives then I used another theorem which I edited in the question. Thank you
$endgroup$
– Dreamer123
Jan 8 at 17:16
1
1
$begingroup$
@Dreamer123 The question here is what results can be used and what cannot be used. You are using two theorems which make the question trivial, but usually a question like this is supposed to be answered without use of such theorems. I have just followed standard practice in my answer.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 23:11
$begingroup$
@Dreamer123 The question here is what results can be used and what cannot be used. You are using two theorems which make the question trivial, but usually a question like this is supposed to be answered without use of such theorems. I have just followed standard practice in my answer.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 23:11
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%2f3059432%2fa-question-on-the-completeness-of-ca-b-sup-where-a-b-subset-mathbb%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
