Show that $f$ is a polynomial if it's the uniform limit of polynomais
Multi tool use
2
$begingroup$
Let $f:Bbb Rto Bbb R$ be a function which is the uniform limit of polynomials. I want to show that $f$ is a polynomial.
I mean this seems a bit trivial...
If it's the uniform limit of the set of polynomials doesn't that guarantee it's a polynomial?
If I define $f$ to be the set of all polynomial do I define a uniform limit $L$ s.t $d(f,L)<epsilon$? or approach via contradiction?
any help would be great
real-analysis
edited Oct 7 '14 at 4:24
leo
6,04553582
asked Oct 7 '14 at 3:47
user146296user146296
448
$endgroup$
|
show 1 more comment
2
$begingroup$
Let $f:Bbb Rto Bbb R$ be a function which is the uniform limit of polynomials. I want to show that $f$ is a polynomial.
I mean this seems a bit trivial...
If it's the uniform limit of the set of polynomials doesn't that guarantee it's a polynomial?
If I define $f$ to be the set of all polynomial do I define a uniform limit $L$ s.t $d(f,L)<epsilon$? or approach via contradiction?
any help would be great
real-analysis
edited Oct 7 '14 at 4:24
leo
6,04553582
asked Oct 7 '14 at 3:47
user146296user146296
448
$endgroup$
4
$begingroup$
It seems trivial if you think about it in the right way. However, $sin x$ is a limit of polynomials and that is not a finite polynomial.
$endgroup$
– user164587
Oct 7 '14 at 3:52
1
$begingroup$
Any continuous function on a closed bounded (thus, compact) interval is the uniform limit of a sequence of polynomials. Thus, the claim you try to prove is false, @user164587
$endgroup$
– Timbuc
Oct 7 '14 at 3:56
$begingroup$
How do you know how to define a limit of polynomials? Is there a different way to phrase this that would make it easier to understand?
$endgroup$
– user146296
Oct 7 '14 at 3:57
3
$begingroup$
@Timbuc The point is that this is on all of $mathbb R$, not on a compact interval.
$endgroup$
– Robert Israel
Oct 7 '14 at 3:59
$begingroup$
I see , @Robert.
$endgroup$
– Timbuc
Oct 7 '14 at 4:38
|
show 1 more comment
2
2
2
1
$begingroup$
Let $f:Bbb Rto Bbb R$ be a function which is the uniform limit of polynomials. I want to show that $f$ is a polynomial.
I mean this seems a bit trivial...
If it's the uniform limit of the set of polynomials doesn't that guarantee it's a polynomial?
If I define $f$ to be the set of all polynomial do I define a uniform limit $L$ s.t $d(f,L)<epsilon$? or approach via contradiction?
any help would be great
real-analysis
edited Oct 7 '14 at 4:24
leo
6,04553582
asked Oct 7 '14 at 3:47
user146296user146296
448
$endgroup$
Let $f:Bbb Rto Bbb R$ be a function which is the uniform limit of polynomials. I want to show that $f$ is a polynomial.
I mean this seems a bit trivial...
If it's the uniform limit of the set of polynomials doesn't that guarantee it's a polynomial?
If I define $f$ to be the set of all polynomial do I define a uniform limit $L$ s.t $d(f,L)<epsilon$? or approach via contradiction?
any help would be great
real-analysis
real-analysis
edited Oct 7 '14 at 4:24
leo
6,04553582
asked Oct 7 '14 at 3:47
user146296user146296
448
edited Oct 7 '14 at 4:24
leo
6,04553582
asked Oct 7 '14 at 3:47
user146296user146296
448
edited Oct 7 '14 at 4:24
leo
6,04553582
edited Oct 7 '14 at 4:24
leo
6,04553582
edited Oct 7 '14 at 4:24
leo
6,04553582
6,04553582
asked Oct 7 '14 at 3:47
user146296user146296
448
asked Oct 7 '14 at 3:47
user146296user146296
448
asked Oct 7 '14 at 3:47
user146296user146296
448
448
4
$begingroup$
It seems trivial if you think about it in the right way. However, $sin x$ is a limit of polynomials and that is not a finite polynomial.
$endgroup$
– user164587
Oct 7 '14 at 3:52
1
$begingroup$
Any continuous function on a closed bounded (thus, compact) interval is the uniform limit of a sequence of polynomials. Thus, the claim you try to prove is false, @user164587
$endgroup$
– Timbuc
Oct 7 '14 at 3:56
$begingroup$
How do you know how to define a limit of polynomials? Is there a different way to phrase this that would make it easier to understand?
$endgroup$
– user146296
Oct 7 '14 at 3:57
3
$begingroup$
@Timbuc The point is that this is on all of $mathbb R$, not on a compact interval.
$endgroup$
– Robert Israel
Oct 7 '14 at 3:59
$begingroup$
I see , @Robert.
$endgroup$
– Timbuc
Oct 7 '14 at 4:38
|
show 1 more comment
4
$begingroup$
It seems trivial if you think about it in the right way. However, $sin x$ is a limit of polynomials and that is not a finite polynomial.
$endgroup$
– user164587
Oct 7 '14 at 3:52
1
$begingroup$
Any continuous function on a closed bounded (thus, compact) interval is the uniform limit of a sequence of polynomials. Thus, the claim you try to prove is false, @user164587
$endgroup$
– Timbuc
Oct 7 '14 at 3:56
$begingroup$
How do you know how to define a limit of polynomials? Is there a different way to phrase this that would make it easier to understand?
$endgroup$
– user146296
Oct 7 '14 at 3:57
3
$begingroup$
@Timbuc The point is that this is on all of $mathbb R$, not on a compact interval.
$endgroup$
– Robert Israel
Oct 7 '14 at 3:59
$begingroup$
I see , @Robert.
$endgroup$
– Timbuc
Oct 7 '14 at 4:38
4
4
$begingroup$
It seems trivial if you think about it in the right way. However, $sin x$ is a limit of polynomials and that is not a finite polynomial.
$endgroup$
– user164587
Oct 7 '14 at 3:52
$begingroup$
It seems trivial if you think about it in the right way. However, $sin x$ is a limit of polynomials and that is not a finite polynomial.
$endgroup$
– user164587
Oct 7 '14 at 3:52
1
1
$begingroup$
Any continuous function on a closed bounded (thus, compact) interval is the uniform limit of a sequence of polynomials. Thus, the claim you try to prove is false, @user164587
$endgroup$
– Timbuc
Oct 7 '14 at 3:56
$begingroup$
Any continuous function on a closed bounded (thus, compact) interval is the uniform limit of a sequence of polynomials. Thus, the claim you try to prove is false, @user164587
$endgroup$
– Timbuc
Oct 7 '14 at 3:56
$begingroup$
How do you know how to define a limit of polynomials? Is there a different way to phrase this that would make it easier to understand?
$endgroup$
– user146296
Oct 7 '14 at 3:57
$begingroup$
How do you know how to define a limit of polynomials? Is there a different way to phrase this that would make it easier to understand?
$endgroup$
– user146296
Oct 7 '14 at 3:57
3
3
$begingroup$
@Timbuc The point is that this is on all of $mathbb R$, not on a compact interval.
$endgroup$
– Robert Israel
Oct 7 '14 at 3:59
$begingroup$
@Timbuc The point is that this is on all of $mathbb R$, not on a compact interval.
$endgroup$
– Robert Israel
Oct 7 '14 at 3:59
$begingroup$
I see , @Robert.
$endgroup$
– Timbuc
Oct 7 '14 at 4:38
$begingroup$
I see , @Robert.
$endgroup$
– Timbuc
Oct 7 '14 at 4:38
|
show 1 more comment
1 Answer
1
active
oldest
votes
3
$begingroup$
Hint: if $f$ is a non-constant polynomial, then $f$ is unbounded. What does this say about $f_n(z) - f_m(z)$, if $f_n$ is a sequence of polynomials that converges uniformly on $mathbb R$?
answered Oct 7 '14 at 4:04
Robert IsraelRobert Israel
322k23212465
$endgroup$
$begingroup$
If fn and fm are both sequences of polynomials then the difference will be a polynomial right. I guess I'm struggling with writing up a proof for this.
$endgroup$
– user146296
Oct 7 '14 at 4:19
$begingroup$
@leo We don't know yet that $f(z)$ is a polynomial. That's what we're trying to prove.
$endgroup$
– Robert Israel
Oct 7 '14 at 6:53
1
$begingroup$
@user146296 It's just one sequence of polynomials, not two. $f_n$ and $f_m$ are two members of the sequence. If it converges uniformly, then for every $epsilon > 0$ there is $N$ such that for all $m$ and $n$ both $> N$, ...
$endgroup$
– Robert Israel
Oct 7 '14 at 6:55
$begingroup$
You are right. ${}$
$endgroup$
– leo
Oct 7 '14 at 15:43
2
$begingroup$
The point is that if the sequence converges uniformly, and $m$ is large enough, then $|f_n - f_m|$ is uniformly bounded for $n > m$, and since $f_n - f_m$ is a polynomial this implies that $f_n - f_m$ is a constant function. Now you take the limit as $n to infty$, and find that $f = f_m + text{constant}$, which is a polynomial.
$endgroup$
– Robert Israel
Oct 7 '14 at 22:27
|
show 1 more 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
});
}
});
draft saved
draft discarded
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%2f961796%2fshow-that-f-is-a-polynomial-if-its-the-uniform-limit-of-polynomais%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
3
$begingroup$
Hint: if $f$ is a non-constant polynomial, then $f$ is unbounded. What does this say about $f_n(z) - f_m(z)$, if $f_n$ is a sequence of polynomials that converges uniformly on $mathbb R$?
answered Oct 7 '14 at 4:04
Robert IsraelRobert Israel
322k23212465
$endgroup$
$begingroup$
If fn and fm are both sequences of polynomials then the difference will be a polynomial right. I guess I'm struggling with writing up a proof for this.
$endgroup$
– user146296
Oct 7 '14 at 4:19
$begingroup$
@leo We don't know yet that $f(z)$ is a polynomial. That's what we're trying to prove.
$endgroup$
– Robert Israel
Oct 7 '14 at 6:53
1
$begingroup$
@user146296 It's just one sequence of polynomials, not two. $f_n$ and $f_m$ are two members of the sequence. If it converges uniformly, then for every $epsilon > 0$ there is $N$ such that for all $m$ and $n$ both $> N$, ...
$endgroup$
– Robert Israel
Oct 7 '14 at 6:55
$begingroup$
You are right. ${}$
$endgroup$
– leo
Oct 7 '14 at 15:43
2
$begingroup$
The point is that if the sequence converges uniformly, and $m$ is large enough, then $|f_n - f_m|$ is uniformly bounded for $n > m$, and since $f_n - f_m$ is a polynomial this implies that $f_n - f_m$ is a constant function. Now you take the limit as $n to infty$, and find that $f = f_m + text{constant}$, which is a polynomial.
$endgroup$
– Robert Israel
Oct 7 '14 at 22:27
|
show 1 more comment
3
$begingroup$
Hint: if $f$ is a non-constant polynomial, then $f$ is unbounded. What does this say about $f_n(z) - f_m(z)$, if $f_n$ is a sequence of polynomials that converges uniformly on $mathbb R$?
answered Oct 7 '14 at 4:04
Robert IsraelRobert Israel
322k23212465
$endgroup$
$begingroup$
If fn and fm are both sequences of polynomials then the difference will be a polynomial right. I guess I'm struggling with writing up a proof for this.
$endgroup$
– user146296
Oct 7 '14 at 4:19
$begingroup$
@leo We don't know yet that $f(z)$ is a polynomial. That's what we're trying to prove.
$endgroup$
– Robert Israel
Oct 7 '14 at 6:53
1
$begingroup$
@user146296 It's just one sequence of polynomials, not two. $f_n$ and $f_m$ are two members of the sequence. If it converges uniformly, then for every $epsilon > 0$ there is $N$ such that for all $m$ and $n$ both $> N$, ...
$endgroup$
– Robert Israel
Oct 7 '14 at 6:55
$begingroup$
You are right. ${}$
$endgroup$
– leo
Oct 7 '14 at 15:43
2
$begingroup$
The point is that if the sequence converges uniformly, and $m$ is large enough, then $|f_n - f_m|$ is uniformly bounded for $n > m$, and since $f_n - f_m$ is a polynomial this implies that $f_n - f_m$ is a constant function. Now you take the limit as $n to infty$, and find that $f = f_m + text{constant}$, which is a polynomial.
$endgroup$
– Robert Israel
Oct 7 '14 at 22:27
|
show 1 more comment
3
3
3
$begingroup$
Hint: if $f$ is a non-constant polynomial, then $f$ is unbounded. What does this say about $f_n(z) - f_m(z)$, if $f_n$ is a sequence of polynomials that converges uniformly on $mathbb R$?
answered Oct 7 '14 at 4:04
Robert IsraelRobert Israel
322k23212465
$endgroup$
Hint: if $f$ is a non-constant polynomial, then $f$ is unbounded. What does this say about $f_n(z) - f_m(z)$, if $f_n$ is a sequence of polynomials that converges uniformly on $mathbb R$?
answered Oct 7 '14 at 4:04
Robert IsraelRobert Israel
322k23212465
answered Oct 7 '14 at 4:04
Robert IsraelRobert Israel
322k23212465
answered Oct 7 '14 at 4:04
Robert IsraelRobert Israel
322k23212465
answered Oct 7 '14 at 4:04
Robert IsraelRobert Israel
322k23212465
322k23212465
$begingroup$
If fn and fm are both sequences of polynomials then the difference will be a polynomial right. I guess I'm struggling with writing up a proof for this.
$endgroup$
– user146296
Oct 7 '14 at 4:19
$begingroup$
@leo We don't know yet that $f(z)$ is a polynomial. That's what we're trying to prove.
$endgroup$
– Robert Israel
Oct 7 '14 at 6:53
1
$begingroup$
@user146296 It's just one sequence of polynomials, not two. $f_n$ and $f_m$ are two members of the sequence. If it converges uniformly, then for every $epsilon > 0$ there is $N$ such that for all $m$ and $n$ both $> N$, ...
$endgroup$
– Robert Israel
Oct 7 '14 at 6:55
$begingroup$
You are right. ${}$
$endgroup$
– leo
Oct 7 '14 at 15:43
2
$begingroup$
The point is that if the sequence converges uniformly, and $m$ is large enough, then $|f_n - f_m|$ is uniformly bounded for $n > m$, and since $f_n - f_m$ is a polynomial this implies that $f_n - f_m$ is a constant function. Now you take the limit as $n to infty$, and find that $f = f_m + text{constant}$, which is a polynomial.
$endgroup$
– Robert Israel
Oct 7 '14 at 22:27
|
show 1 more comment
$begingroup$
If fn and fm are both sequences of polynomials then the difference will be a polynomial right. I guess I'm struggling with writing up a proof for this.
$endgroup$
– user146296
Oct 7 '14 at 4:19
$begingroup$
@leo We don't know yet that $f(z)$ is a polynomial. That's what we're trying to prove.
$endgroup$
– Robert Israel
Oct 7 '14 at 6:53
1
$begingroup$
@user146296 It's just one sequence of polynomials, not two. $f_n$ and $f_m$ are two members of the sequence. If it converges uniformly, then for every $epsilon > 0$ there is $N$ such that for all $m$ and $n$ both $> N$, ...
$endgroup$
– Robert Israel
Oct 7 '14 at 6:55
$begingroup$
You are right. ${}$
$endgroup$
– leo
Oct 7 '14 at 15:43
2
$begingroup$
The point is that if the sequence converges uniformly, and $m$ is large enough, then $|f_n - f_m|$ is uniformly bounded for $n > m$, and since $f_n - f_m$ is a polynomial this implies that $f_n - f_m$ is a constant function. Now you take the limit as $n to infty$, and find that $f = f_m + text{constant}$, which is a polynomial.
$endgroup$
– Robert Israel
Oct 7 '14 at 22:27
$begingroup$
If fn and fm are both sequences of polynomials then the difference will be a polynomial right. I guess I'm struggling with writing up a proof for this.
$endgroup$
– user146296
Oct 7 '14 at 4:19
$begingroup$
If fn and fm are both sequences of polynomials then the difference will be a polynomial right. I guess I'm struggling with writing up a proof for this.
$endgroup$
– user146296
Oct 7 '14 at 4:19
$begingroup$
@leo We don't know yet that $f(z)$ is a polynomial. That's what we're trying to prove.
$endgroup$
– Robert Israel
Oct 7 '14 at 6:53
$begingroup$
@leo We don't know yet that $f(z)$ is a polynomial. That's what we're trying to prove.
$endgroup$
– Robert Israel
Oct 7 '14 at 6:53
1
1
$begingroup$
@user146296 It's just one sequence of polynomials, not two. $f_n$ and $f_m$ are two members of the sequence. If it converges uniformly, then for every $epsilon > 0$ there is $N$ such that for all $m$ and $n$ both $> N$, ...
$endgroup$
– Robert Israel
Oct 7 '14 at 6:55
$begingroup$
@user146296 It's just one sequence of polynomials, not two. $f_n$ and $f_m$ are two members of the sequence. If it converges uniformly, then for every $epsilon > 0$ there is $N$ such that for all $m$ and $n$ both $> N$, ...
$endgroup$
– Robert Israel
Oct 7 '14 at 6:55
$begingroup$
You are right. ${}$
$endgroup$
– leo
Oct 7 '14 at 15:43
$begingroup$
You are right. ${}$
$endgroup$
– leo
Oct 7 '14 at 15:43
2
2
$begingroup$
The point is that if the sequence converges uniformly, and $m$ is large enough, then $|f_n - f_m|$ is uniformly bounded for $n > m$, and since $f_n - f_m$ is a polynomial this implies that $f_n - f_m$ is a constant function. Now you take the limit as $n to infty$, and find that $f = f_m + text{constant}$, which is a polynomial.
$endgroup$
– Robert Israel
Oct 7 '14 at 22:27
$begingroup$
The point is that if the sequence converges uniformly, and $m$ is large enough, then $|f_n - f_m|$ is uniformly bounded for $n > m$, and since $f_n - f_m$ is a polynomial this implies that $f_n - f_m$ is a constant function. Now you take the limit as $n to infty$, and find that $f = f_m + text{constant}$, which is a polynomial.
$endgroup$
– Robert Israel
Oct 7 '14 at 22:27
|
show 1 more comment
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
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%2f961796%2fshow-that-f-is-a-polynomial-if-its-the-uniform-limit-of-polynomais%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
U5hUp,phNfAHrMqDg3ZEBEcCRj3X1
4
$begingroup$
It seems trivial if you think about it in the right way. However, $sin x$ is a limit of polynomials and that is not a finite polynomial.
$endgroup$
– user164587
Oct 7 '14 at 3:52
1
$begingroup$
Any continuous function on a closed bounded (thus, compact) interval is the uniform limit of a sequence of polynomials. Thus, the claim you try to prove is false, @user164587
$endgroup$
– Timbuc
Oct 7 '14 at 3:56
$begingroup$
How do you know how to define a limit of polynomials? Is there a different way to phrase this that would make it easier to understand?
$endgroup$
– user146296
Oct 7 '14 at 3:57
3
$begingroup$
@Timbuc The point is that this is on all of $mathbb R$, not on a compact interval.
$endgroup$
– Robert Israel
Oct 7 '14 at 3:59
$begingroup$
I see , @Robert.
$endgroup$
– Timbuc
Oct 7 '14 at 4:38