Show that $f$ is a polynomial if it's the uniform limit of polynomais
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
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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
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1 Answer
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1 Answer
1
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oldest
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active
oldest
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active
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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
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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