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Can we approximate continuous functions arbitrarily well with polynomials? (beyond Weierstrass )

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0

$begingroup$

Let $f:(0,1) to mathbbR$ be continuous, and let $delta:(0,1) to mathbbR$ be continuous and positive.

Does there always exist a polynomial $p(x)$ satisfying $|f(x)-p(x)| < delta(x)$ for every $x in (0,1)$?

Edit: I should have written $[0,1]$ (the closed interval) as the domain instead of $(0,1)$. (to rule out problems which come from the fact $f$ is not bounded, or uniformly continuous; if $f$ is not bounded, then it cannot be approximated by polynomials).

---

I guess that the answer is negative, but I don't see how to build a "sufficiently bad" $delta$.

When $delta$ is constant, this is just the Weierstrass approximation theorem.

Moreover, if we allow $p(x)$ to be an arbitrary smooth function, then we can always achieve a $delta$-approximation, via a partition of unity argument.

real-analysis calculus polynomials approximation approximation-theory

share|cite|improve this question

edited Mar 28 at 15:02

Asaf Shachar

asked Mar 28 at 14:54

Asaf ShacharAsaf Shachar

5,79231145

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can do Weierstrass again provided that $f$ is uniformly continuous and $delta$ is bounded below away from zero (because in this case you're really just applying Weierstrass to the extension of $f$ to $[0,1]$ and using $inf delta$ as your tolerance). If $delta$ is not bounded below away from zero (and $f$ is still uniformly continuous) then I think you can do it again, by splitting $p$ into an interpolant of $f$ at the endpoints plus a corrector, but I'd have to mess with the estimates to make sure things still work out in a neighborhood of the endpoints.
  $endgroup$
  – Ian
  Mar 28 at 14:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $f$ is not uniformly continuous then things can certainly break when $f$ is not bounded, and I suspect they can break when $f$ has an oscillatory singularity at one of the endpoints as well.
  $endgroup$
  – Ian
  Mar 28 at 14:59
  

  
  
  
  


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  $begingroup$
  It's easy to show that $sin(1/x)$ can't be approximated within $delta$ on $(0,1)$ by a polynomial if $delta < 1/2$.
  $endgroup$
  – Robert Israel
  Mar 28 at 15:02
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $f:(0,1) to mathbbR$ be continuous, and let $delta:(0,1) to mathbbR$ be continuous and positive.

Does there always exist a polynomial $p(x)$ satisfying $|f(x)-p(x)| < delta(x)$ for every $x in (0,1)$?

Edit: I should have written $[0,1]$ (the closed interval) as the domain instead of $(0,1)$. (to rule out problems which come from the fact $f$ is not bounded, or uniformly continuous; if $f$ is not bounded, then it cannot be approximated by polynomials).

---

I guess that the answer is negative, but I don't see how to build a "sufficiently bad" $delta$.

When $delta$ is constant, this is just the Weierstrass approximation theorem.

Moreover, if we allow $p(x)$ to be an arbitrary smooth function, then we can always achieve a $delta$-approximation, via a partition of unity argument.

real-analysis calculus polynomials approximation approximation-theory

share|cite|improve this question

edited Mar 28 at 15:02

Asaf Shachar

asked Mar 28 at 14:54

Asaf ShacharAsaf Shachar

5,79231145

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can do Weierstrass again provided that $f$ is uniformly continuous and $delta$ is bounded below away from zero (because in this case you're really just applying Weierstrass to the extension of $f$ to $[0,1]$ and using $inf delta$ as your tolerance). If $delta$ is not bounded below away from zero (and $f$ is still uniformly continuous) then I think you can do it again, by splitting $p$ into an interpolant of $f$ at the endpoints plus a corrector, but I'd have to mess with the estimates to make sure things still work out in a neighborhood of the endpoints.
  $endgroup$
  – Ian
  Mar 28 at 14:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $f$ is not uniformly continuous then things can certainly break when $f$ is not bounded, and I suspect they can break when $f$ has an oscillatory singularity at one of the endpoints as well.
  $endgroup$
  – Ian
  Mar 28 at 14:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's easy to show that $sin(1/x)$ can't be approximated within $delta$ on $(0,1)$ by a polynomial if $delta < 1/2$.
  $endgroup$
  – Robert Israel
  Mar 28 at 15:02
  

  
  
  
  

add a comment | 

$begingroup$

Let $f:(0,1) to mathbbR$ be continuous, and let $delta:(0,1) to mathbbR$ be continuous and positive.

Does there always exist a polynomial $p(x)$ satisfying $|f(x)-p(x)| < delta(x)$ for every $x in (0,1)$?

Edit: I should have written $[0,1]$ (the closed interval) as the domain instead of $(0,1)$. (to rule out problems which come from the fact $f$ is not bounded, or uniformly continuous; if $f$ is not bounded, then it cannot be approximated by polynomials).

---

I guess that the answer is negative, but I don't see how to build a "sufficiently bad" $delta$.

When $delta$ is constant, this is just the Weierstrass approximation theorem.

Moreover, if we allow $p(x)$ to be an arbitrary smooth function, then we can always achieve a $delta$-approximation, via a partition of unity argument.

real-analysis calculus polynomials approximation approximation-theory

share|cite|improve this question

edited Mar 28 at 15:02

Asaf Shachar

asked Mar 28 at 14:54

Asaf ShacharAsaf Shachar

5,79231145

$endgroup$

share|cite|improve this question

edited Mar 28 at 15:02

Asaf Shachar

asked Mar 28 at 14:54

Asaf ShacharAsaf Shachar

5,79231145

share|cite|improve this question

edited Mar 28 at 15:02

Asaf Shachar

asked Mar 28 at 14:54

Asaf ShacharAsaf Shachar

5,79231145

asked Mar 28 at 14:54

Asaf ShacharAsaf Shachar

5,79231145

asked Mar 28 at 14:54

Asaf ShacharAsaf Shachar

5,79231145

1 Answer
1

active

oldest

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4

$begingroup$

On $(0,1)$, no, even if $delta$ is constant, because $f$ might not be bounded while polynomials are. Weierstrass requires a closed, bounded interval.

EDIT: With the interval as $[0,1]$, $min_x in [0,1] delta(x)$ exists and is positive, so we might as well replace $delta$ by that constant, and then we can use the Weierstrass theorem.

share|cite|improve this answer

edited Mar 28 at 15:05

answered Mar 28 at 14:59

Robert IsraelRobert Israel

330k23219473

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, you are right. I really should have written the closed interval as the domain, not the open one. (The point was not to see that Weierstrass theorem fails on non-compact domains, but to investigate how power polynomials have to approximate-better than uniformly, that is).
  $endgroup$
  – Asaf Shachar
  Mar 28 at 15:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AsafShachar Unless $delta$ on $[0,1]$ has zeros, it becomes sufficient to just do Weierstrass again, once you switch over to the compact case.
  $endgroup$
  – Ian
  Mar 28 at 15:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But if "positive" means $> 0$, $delta$ can't have zeros.
  $endgroup$
  – Robert Israel
  Mar 28 at 15:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RobertIsrael Indeed; I'm just pointing out a possible generalization that is actually nontrivial. Still, I think given an interpolant $q$ at the zeros of $delta$ you can then take $p=q(1+r)$ where $r$ is a Weierstrass-type approximation of $fracf-qq$.
  $endgroup$
  – Ian
  Mar 28 at 15:08
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Ian Thanks, you are right. I should have thought more about a careful formulation of the question. I guess that we could try to make the question less trivial in one of two ways: (1) allow $delta$ to have finitely many zeros. (2) Keep the domain open, but assume that $f$ is bounded.
  $endgroup$
  – Asaf Shachar
  Mar 28 at 15:12
  
  

  
  
  
  

 | 
show 3 more comments

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4

$begingroup$

On $(0,1)$, no, even if $delta$ is constant, because $f$ might not be bounded while polynomials are. Weierstrass requires a closed, bounded interval.

EDIT: With the interval as $[0,1]$, $min_x in [0,1] delta(x)$ exists and is positive, so we might as well replace $delta$ by that constant, and then we can use the Weierstrass theorem.

share|cite|improve this answer

edited Mar 28 at 15:05

answered Mar 28 at 14:59

Robert IsraelRobert Israel

330k23219473

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, you are right. I really should have written the closed interval as the domain, not the open one. (The point was not to see that Weierstrass theorem fails on non-compact domains, but to investigate how power polynomials have to approximate-better than uniformly, that is).
  $endgroup$
  – Asaf Shachar
  Mar 28 at 15:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AsafShachar Unless $delta$ on $[0,1]$ has zeros, it becomes sufficient to just do Weierstrass again, once you switch over to the compact case.
  $endgroup$
  – Ian
  Mar 28 at 15:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But if "positive" means $> 0$, $delta$ can't have zeros.
  $endgroup$
  – Robert Israel
  Mar 28 at 15:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RobertIsrael Indeed; I'm just pointing out a possible generalization that is actually nontrivial. Still, I think given an interpolant $q$ at the zeros of $delta$ you can then take $p=q(1+r)$ where $r$ is a Weierstrass-type approximation of $fracf-qq$.
  $endgroup$
  – Ian
  Mar 28 at 15:08
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Ian Thanks, you are right. I should have thought more about a careful formulation of the question. I guess that we could try to make the question less trivial in one of two ways: (1) allow $delta$ to have finitely many zeros. (2) Keep the domain open, but assume that $f$ is bounded.
  $endgroup$
  – Asaf Shachar
  Mar 28 at 15:12
  
  

  
  
  
  

 | 
show 3 more comments

4

$begingroup$

On $(0,1)$, no, even if $delta$ is constant, because $f$ might not be bounded while polynomials are. Weierstrass requires a closed, bounded interval.

EDIT: With the interval as $[0,1]$, $min_x in [0,1] delta(x)$ exists and is positive, so we might as well replace $delta$ by that constant, and then we can use the Weierstrass theorem.

share|cite|improve this answer

edited Mar 28 at 15:05

answered Mar 28 at 14:59

Robert IsraelRobert Israel

330k23219473

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, you are right. I really should have written the closed interval as the domain, not the open one. (The point was not to see that Weierstrass theorem fails on non-compact domains, but to investigate how power polynomials have to approximate-better than uniformly, that is).
  $endgroup$
  – Asaf Shachar
  Mar 28 at 15:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AsafShachar Unless $delta$ on $[0,1]$ has zeros, it becomes sufficient to just do Weierstrass again, once you switch over to the compact case.
  $endgroup$
  – Ian
  Mar 28 at 15:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But if "positive" means $> 0$, $delta$ can't have zeros.
  $endgroup$
  – Robert Israel
  Mar 28 at 15:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RobertIsrael Indeed; I'm just pointing out a possible generalization that is actually nontrivial. Still, I think given an interpolant $q$ at the zeros of $delta$ you can then take $p=q(1+r)$ where $r$ is a Weierstrass-type approximation of $fracf-qq$.
  $endgroup$
  – Ian
  Mar 28 at 15:08
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Ian Thanks, you are right. I should have thought more about a careful formulation of the question. I guess that we could try to make the question less trivial in one of two ways: (1) allow $delta$ to have finitely many zeros. (2) Keep the domain open, but assume that $f$ is bounded.
  $endgroup$
  – Asaf Shachar
  Mar 28 at 15:12
  
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

On $(0,1)$, no, even if $delta$ is constant, because $f$ might not be bounded while polynomials are. Weierstrass requires a closed, bounded interval.

EDIT: With the interval as $[0,1]$, $min_x in [0,1] delta(x)$ exists and is positive, so we might as well replace $delta$ by that constant, and then we can use the Weierstrass theorem.

share|cite|improve this answer

edited Mar 28 at 15:05

answered Mar 28 at 14:59

Robert IsraelRobert Israel

330k23219473

$endgroup$

share|cite|improve this answer

edited Mar 28 at 15:05

answered Mar 28 at 14:59

Robert IsraelRobert Israel

330k23219473

answered Mar 28 at 14:59

Robert IsraelRobert Israel

330k23219473

answered Mar 28 at 14:59

Robert IsraelRobert Israel

330k23219473