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Question regarding Weierstrass theorem generalized to Stone-Weierstrass

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1

$begingroup$

Weierstrass theorem.

Lef $f$ be a defined and continuous function in $[a,b]$. Given $epsilon>0,$ there exists a polynomial $P$ such that $vert f(x)-P(x)vert<epsilon,$ for all $xin[a,b].$

Stone-Weierstrass theorem.

Let $X$ be a topological compact space. If $F$ is an algebra of $C(X)$ that separate points and contains the constant functions then $F$ is dense with respect to the uniform convergence in $C(X).$

How is that an algebra $F$ of $C(X)$ that separate points and contains the constant functions is the generalization of $P(x)$ ?

Can someone shed some light on this?

real-analysis functional-analysis analysis weierstrass-approximation

share|cite|improve this question

edited Mar 16 at 20:28

Sambo

2,3012532

asked Mar 16 at 20:01

IsaIsa

295519

$endgroup$

add a comment | 

1

$begingroup$

Weierstrass theorem.

Lef $f$ be a defined and continuous function in $[a,b]$. Given $epsilon>0,$ there exists a polynomial $P$ such that $vert f(x)-P(x)vert<epsilon,$ for all $xin[a,b].$

Stone-Weierstrass theorem.

Let $X$ be a topological compact space. If $F$ is an algebra of $C(X)$ that separate points and contains the constant functions then $F$ is dense with respect to the uniform convergence in $C(X).$

How is that an algebra $F$ of $C(X)$ that separate points and contains the constant functions is the generalization of $P(x)$ ?

Can someone shed some light on this?

real-analysis functional-analysis analysis weierstrass-approximation

share|cite|improve this question

edited Mar 16 at 20:28

Sambo

2,3012532

asked Mar 16 at 20:01

IsaIsa

295519

$endgroup$

add a comment | 

$begingroup$

Weierstrass theorem.

Lef $f$ be a defined and continuous function in $[a,b]$. Given $epsilon>0,$ there exists a polynomial $P$ such that $vert f(x)-P(x)vert<epsilon,$ for all $xin[a,b].$

Stone-Weierstrass theorem.

Let $X$ be a topological compact space. If $F$ is an algebra of $C(X)$ that separate points and contains the constant functions then $F$ is dense with respect to the uniform convergence in $C(X).$

How is that an algebra $F$ of $C(X)$ that separate points and contains the constant functions is the generalization of $P(x)$ ?

Can someone shed some light on this?

real-analysis functional-analysis analysis weierstrass-approximation

share|cite|improve this question

edited Mar 16 at 20:28

Sambo

2,3012532

asked Mar 16 at 20:01

IsaIsa

295519

$endgroup$

share|cite|improve this question

edited Mar 16 at 20:28

Sambo

2,3012532

asked Mar 16 at 20:01

IsaIsa

295519

share|cite|improve this question

edited Mar 16 at 20:28

Sambo

2,3012532

asked Mar 16 at 20:01

IsaIsa

295519

edited Mar 16 at 20:28

Sambo

2,3012532

edited Mar 16 at 20:28

Sambo

2,3012532

asked Mar 16 at 20:01

IsaIsa

295519

asked Mar 16 at 20:01

IsaIsa

295519

2 Answers
2

active

oldest

votes

2

$begingroup$

The set of polynomials defined on $[a,b]$ is an algebra of $C([a,b])$ that separate points (for all $x,yin[a,b]$ such that $xneq y$, there exists a polynomial $P$ such that $P(x)neq P(y)$) and contains the constant functions.

share|cite|improve this answer

answered Mar 16 at 20:06

WillWill

5115

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your answer. In which part will this with respect to the uniform convergence in 𝐶(𝑋) be included?
  $endgroup$
  – Isa
  Mar 17 at 0:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The Weierstraß theorem states that there exists a sequence of polynomials which converges uniformly on $[a,b]$ to $f$. Indeed, you can rewrite your statement in this way: $exists$ polynomial $P$ such that $sup_xin[a,b]vert f(x)-P(x)vertlevarepsilon$.
  $endgroup$
  – Will
  Mar 17 at 8:54
  

  
  
  
  

add a comment | 

2

$begingroup$

Hint:

For $x_0, y_0 in [a,b]$ such that $x_0 ne y_0$ we consider the polynomial $p(t) = t-x_0$. Then $p(x_0) = 0$ but $p(y_0) = y_0 - x_0 ne 0$.

Hence polynomials on $[a,b]$ separate the points of $[a,b]$.

Also, constant functions are polynomials so...

share|cite|improve this answer

answered Mar 16 at 20:25

mechanodroidmechanodroid

29k62648

$endgroup$

add a comment | 

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2

$begingroup$

The set of polynomials defined on $[a,b]$ is an algebra of $C([a,b])$ that separate points (for all $x,yin[a,b]$ such that $xneq y$, there exists a polynomial $P$ such that $P(x)neq P(y)$) and contains the constant functions.

share|cite|improve this answer

answered Mar 16 at 20:06

WillWill

5115

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your answer. In which part will this with respect to the uniform convergence in 𝐶(𝑋) be included?
  $endgroup$
  – Isa
  Mar 17 at 0:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The Weierstraß theorem states that there exists a sequence of polynomials which converges uniformly on $[a,b]$ to $f$. Indeed, you can rewrite your statement in this way: $exists$ polynomial $P$ such that $sup_xin[a,b]vert f(x)-P(x)vertlevarepsilon$.
  $endgroup$
  – Will
  Mar 17 at 8:54
  

  
  
  
  

add a comment | 

2

$begingroup$

The set of polynomials defined on $[a,b]$ is an algebra of $C([a,b])$ that separate points (for all $x,yin[a,b]$ such that $xneq y$, there exists a polynomial $P$ such that $P(x)neq P(y)$) and contains the constant functions.

share|cite|improve this answer

answered Mar 16 at 20:06

WillWill

5115

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your answer. In which part will this with respect to the uniform convergence in 𝐶(𝑋) be included?
  $endgroup$
  – Isa
  Mar 17 at 0:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The Weierstraß theorem states that there exists a sequence of polynomials which converges uniformly on $[a,b]$ to $f$. Indeed, you can rewrite your statement in this way: $exists$ polynomial $P$ such that $sup_xin[a,b]vert f(x)-P(x)vertlevarepsilon$.
  $endgroup$
  – Will
  Mar 17 at 8:54
  

  
  
  
  

add a comment | 

$begingroup$

The set of polynomials defined on $[a,b]$ is an algebra of $C([a,b])$ that separate points (for all $x,yin[a,b]$ such that $xneq y$, there exists a polynomial $P$ such that $P(x)neq P(y)$) and contains the constant functions.

share|cite|improve this answer

answered Mar 16 at 20:06

WillWill

5115

$endgroup$

share|cite|improve this answer

answered Mar 16 at 20:06

WillWill

5115

answered Mar 16 at 20:06

WillWill

5115

answered Mar 16 at 20:06

WillWill

5115

2

$begingroup$

Hint:

For $x_0, y_0 in [a,b]$ such that $x_0 ne y_0$ we consider the polynomial $p(t) = t-x_0$. Then $p(x_0) = 0$ but $p(y_0) = y_0 - x_0 ne 0$.

Hence polynomials on $[a,b]$ separate the points of $[a,b]$.

Also, constant functions are polynomials so...

share|cite|improve this answer

answered Mar 16 at 20:25

mechanodroidmechanodroid

29k62648

$endgroup$

add a comment | 

2

$begingroup$

Hint:

For $x_0, y_0 in [a,b]$ such that $x_0 ne y_0$ we consider the polynomial $p(t) = t-x_0$. Then $p(x_0) = 0$ but $p(y_0) = y_0 - x_0 ne 0$.

Hence polynomials on $[a,b]$ separate the points of $[a,b]$.

Also, constant functions are polynomials so...

share|cite|improve this answer

answered Mar 16 at 20:25

mechanodroidmechanodroid

29k62648

$endgroup$

add a comment | 

$begingroup$

Hint:

For $x_0, y_0 in [a,b]$ such that $x_0 ne y_0$ we consider the polynomial $p(t) = t-x_0$. Then $p(x_0) = 0$ but $p(y_0) = y_0 - x_0 ne 0$.

Hence polynomials on $[a,b]$ separate the points of $[a,b]$.

Also, constant functions are polynomials so...

share|cite|improve this answer

answered Mar 16 at 20:25

mechanodroidmechanodroid

29k62648

$endgroup$

share|cite|improve this answer

answered Mar 16 at 20:25

mechanodroidmechanodroid

29k62648

answered Mar 16 at 20:25

mechanodroidmechanodroid

29k62648

answered Mar 16 at 20:25

mechanodroidmechanodroid

29k62648