Stone -Weierstrass theorem
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let $A$ be a set of continuous functions over the closed interval $[0,1]$, which applies to the following conditions:
$ 1: forall f in A , forall x in [0,1] , f(x) geq 0 $
$2: forall f, g in A , f + g in A $
$3: forall x in [0,1], textbf{there is} f in A quad textbf{so that} f(x) > 0 $
By Stone -Weierstrass theorem , How can I Prove "that there is $ h in A$ so that $ forall x in [0,1], h(x) > 0$" ?.
Should I use the below theorem?
THEOREM:
Let A $subset C(K)$ such that
1) A is a subalgebra with unity 1
2) For each $ x_1, x_2 in K $ with $ x_1 neq x_2 $, exist $f in A$ such that f($ x_1$) $ neq $ f($ x_2$).
Then $ overline A = C(K)$, where C(K) is the space of continuous functions over a compact space
real-analysis functional-analysis
asked Nov 18 at 19:34
joe
874
add a comment |
up vote
0
down vote
favorite
let $A$ be a set of continuous functions over the closed interval $[0,1]$, which applies to the following conditions:
$ 1: forall f in A , forall x in [0,1] , f(x) geq 0 $
$2: forall f, g in A , f + g in A $
$3: forall x in [0,1], textbf{there is} f in A quad textbf{so that} f(x) > 0 $
By Stone -Weierstrass theorem , How can I Prove "that there is $ h in A$ so that $ forall x in [0,1], h(x) > 0$" ?.
Should I use the below theorem?
THEOREM:
Let A $subset C(K)$ such that
1) A is a subalgebra with unity 1
2) For each $ x_1, x_2 in K $ with $ x_1 neq x_2 $, exist $f in A$ such that f($ x_1$) $ neq $ f($ x_2$).
Then $ overline A = C(K)$, where C(K) is the space of continuous functions over a compact space
real-analysis functional-analysis
asked Nov 18 at 19:34
joe
874
1
It does not seem like you can use the theorem. Consider $A := {f: [0,1] to mathbb R|, forall x: f(x) = c geq 0}$. This $A$ does not satisfy condition 2) of your theorem, but it does satisfy 1,2 and 3 above.
– N.Beck
Nov 18 at 19:47
You recieved an answer to your question. Is it what you needed? If so, you should upvote and accept it.
– 5xum
Nov 27 at 10:51
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
let $A$ be a set of continuous functions over the closed interval $[0,1]$, which applies to the following conditions:
$ 1: forall f in A , forall x in [0,1] , f(x) geq 0 $
$2: forall f, g in A , f + g in A $
$3: forall x in [0,1], textbf{there is} f in A quad textbf{so that} f(x) > 0 $
By Stone -Weierstrass theorem , How can I Prove "that there is $ h in A$ so that $ forall x in [0,1], h(x) > 0$" ?.
Should I use the below theorem?
THEOREM:
Let A $subset C(K)$ such that
1) A is a subalgebra with unity 1
2) For each $ x_1, x_2 in K $ with $ x_1 neq x_2 $, exist $f in A$ such that f($ x_1$) $ neq $ f($ x_2$).
Then $ overline A = C(K)$, where C(K) is the space of continuous functions over a compact space
real-analysis functional-analysis
asked Nov 18 at 19:34
joe
874
let $A$ be a set of continuous functions over the closed interval $[0,1]$, which applies to the following conditions:
$ 1: forall f in A , forall x in [0,1] , f(x) geq 0 $
$2: forall f, g in A , f + g in A $
$3: forall x in [0,1], textbf{there is} f in A quad textbf{so that} f(x) > 0 $
By Stone -Weierstrass theorem , How can I Prove "that there is $ h in A$ so that $ forall x in [0,1], h(x) > 0$" ?.
Should I use the below theorem?
THEOREM:
Let A $subset C(K)$ such that
1) A is a subalgebra with unity 1
2) For each $ x_1, x_2 in K $ with $ x_1 neq x_2 $, exist $f in A$ such that f($ x_1$) $ neq $ f($ x_2$).
Then $ overline A = C(K)$, where C(K) is the space of continuous functions over a compact space
real-analysis functional-analysis
real-analysis functional-analysis
asked Nov 18 at 19:34
joe
874
asked Nov 18 at 19:34
joe
874
edited Nov 18 at 19:39
asked Nov 18 at 19:34
joe
874
asked Nov 18 at 19:34
joe
874
asked Nov 18 at 19:34
joe
874
874
1
It does not seem like you can use the theorem. Consider $A := {f: [0,1] to mathbb R|, forall x: f(x) = c geq 0}$. This $A$ does not satisfy condition 2) of your theorem, but it does satisfy 1,2 and 3 above.
– N.Beck
Nov 18 at 19:47
You recieved an answer to your question. Is it what you needed? If so, you should upvote and accept it.
– 5xum
Nov 27 at 10:51
add a comment |
1
It does not seem like you can use the theorem. Consider $A := {f: [0,1] to mathbb R|, forall x: f(x) = c geq 0}$. This $A$ does not satisfy condition 2) of your theorem, but it does satisfy 1,2 and 3 above.
– N.Beck
Nov 18 at 19:47
You recieved an answer to your question. Is it what you needed? If so, you should upvote and accept it.
– 5xum
Nov 27 at 10:51
1
1
It does not seem like you can use the theorem. Consider $A := {f: [0,1] to mathbb R|, forall x: f(x) = c geq 0}$. This $A$ does not satisfy condition 2) of your theorem, but it does satisfy 1,2 and 3 above.
– N.Beck
Nov 18 at 19:47
It does not seem like you can use the theorem. Consider $A := {f: [0,1] to mathbb R|, forall x: f(x) = c geq 0}$. This $A$ does not satisfy condition 2) of your theorem, but it does satisfy 1,2 and 3 above.
– N.Beck
Nov 18 at 19:47
You recieved an answer to your question. Is it what you needed? If so, you should upvote and accept it.
– 5xum
Nov 27 at 10:51
You recieved an answer to your question. Is it what you needed? If so, you should upvote and accept it.
– 5xum
Nov 27 at 10:51
add a comment |
1 Answer
1
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2
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I would use compactness: find for every $x$ a function $f_x$ as in 3), also take $delta_x>0$ such that $f_x$ is positive inside the $delta_x$-nbd of $x$. Now apply compactness to reduce to finitely many function and add them.
Addendum: there are finitely many $x_1$, ..., $x_n$ such that $[0,1]$ is covered by the corresponding $delta_{x_i}$-nbds. Now take $f=f_{x_1}+cdots+f_{x_n}$.
answered Nov 18 at 19:58
hartkp
1,27965
please, give me more guidance about "apply compactness to reduce to finitely many function and add them. "? Thanks.
– joe
Nov 19 at 5:05
I have expanded the answer a bit.
– hartkp
Nov 19 at 9:00
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
I would use compactness: find for every $x$ a function $f_x$ as in 3), also take $delta_x>0$ such that $f_x$ is positive inside the $delta_x$-nbd of $x$. Now apply compactness to reduce to finitely many function and add them.
Addendum: there are finitely many $x_1$, ..., $x_n$ such that $[0,1]$ is covered by the corresponding $delta_{x_i}$-nbds. Now take $f=f_{x_1}+cdots+f_{x_n}$.
answered Nov 18 at 19:58
hartkp
1,27965
please, give me more guidance about "apply compactness to reduce to finitely many function and add them. "? Thanks.
– joe
Nov 19 at 5:05
I have expanded the answer a bit.
– hartkp
Nov 19 at 9:00
add a comment |
up vote
2
down vote
I would use compactness: find for every $x$ a function $f_x$ as in 3), also take $delta_x>0$ such that $f_x$ is positive inside the $delta_x$-nbd of $x$. Now apply compactness to reduce to finitely many function and add them.
Addendum: there are finitely many $x_1$, ..., $x_n$ such that $[0,1]$ is covered by the corresponding $delta_{x_i}$-nbds. Now take $f=f_{x_1}+cdots+f_{x_n}$.
answered Nov 18 at 19:58
hartkp
1,27965
please, give me more guidance about "apply compactness to reduce to finitely many function and add them. "? Thanks.
– joe
Nov 19 at 5:05
I have expanded the answer a bit.
– hartkp
Nov 19 at 9:00
add a comment |
up vote
2
down vote
up vote
2
down vote
I would use compactness: find for every $x$ a function $f_x$ as in 3), also take $delta_x>0$ such that $f_x$ is positive inside the $delta_x$-nbd of $x$. Now apply compactness to reduce to finitely many function and add them.
Addendum: there are finitely many $x_1$, ..., $x_n$ such that $[0,1]$ is covered by the corresponding $delta_{x_i}$-nbds. Now take $f=f_{x_1}+cdots+f_{x_n}$.
answered Nov 18 at 19:58
hartkp
1,27965
I would use compactness: find for every $x$ a function $f_x$ as in 3), also take $delta_x>0$ such that $f_x$ is positive inside the $delta_x$-nbd of $x$. Now apply compactness to reduce to finitely many function and add them.
Addendum: there are finitely many $x_1$, ..., $x_n$ such that $[0,1]$ is covered by the corresponding $delta_{x_i}$-nbds. Now take $f=f_{x_1}+cdots+f_{x_n}$.
answered Nov 18 at 19:58
hartkp
1,27965
edited Nov 19 at 9:00
answered Nov 18 at 19:58
hartkp
1,27965
answered Nov 18 at 19:58
hartkp
1,27965
answered Nov 18 at 19:58
hartkp
1,27965
1,27965
please, give me more guidance about "apply compactness to reduce to finitely many function and add them. "? Thanks.
– joe
Nov 19 at 5:05
I have expanded the answer a bit.
– hartkp
Nov 19 at 9:00
add a comment |
please, give me more guidance about "apply compactness to reduce to finitely many function and add them. "? Thanks.
– joe
Nov 19 at 5:05
I have expanded the answer a bit.
– hartkp
Nov 19 at 9:00
please, give me more guidance about "apply compactness to reduce to finitely many function and add them. "? Thanks.
– joe
Nov 19 at 5:05
please, give me more guidance about "apply compactness to reduce to finitely many function and add them. "? Thanks.
– joe
Nov 19 at 5:05
I have expanded the answer a bit.
– hartkp
Nov 19 at 9:00
I have expanded the answer a bit.
– hartkp
Nov 19 at 9:00
add a comment |
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1
It does not seem like you can use the theorem. Consider $A := {f: [0,1] to mathbb R|, forall x: f(x) = c geq 0}$. This $A$ does not satisfy condition 2) of your theorem, but it does satisfy 1,2 and 3 above.
– N.Beck
Nov 18 at 19:47
You recieved an answer to your question. Is it what you needed? If so, you should upvote and accept it.
– 5xum
Nov 27 at 10:51