Mixing
Lemma for Stone-Weierstrass
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During the proof of the Stone-Weierstrass Theorem there is a lemma given:
Lemma: Let $g_1$ and $g_2$ be in $mathcalA$ (an algebra of real-valued continuous functions which strongly separates points in an arbitrary compact metric space $M$). Then for every error $1/m$ there exist functions $g_3$ and $g_4$ in $mathcalA$ such that $|g_3-max(g_1,g_2)| le 1/m$ and $|g_4-min(g_1,g_2)| le 1/m$ for all points of $M$.
The proof of the lemma starts of by saying that it suffices to prove the following: if $g$ is in $mathcalA$ then there exists $g_1$ in $mathcalA$ with $|g_1-|g|| le 1/m$. I'm sure this is quite trivial but I can't justify why it's sufficient to prove this statement. Could someone explain why?
real-analysis analysis
asked Jul 16 at 15:46
Twiss013
11912
add a comment |Â
up vote
1
down vote
favorite
During the proof of the Stone-Weierstrass Theorem there is a lemma given:
Lemma: Let $g_1$ and $g_2$ be in $mathcalA$ (an algebra of real-valued continuous functions which strongly separates points in an arbitrary compact metric space $M$). Then for every error $1/m$ there exist functions $g_3$ and $g_4$ in $mathcalA$ such that $|g_3-max(g_1,g_2)| le 1/m$ and $|g_4-min(g_1,g_2)| le 1/m$ for all points of $M$.
The proof of the lemma starts of by saying that it suffices to prove the following: if $g$ is in $mathcalA$ then there exists $g_1$ in $mathcalA$ with $|g_1-|g|| le 1/m$. I'm sure this is quite trivial but I can't justify why it's sufficient to prove this statement. Could someone explain why?
real-analysis analysis
asked Jul 16 at 15:46
Twiss013
11912
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
During the proof of the Stone-Weierstrass Theorem there is a lemma given:
Lemma: Let $g_1$ and $g_2$ be in $mathcalA$ (an algebra of real-valued continuous functions which strongly separates points in an arbitrary compact metric space $M$). Then for every error $1/m$ there exist functions $g_3$ and $g_4$ in $mathcalA$ such that $|g_3-max(g_1,g_2)| le 1/m$ and $|g_4-min(g_1,g_2)| le 1/m$ for all points of $M$.
The proof of the lemma starts of by saying that it suffices to prove the following: if $g$ is in $mathcalA$ then there exists $g_1$ in $mathcalA$ with $|g_1-|g|| le 1/m$. I'm sure this is quite trivial but I can't justify why it's sufficient to prove this statement. Could someone explain why?
real-analysis analysis
asked Jul 16 at 15:46
Twiss013
11912
During the proof of the Stone-Weierstrass Theorem there is a lemma given:
Lemma: Let $g_1$ and $g_2$ be in $mathcalA$ (an algebra of real-valued continuous functions which strongly separates points in an arbitrary compact metric space $M$). Then for every error $1/m$ there exist functions $g_3$ and $g_4$ in $mathcalA$ such that $|g_3-max(g_1,g_2)| le 1/m$ and $|g_4-min(g_1,g_2)| le 1/m$ for all points of $M$.
The proof of the lemma starts of by saying that it suffices to prove the following: if $g$ is in $mathcalA$ then there exists $g_1$ in $mathcalA$ with $|g_1-|g|| le 1/m$. I'm sure this is quite trivial but I can't justify why it's sufficient to prove this statement. Could someone explain why?
real-analysis analysis
asked Jul 16 at 15:46
Twiss013
11912
asked Jul 16 at 15:46
Twiss013
11912
asked Jul 16 at 15:46
Twiss013
11912
asked Jul 16 at 15:46
Twiss013
11912
11912
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
You have $max(f,g)=frac12(f+g+|f-g|)$ and $min(f,g)=frac12(f+g-|f-g|)$. So if you can approximate $|f-g|$ well, you can approximate $min$ and $max$.
answered Jul 16 at 16:07
Kusma
1,127112
Is the absolute value of g unnecessary in the inequality then?
– Twiss013
Jul 16 at 16:24
It is necessary. You start with $h_1,h_2incal A$, then you find $Hin cal A$ such that $|H-|h_1-h_2||<frac1m$, then you have for $min(h_1,h_2)$ that $|frac12(h_1+h_2+H)-min(h_1,h_2)|=frac12|H-|h_1-h_2||<frac12m$ and (as $cal A$ is a vector space) $frac12(h+1+h_2+H)in cal A$.
– Kusma
Jul 16 at 16:36
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
You have $max(f,g)=frac12(f+g+|f-g|)$ and $min(f,g)=frac12(f+g-|f-g|)$. So if you can approximate $|f-g|$ well, you can approximate $min$ and $max$.
answered Jul 16 at 16:07
Kusma
1,127112
Is the absolute value of g unnecessary in the inequality then?
– Twiss013
Jul 16 at 16:24
It is necessary. You start with $h_1,h_2incal A$, then you find $Hin cal A$ such that $|H-|h_1-h_2||<frac1m$, then you have for $min(h_1,h_2)$ that $|frac12(h_1+h_2+H)-min(h_1,h_2)|=frac12|H-|h_1-h_2||<frac12m$ and (as $cal A$ is a vector space) $frac12(h+1+h_2+H)in cal A$.
– Kusma
Jul 16 at 16:36
add a comment |Â
up vote
3
down vote
accepted
You have $max(f,g)=frac12(f+g+|f-g|)$ and $min(f,g)=frac12(f+g-|f-g|)$. So if you can approximate $|f-g|$ well, you can approximate $min$ and $max$.
answered Jul 16 at 16:07
Kusma
1,127112
Is the absolute value of g unnecessary in the inequality then?
– Twiss013
Jul 16 at 16:24
It is necessary. You start with $h_1,h_2incal A$, then you find $Hin cal A$ such that $|H-|h_1-h_2||<frac1m$, then you have for $min(h_1,h_2)$ that $|frac12(h_1+h_2+H)-min(h_1,h_2)|=frac12|H-|h_1-h_2||<frac12m$ and (as $cal A$ is a vector space) $frac12(h+1+h_2+H)in cal A$.
– Kusma
Jul 16 at 16:36
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
You have $max(f,g)=frac12(f+g+|f-g|)$ and $min(f,g)=frac12(f+g-|f-g|)$. So if you can approximate $|f-g|$ well, you can approximate $min$ and $max$.
answered Jul 16 at 16:07
Kusma
1,127112
You have $max(f,g)=frac12(f+g+|f-g|)$ and $min(f,g)=frac12(f+g-|f-g|)$. So if you can approximate $|f-g|$ well, you can approximate $min$ and $max$.
answered Jul 16 at 16:07
Kusma
1,127112
answered Jul 16 at 16:07
Kusma
1,127112
answered Jul 16 at 16:07
Kusma
1,127112
answered Jul 16 at 16:07
Kusma
1,127112
1,127112
Is the absolute value of g unnecessary in the inequality then?
– Twiss013
Jul 16 at 16:24
It is necessary. You start with $h_1,h_2incal A$, then you find $Hin cal A$ such that $|H-|h_1-h_2||<frac1m$, then you have for $min(h_1,h_2)$ that $|frac12(h_1+h_2+H)-min(h_1,h_2)|=frac12|H-|h_1-h_2||<frac12m$ and (as $cal A$ is a vector space) $frac12(h+1+h_2+H)in cal A$.
– Kusma
Jul 16 at 16:36
add a comment |Â
Is the absolute value of g unnecessary in the inequality then?
– Twiss013
Jul 16 at 16:24
It is necessary. You start with $h_1,h_2incal A$, then you find $Hin cal A$ such that $|H-|h_1-h_2||<frac1m$, then you have for $min(h_1,h_2)$ that $|frac12(h_1+h_2+H)-min(h_1,h_2)|=frac12|H-|h_1-h_2||<frac12m$ and (as $cal A$ is a vector space) $frac12(h+1+h_2+H)in cal A$.
– Kusma
Jul 16 at 16:36
Is the absolute value of g unnecessary in the inequality then?
– Twiss013
Jul 16 at 16:24
Is the absolute value of g unnecessary in the inequality then?
– Twiss013
Jul 16 at 16:24
It is necessary. You start with $h_1,h_2incal A$, then you find $Hin cal A$ such that $|H-|h_1-h_2||<frac1m$, then you have for $min(h_1,h_2)$ that $|frac12(h_1+h_2+H)-min(h_1,h_2)|=frac12|H-|h_1-h_2||<frac12m$ and (as $cal A$ is a vector space) $frac12(h+1+h_2+H)in cal A$.
– Kusma
Jul 16 at 16:36
It is necessary. You start with $h_1,h_2incal A$, then you find $Hin cal A$ such that $|H-|h_1-h_2||<frac1m$, then you have for $min(h_1,h_2)$ that $|frac12(h_1+h_2+H)-min(h_1,h_2)|=frac12|H-|h_1-h_2||<frac12m$ and (as $cal A$ is a vector space) $frac12(h+1+h_2+H)in cal A$.
– Kusma
Jul 16 at 16:36
add a comment |Â
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