Hahn Banach Theorem implying existence of a nonzero linear functional taking 0 in a linear subspace Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Hahn Banach theorem with no dominating sublinear functionalWhy do we need the Hahn-Banach Theorem to extend a bounded linear functional?Incorrect proof of Hahn Banach TheoremTrouble Applying Hahn-Banach TheoremWhat kind of assumptions can be made from the Hahn-Banach theorem?Hahn Banach extension of linear functional $f$Consequence of the Hahn-Banach TheoremConfusion on application of Hahn Banach Theoremconsequence of Hahn-Banach theorem$F(y) = F(x)$ for aribtrary continuous linear functional $F$, then by Hahn-Banach $y=x$?
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Hahn Banach Theorem implying existence of a nonzero linear functional taking 0 in a linear subspace
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Hahn Banach theorem with no dominating sublinear functionalWhy do we need the Hahn-Banach Theorem to extend a bounded linear functional?Incorrect proof of Hahn Banach TheoremTrouble Applying Hahn-Banach TheoremWhat kind of assumptions can be made from the Hahn-Banach theorem?Hahn Banach extension of linear functional $f$Consequence of the Hahn-Banach TheoremConfusion on application of Hahn Banach Theoremconsequence of Hahn-Banach theorem$F(y) = F(x)$ for aribtrary continuous linear functional $F$, then by Hahn-Banach $y=x$?
$begingroup$
I am reading this paper. In the proof of theorem 1, it is stated
By the Hahn-Banach theorem, there is a bounded linear functional on $C(I_n)$, call it $L$, with the property that $Lne 0$ but $L(R) = L(S) = 0$.
$C(I_n)$ is space of continuous functions on $[0,1]^n$. $S$ is a linear subspace in it. $R$ is the closure of $S$.
Can you explain to me why this statement is true?
functional-analysis continuity unbounded-operators hahn-banach-theorem geometric-functional-analysis
asked Apr 2 at 1:33
ztyhztyh
468
$endgroup$
add a comment |
$begingroup$
I am reading this paper. In the proof of theorem 1, it is stated
By the Hahn-Banach theorem, there is a bounded linear functional on $C(I_n)$, call it $L$, with the property that $Lne 0$ but $L(R) = L(S) = 0$.
$C(I_n)$ is space of continuous functions on $[0,1]^n$. $S$ is a linear subspace in it. $R$ is the closure of $S$.
Can you explain to me why this statement is true?
functional-analysis continuity unbounded-operators hahn-banach-theorem geometric-functional-analysis
asked Apr 2 at 1:33
ztyhztyh
468
$endgroup$
$begingroup$
This is true as long as $R$ is not the entire space. What version of the Hahn-Banach Theorem do you know? There's one that's almost exactly this statement.
$endgroup$
– Jose27
Apr 2 at 1:59
$begingroup$
I was looking at Rudin 1991. Also Friedman's lemma and theorem. I see on Wikipedia they list something very similar as an important consequence.
$endgroup$
– ztyh
Apr 2 at 2:11
$begingroup$
They do assume $R$ is not all of $C(I_n)$.
$endgroup$
– ztyh
Apr 2 at 2:33
add a comment |
$begingroup$
I am reading this paper. In the proof of theorem 1, it is stated
By the Hahn-Banach theorem, there is a bounded linear functional on $C(I_n)$, call it $L$, with the property that $Lne 0$ but $L(R) = L(S) = 0$.
$C(I_n)$ is space of continuous functions on $[0,1]^n$. $S$ is a linear subspace in it. $R$ is the closure of $S$.
Can you explain to me why this statement is true?
functional-analysis continuity unbounded-operators hahn-banach-theorem geometric-functional-analysis
asked Apr 2 at 1:33
ztyhztyh
468
$endgroup$
I am reading this paper. In the proof of theorem 1, it is stated
By the Hahn-Banach theorem, there is a bounded linear functional on $C(I_n)$, call it $L$, with the property that $Lne 0$ but $L(R) = L(S) = 0$.
$C(I_n)$ is space of continuous functions on $[0,1]^n$. $S$ is a linear subspace in it. $R$ is the closure of $S$.
Can you explain to me why this statement is true?
functional-analysis continuity unbounded-operators hahn-banach-theorem geometric-functional-analysis
functional-analysis continuity unbounded-operators hahn-banach-theorem geometric-functional-analysis
asked Apr 2 at 1:33
ztyhztyh
468
asked Apr 2 at 1:33
ztyhztyh
468
asked Apr 2 at 1:33
ztyhztyh
468
asked Apr 2 at 1:33
ztyhztyh
468
asked Apr 2 at 1:33
ztyhztyh
468
468
$begingroup$
This is true as long as $R$ is not the entire space. What version of the Hahn-Banach Theorem do you know? There's one that's almost exactly this statement.
$endgroup$
– Jose27
Apr 2 at 1:59
$begingroup$
I was looking at Rudin 1991. Also Friedman's lemma and theorem. I see on Wikipedia they list something very similar as an important consequence.
$endgroup$
– ztyh
Apr 2 at 2:11
$begingroup$
They do assume $R$ is not all of $C(I_n)$.
$endgroup$
– ztyh
Apr 2 at 2:33
add a comment |
$begingroup$
This is true as long as $R$ is not the entire space. What version of the Hahn-Banach Theorem do you know? There's one that's almost exactly this statement.
$endgroup$
– Jose27
Apr 2 at 1:59
$begingroup$
I was looking at Rudin 1991. Also Friedman's lemma and theorem. I see on Wikipedia they list something very similar as an important consequence.
$endgroup$
– ztyh
Apr 2 at 2:11
$begingroup$
They do assume $R$ is not all of $C(I_n)$.
$endgroup$
– ztyh
Apr 2 at 2:33
$begingroup$
This is true as long as $R$ is not the entire space. What version of the Hahn-Banach Theorem do you know? There's one that's almost exactly this statement.
$endgroup$
– Jose27
Apr 2 at 1:59
$begingroup$
This is true as long as $R$ is not the entire space. What version of the Hahn-Banach Theorem do you know? There's one that's almost exactly this statement.
$endgroup$
– Jose27
Apr 2 at 1:59
$begingroup$
I was looking at Rudin 1991. Also Friedman's lemma and theorem. I see on Wikipedia they list something very similar as an important consequence.
$endgroup$
– ztyh
Apr 2 at 2:11
$begingroup$
I was looking at Rudin 1991. Also Friedman's lemma and theorem. I see on Wikipedia they list something very similar as an important consequence.
$endgroup$
– ztyh
Apr 2 at 2:11
$begingroup$
They do assume $R$ is not all of $C(I_n)$.
$endgroup$
– ztyh
Apr 2 at 2:33
$begingroup$
They do assume $R$ is not all of $C(I_n)$.
$endgroup$
– ztyh
Apr 2 at 2:33
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $M=f+af_0:ain mathbb R$ where $f_0$ is any fixed element not in $R$. Define $T(f+af_0)=a$. If we show that this is continuous on the space spanned by $R cup f_0$ we can use Hahn Banach Theorem to get a continuous linear functional which is $0$ on $R$ and has the value $1$ at $f_0$. I will let you verify that $T$ is well defined. Suppose $f_n+a_nf_0 to g$. If $(a_n)$ is unbounded it has a subsequence $a_n'$ converging to $pm infty$. Dividing by this we get $frac f_n' a_n' +f_0=0$ which shows that $-f_0$ is the limit of sequence from $R$ which is a contardiction. Hence $(a_n)$ is bounded and it has subsequence converging to some $a$. we then get $f_n+a_nf_0 to g=f+af_0$ for some $f in R$ and $a=T(g)=lim a_n' =lim T(f_n'+a_n'f_0)$. By arguing with subsequences we see that $T$ is continuous.
answered Apr 2 at 6:22
Kavi Rama MurthyKavi Rama Murthy
76.1k53370
$endgroup$
$begingroup$
What exactly do I have to say about $T$ being well defined?
$endgroup$
– ztyh
Apr 2 at 14:47
$begingroup$
Also can I just say $T(lim [f_n+a_nf_0])=lim a_n=lim T(f_n+a_nf_0)$ to show that it is continuous and then say $T$ is linear?
$endgroup$
– ztyh
Apr 2 at 15:31
1
$begingroup$
@ztyh No you cannot show continuity this way because we don't know that $lim a_n$ exists.
$endgroup$
– Kavi Rama Murthy
Apr 2 at 23:06
$begingroup$
Can I just use bounded iff continuous? When you prove $(a_n)$ cannot tend to $infty$, you can say $T$ is bounded and that is the end of the proof?
$endgroup$
– ztyh
Apr 3 at 22:32
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
Let $M=f+af_0:ain mathbb R$ where $f_0$ is any fixed element not in $R$. Define $T(f+af_0)=a$. If we show that this is continuous on the space spanned by $R cup f_0$ we can use Hahn Banach Theorem to get a continuous linear functional which is $0$ on $R$ and has the value $1$ at $f_0$. I will let you verify that $T$ is well defined. Suppose $f_n+a_nf_0 to g$. If $(a_n)$ is unbounded it has a subsequence $a_n'$ converging to $pm infty$. Dividing by this we get $frac f_n' a_n' +f_0=0$ which shows that $-f_0$ is the limit of sequence from $R$ which is a contardiction. Hence $(a_n)$ is bounded and it has subsequence converging to some $a$. we then get $f_n+a_nf_0 to g=f+af_0$ for some $f in R$ and $a=T(g)=lim a_n' =lim T(f_n'+a_n'f_0)$. By arguing with subsequences we see that $T$ is continuous.
answered Apr 2 at 6:22
Kavi Rama MurthyKavi Rama Murthy
76.1k53370
$endgroup$
$begingroup$
What exactly do I have to say about $T$ being well defined?
$endgroup$
– ztyh
Apr 2 at 14:47
$begingroup$
Also can I just say $T(lim [f_n+a_nf_0])=lim a_n=lim T(f_n+a_nf_0)$ to show that it is continuous and then say $T$ is linear?
$endgroup$
– ztyh
Apr 2 at 15:31
1
$begingroup$
@ztyh No you cannot show continuity this way because we don't know that $lim a_n$ exists.
$endgroup$
– Kavi Rama Murthy
Apr 2 at 23:06
$begingroup$
Can I just use bounded iff continuous? When you prove $(a_n)$ cannot tend to $infty$, you can say $T$ is bounded and that is the end of the proof?
$endgroup$
– ztyh
Apr 3 at 22:32
add a comment |
$begingroup$
Let $M=f+af_0:ain mathbb R$ where $f_0$ is any fixed element not in $R$. Define $T(f+af_0)=a$. If we show that this is continuous on the space spanned by $R cup f_0$ we can use Hahn Banach Theorem to get a continuous linear functional which is $0$ on $R$ and has the value $1$ at $f_0$. I will let you verify that $T$ is well defined. Suppose $f_n+a_nf_0 to g$. If $(a_n)$ is unbounded it has a subsequence $a_n'$ converging to $pm infty$. Dividing by this we get $frac f_n' a_n' +f_0=0$ which shows that $-f_0$ is the limit of sequence from $R$ which is a contardiction. Hence $(a_n)$ is bounded and it has subsequence converging to some $a$. we then get $f_n+a_nf_0 to g=f+af_0$ for some $f in R$ and $a=T(g)=lim a_n' =lim T(f_n'+a_n'f_0)$. By arguing with subsequences we see that $T$ is continuous.
answered Apr 2 at 6:22
Kavi Rama MurthyKavi Rama Murthy
76.1k53370
$endgroup$
$begingroup$
What exactly do I have to say about $T$ being well defined?
$endgroup$
– ztyh
Apr 2 at 14:47
$begingroup$
Also can I just say $T(lim [f_n+a_nf_0])=lim a_n=lim T(f_n+a_nf_0)$ to show that it is continuous and then say $T$ is linear?
$endgroup$
– ztyh
Apr 2 at 15:31
1
$begingroup$
@ztyh No you cannot show continuity this way because we don't know that $lim a_n$ exists.
$endgroup$
– Kavi Rama Murthy
Apr 2 at 23:06
$begingroup$
Can I just use bounded iff continuous? When you prove $(a_n)$ cannot tend to $infty$, you can say $T$ is bounded and that is the end of the proof?
$endgroup$
– ztyh
Apr 3 at 22:32
add a comment |
$begingroup$
Let $M=f+af_0:ain mathbb R$ where $f_0$ is any fixed element not in $R$. Define $T(f+af_0)=a$. If we show that this is continuous on the space spanned by $R cup f_0$ we can use Hahn Banach Theorem to get a continuous linear functional which is $0$ on $R$ and has the value $1$ at $f_0$. I will let you verify that $T$ is well defined. Suppose $f_n+a_nf_0 to g$. If $(a_n)$ is unbounded it has a subsequence $a_n'$ converging to $pm infty$. Dividing by this we get $frac f_n' a_n' +f_0=0$ which shows that $-f_0$ is the limit of sequence from $R$ which is a contardiction. Hence $(a_n)$ is bounded and it has subsequence converging to some $a$. we then get $f_n+a_nf_0 to g=f+af_0$ for some $f in R$ and $a=T(g)=lim a_n' =lim T(f_n'+a_n'f_0)$. By arguing with subsequences we see that $T$ is continuous.
answered Apr 2 at 6:22
Kavi Rama MurthyKavi Rama Murthy
76.1k53370
$endgroup$
Let $M=f+af_0:ain mathbb R$ where $f_0$ is any fixed element not in $R$. Define $T(f+af_0)=a$. If we show that this is continuous on the space spanned by $R cup f_0$ we can use Hahn Banach Theorem to get a continuous linear functional which is $0$ on $R$ and has the value $1$ at $f_0$. I will let you verify that $T$ is well defined. Suppose $f_n+a_nf_0 to g$. If $(a_n)$ is unbounded it has a subsequence $a_n'$ converging to $pm infty$. Dividing by this we get $frac f_n' a_n' +f_0=0$ which shows that $-f_0$ is the limit of sequence from $R$ which is a contardiction. Hence $(a_n)$ is bounded and it has subsequence converging to some $a$. we then get $f_n+a_nf_0 to g=f+af_0$ for some $f in R$ and $a=T(g)=lim a_n' =lim T(f_n'+a_n'f_0)$. By arguing with subsequences we see that $T$ is continuous.
answered Apr 2 at 6:22
Kavi Rama MurthyKavi Rama Murthy
76.1k53370
answered Apr 2 at 6:22
Kavi Rama MurthyKavi Rama Murthy
76.1k53370
answered Apr 2 at 6:22
Kavi Rama MurthyKavi Rama Murthy
76.1k53370
answered Apr 2 at 6:22
Kavi Rama MurthyKavi Rama Murthy
76.1k53370
76.1k53370
$begingroup$
What exactly do I have to say about $T$ being well defined?
$endgroup$
– ztyh
Apr 2 at 14:47
$begingroup$
Also can I just say $T(lim [f_n+a_nf_0])=lim a_n=lim T(f_n+a_nf_0)$ to show that it is continuous and then say $T$ is linear?
$endgroup$
– ztyh
Apr 2 at 15:31
1
$begingroup$
@ztyh No you cannot show continuity this way because we don't know that $lim a_n$ exists.
$endgroup$
– Kavi Rama Murthy
Apr 2 at 23:06
$begingroup$
Can I just use bounded iff continuous? When you prove $(a_n)$ cannot tend to $infty$, you can say $T$ is bounded and that is the end of the proof?
$endgroup$
– ztyh
Apr 3 at 22:32
add a comment |
$begingroup$
What exactly do I have to say about $T$ being well defined?
$endgroup$
– ztyh
Apr 2 at 14:47
$begingroup$
Also can I just say $T(lim [f_n+a_nf_0])=lim a_n=lim T(f_n+a_nf_0)$ to show that it is continuous and then say $T$ is linear?
$endgroup$
– ztyh
Apr 2 at 15:31
1
$begingroup$
@ztyh No you cannot show continuity this way because we don't know that $lim a_n$ exists.
$endgroup$
– Kavi Rama Murthy
Apr 2 at 23:06
$begingroup$
Can I just use bounded iff continuous? When you prove $(a_n)$ cannot tend to $infty$, you can say $T$ is bounded and that is the end of the proof?
$endgroup$
– ztyh
Apr 3 at 22:32
$begingroup$
What exactly do I have to say about $T$ being well defined?
$endgroup$
– ztyh
Apr 2 at 14:47
$begingroup$
What exactly do I have to say about $T$ being well defined?
$endgroup$
– ztyh
Apr 2 at 14:47
$begingroup$
Also can I just say $T(lim [f_n+a_nf_0])=lim a_n=lim T(f_n+a_nf_0)$ to show that it is continuous and then say $T$ is linear?
$endgroup$
– ztyh
Apr 2 at 15:31
$begingroup$
Also can I just say $T(lim [f_n+a_nf_0])=lim a_n=lim T(f_n+a_nf_0)$ to show that it is continuous and then say $T$ is linear?
$endgroup$
– ztyh
Apr 2 at 15:31
1
1
$begingroup$
@ztyh No you cannot show continuity this way because we don't know that $lim a_n$ exists.
$endgroup$
– Kavi Rama Murthy
Apr 2 at 23:06
$begingroup$
@ztyh No you cannot show continuity this way because we don't know that $lim a_n$ exists.
$endgroup$
– Kavi Rama Murthy
Apr 2 at 23:06
$begingroup$
Can I just use bounded iff continuous? When you prove $(a_n)$ cannot tend to $infty$, you can say $T$ is bounded and that is the end of the proof?
$endgroup$
– ztyh
Apr 3 at 22:32
$begingroup$
Can I just use bounded iff continuous? When you prove $(a_n)$ cannot tend to $infty$, you can say $T$ is bounded and that is the end of the proof?
$endgroup$
– ztyh
Apr 3 at 22:32
add a comment |
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$begingroup$
This is true as long as $R$ is not the entire space. What version of the Hahn-Banach Theorem do you know? There's one that's almost exactly this statement.
$endgroup$
– Jose27
Apr 2 at 1:59
$begingroup$
I was looking at Rudin 1991. Also Friedman's lemma and theorem. I see on Wikipedia they list something very similar as an important consequence.
$endgroup$
– ztyh
Apr 2 at 2:11
$begingroup$
They do assume $R$ is not all of $C(I_n)$.
$endgroup$
– ztyh
Apr 2 at 2:33