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Linear functional $psi:X to mathbbR$ is $W$-weakly continuous if and only if it belongs to $W$
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I'm trying to understand the proof of the following proposition from Royden and Fitzpatrick (Chapter 14) - the step that I highlight is in which I would like confirmation on my justification. I provide the full proof for completeness.
Proposition 5: Let $X$ be a linear space and W a subspace of $X^#$ (space of linear functionals on X). Then a linear functional $psi: X rightarrow mathbbR$ is W-weakly continuous if and only if it belongs to W.
Definition of "$mathcalF$-weakly continuous":
If $mathcalF$ is any collection of real-valued functions on a set $X$, the $mathcalF$-weak topology is defined to be the weakest topology on $X$ for which all the functions in $mathcalF$ are continuous. A function that is continuous with respect to the $mathcalF$-weak topology is called $mathcalF$-weakly continuous.
Proof:
By definition of the W-weak topology, each linear functional in $W$ is $W$-weakly continuous. It remains to prove the converse. Suppose the linear functional $psi: X rightarrow mathbbR$ is $W$-weakly continuous . By the continuity of $psi$ at $0$, there is a neighborhood $mathcalN$ of $0$ for which $|psi(x)| = |psi(x)- psi(0)| < 1$ if $x in mathcalN$. There is a neighborhood in the base for the $W$-topology at $0$ that is contained in $mathcalN$. Choose $epsilon > 0$ and $psi_1, ...,psi_n$ in $W$ for which $mathcalN_epsilon,psi_1,...,psi_n subset mathcalN$. Thus
$$ |psi(x)| < 1 text if |psi_k(x)|<epsilon text for all 1le k le n$$
By the linearity of $psi$, and $psi_k$'s, we have the inclusion $bigcap_k=1^n ker psi_k subset ker psi$.
By Proposition 4, $psi$ is a linear combination of $psi_k$. Therefore, since W is a linear subspace, $psi in W$. $$tag*$Box$$$
How I understand this is as follows: The step I highlight implies $bigcap_k=1^n x: psi_k(x) < epsilon subset x: psi(x) < 1$. The $epsilon$ has been fixed, and the $psi_k$ rely on said epsilon. Is the idea that by linearity, for any $c in mathbbR$,
beginalign*
bigcap_k=1^n x: psi_k(x) < epsilon subset x: psi(x) < 1 Leftrightarrow& bigcap_k=1^n x: psi_k(cx) < epsilon subset x: psi(cx) < 1\ Leftrightarrow& bigcap_k=1^n x: psi_k(x) < fracepsilonc subset x: psi(x) < frac1c
endalign*
Taking c arbitrarily large will yield $bigcap_k=1^n ker psi_k subset ker psi$ ?
general-topology functional-analysis dual-spaces weak-topology
edited Aug 1 at 12:26
anomaly
16.3k42561
asked Jul 31 at 15:48
ertl
427110
add a comment |Â
up vote
0
down vote
favorite
I'm trying to understand the proof of the following proposition from Royden and Fitzpatrick (Chapter 14) - the step that I highlight is in which I would like confirmation on my justification. I provide the full proof for completeness.
Proposition 5: Let $X$ be a linear space and W a subspace of $X^#$ (space of linear functionals on X). Then a linear functional $psi: X rightarrow mathbbR$ is W-weakly continuous if and only if it belongs to W.
Definition of "$mathcalF$-weakly continuous":
If $mathcalF$ is any collection of real-valued functions on a set $X$, the $mathcalF$-weak topology is defined to be the weakest topology on $X$ for which all the functions in $mathcalF$ are continuous. A function that is continuous with respect to the $mathcalF$-weak topology is called $mathcalF$-weakly continuous.
Proof:
By definition of the W-weak topology, each linear functional in $W$ is $W$-weakly continuous. It remains to prove the converse. Suppose the linear functional $psi: X rightarrow mathbbR$ is $W$-weakly continuous . By the continuity of $psi$ at $0$, there is a neighborhood $mathcalN$ of $0$ for which $|psi(x)| = |psi(x)- psi(0)| < 1$ if $x in mathcalN$. There is a neighborhood in the base for the $W$-topology at $0$ that is contained in $mathcalN$. Choose $epsilon > 0$ and $psi_1, ...,psi_n$ in $W$ for which $mathcalN_epsilon,psi_1,...,psi_n subset mathcalN$. Thus
$$ |psi(x)| < 1 text if |psi_k(x)|<epsilon text for all 1le k le n$$
By the linearity of $psi$, and $psi_k$'s, we have the inclusion $bigcap_k=1^n ker psi_k subset ker psi$.
By Proposition 4, $psi$ is a linear combination of $psi_k$. Therefore, since W is a linear subspace, $psi in W$. $$tag*$Box$$$
How I understand this is as follows: The step I highlight implies $bigcap_k=1^n x: psi_k(x) < epsilon subset x: psi(x) < 1$. The $epsilon$ has been fixed, and the $psi_k$ rely on said epsilon. Is the idea that by linearity, for any $c in mathbbR$,
beginalign*
bigcap_k=1^n x: psi_k(x) < epsilon subset x: psi(x) < 1 Leftrightarrow& bigcap_k=1^n x: psi_k(cx) < epsilon subset x: psi(cx) < 1\ Leftrightarrow& bigcap_k=1^n x: psi_k(x) < fracepsilonc subset x: psi(x) < frac1c
endalign*
Taking c arbitrarily large will yield $bigcap_k=1^n ker psi_k subset ker psi$ ?
general-topology functional-analysis dual-spaces weak-topology
edited Aug 1 at 12:26
anomaly
16.3k42561
asked Jul 31 at 15:48
ertl
427110
I've made some edits to improve formatting. In particular, you should be careful about writing lines that are too long in mathmode (especially with the way you had formatted Proposition 5). You might also want to define $mathcalN_epsilon,psi_1,...,psi_n$. I can guess what it should be but it would help other readers if the notation used is defined.
– Rhys Steele
Aug 1 at 11:45
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm trying to understand the proof of the following proposition from Royden and Fitzpatrick (Chapter 14) - the step that I highlight is in which I would like confirmation on my justification. I provide the full proof for completeness.
Proposition 5: Let $X$ be a linear space and W a subspace of $X^#$ (space of linear functionals on X). Then a linear functional $psi: X rightarrow mathbbR$ is W-weakly continuous if and only if it belongs to W.
Definition of "$mathcalF$-weakly continuous":
If $mathcalF$ is any collection of real-valued functions on a set $X$, the $mathcalF$-weak topology is defined to be the weakest topology on $X$ for which all the functions in $mathcalF$ are continuous. A function that is continuous with respect to the $mathcalF$-weak topology is called $mathcalF$-weakly continuous.
Proof:
By definition of the W-weak topology, each linear functional in $W$ is $W$-weakly continuous. It remains to prove the converse. Suppose the linear functional $psi: X rightarrow mathbbR$ is $W$-weakly continuous . By the continuity of $psi$ at $0$, there is a neighborhood $mathcalN$ of $0$ for which $|psi(x)| = |psi(x)- psi(0)| < 1$ if $x in mathcalN$. There is a neighborhood in the base for the $W$-topology at $0$ that is contained in $mathcalN$. Choose $epsilon > 0$ and $psi_1, ...,psi_n$ in $W$ for which $mathcalN_epsilon,psi_1,...,psi_n subset mathcalN$. Thus
$$ |psi(x)| < 1 text if |psi_k(x)|<epsilon text for all 1le k le n$$
By the linearity of $psi$, and $psi_k$'s, we have the inclusion $bigcap_k=1^n ker psi_k subset ker psi$.
By Proposition 4, $psi$ is a linear combination of $psi_k$. Therefore, since W is a linear subspace, $psi in W$. $$tag*$Box$$$
How I understand this is as follows: The step I highlight implies $bigcap_k=1^n x: psi_k(x) < epsilon subset x: psi(x) < 1$. The $epsilon$ has been fixed, and the $psi_k$ rely on said epsilon. Is the idea that by linearity, for any $c in mathbbR$,
beginalign*
bigcap_k=1^n x: psi_k(x) < epsilon subset x: psi(x) < 1 Leftrightarrow& bigcap_k=1^n x: psi_k(cx) < epsilon subset x: psi(cx) < 1\ Leftrightarrow& bigcap_k=1^n x: psi_k(x) < fracepsilonc subset x: psi(x) < frac1c
endalign*
Taking c arbitrarily large will yield $bigcap_k=1^n ker psi_k subset ker psi$ ?
general-topology functional-analysis dual-spaces weak-topology
edited Aug 1 at 12:26
anomaly
16.3k42561
asked Jul 31 at 15:48
ertl
427110
I'm trying to understand the proof of the following proposition from Royden and Fitzpatrick (Chapter 14) - the step that I highlight is in which I would like confirmation on my justification. I provide the full proof for completeness.
Proposition 5: Let $X$ be a linear space and W a subspace of $X^#$ (space of linear functionals on X). Then a linear functional $psi: X rightarrow mathbbR$ is W-weakly continuous if and only if it belongs to W.
Definition of "$mathcalF$-weakly continuous":
If $mathcalF$ is any collection of real-valued functions on a set $X$, the $mathcalF$-weak topology is defined to be the weakest topology on $X$ for which all the functions in $mathcalF$ are continuous. A function that is continuous with respect to the $mathcalF$-weak topology is called $mathcalF$-weakly continuous.
Proof:
By definition of the W-weak topology, each linear functional in $W$ is $W$-weakly continuous. It remains to prove the converse. Suppose the linear functional $psi: X rightarrow mathbbR$ is $W$-weakly continuous . By the continuity of $psi$ at $0$, there is a neighborhood $mathcalN$ of $0$ for which $|psi(x)| = |psi(x)- psi(0)| < 1$ if $x in mathcalN$. There is a neighborhood in the base for the $W$-topology at $0$ that is contained in $mathcalN$. Choose $epsilon > 0$ and $psi_1, ...,psi_n$ in $W$ for which $mathcalN_epsilon,psi_1,...,psi_n subset mathcalN$. Thus
$$ |psi(x)| < 1 text if |psi_k(x)|<epsilon text for all 1le k le n$$
By the linearity of $psi$, and $psi_k$'s, we have the inclusion $bigcap_k=1^n ker psi_k subset ker psi$.
By Proposition 4, $psi$ is a linear combination of $psi_k$. Therefore, since W is a linear subspace, $psi in W$. $$tag*$Box$$$
How I understand this is as follows: The step I highlight implies $bigcap_k=1^n x: psi_k(x) < epsilon subset x: psi(x) < 1$. The $epsilon$ has been fixed, and the $psi_k$ rely on said epsilon. Is the idea that by linearity, for any $c in mathbbR$,
beginalign*
bigcap_k=1^n x: psi_k(x) < epsilon subset x: psi(x) < 1 Leftrightarrow& bigcap_k=1^n x: psi_k(cx) < epsilon subset x: psi(cx) < 1\ Leftrightarrow& bigcap_k=1^n x: psi_k(x) < fracepsilonc subset x: psi(x) < frac1c
endalign*
Taking c arbitrarily large will yield $bigcap_k=1^n ker psi_k subset ker psi$ ?
general-topology functional-analysis dual-spaces weak-topology
edited Aug 1 at 12:26
anomaly
16.3k42561
asked Jul 31 at 15:48
ertl
427110
edited Aug 1 at 12:26
anomaly
16.3k42561
edited Aug 1 at 12:26
anomaly
16.3k42561
edited Aug 1 at 12:26
anomaly
16.3k42561
16.3k42561
asked Jul 31 at 15:48
ertl
427110
asked Jul 31 at 15:48
ertl
427110
asked Jul 31 at 15:48
ertl
427110
427110
I've made some edits to improve formatting. In particular, you should be careful about writing lines that are too long in mathmode (especially with the way you had formatted Proposition 5). You might also want to define $mathcalN_epsilon,psi_1,...,psi_n$. I can guess what it should be but it would help other readers if the notation used is defined.
– Rhys Steele
Aug 1 at 11:45
add a comment |Â
I've made some edits to improve formatting. In particular, you should be careful about writing lines that are too long in mathmode (especially with the way you had formatted Proposition 5). You might also want to define $mathcalN_epsilon,psi_1,...,psi_n$. I can guess what it should be but it would help other readers if the notation used is defined.
– Rhys Steele
Aug 1 at 11:45
I've made some edits to improve formatting. In particular, you should be careful about writing lines that are too long in mathmode (especially with the way you had formatted Proposition 5). You might also want to define $mathcalN_epsilon,psi_1,...,psi_n$. I can guess what it should be but it would help other readers if the notation used is defined.
– Rhys Steele
Aug 1 at 11:45
I've made some edits to improve formatting. In particular, you should be careful about writing lines that are too long in mathmode (especially with the way you had formatted Proposition 5). You might also want to define $mathcalN_epsilon,psi_1,...,psi_n$. I can guess what it should be but it would help other readers if the notation used is defined.
– Rhys Steele
Aug 1 at 11:45
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
You have the right kind of idea but I think that it is somewhat unclear from your argument why taking $c$ arbitrarily large will give the desired result (since this involves taking a limit of sets). One way to justify this is to conclude from what you've written that
beginalign*
bigcap_c in mathbbN bigg(bigcap_k=1^n x: psi_k(x) < fracepsiloncbigg) subset bigcap_c in mathbbN bigg(x: psi(x) < frac1c bigg)
endalign*
and then convince yourself that the left hand side is $bigcap_k=1^n ker psi_k$ and the right hand side is $ker psi$. If I'm honest, this argument feels unnecessarily complicated to me.
A much simpler way to do this kind of argument (that avoids set builder notations entirely) is to just take $x in bigcap_k=1^n ker psi_k$ and show it's in $ker psi$, using essentially the same idea as is used in your argument. Indeed, since $0 = |psi_k(cx)| < epsilon$ for all $c > 0$, $|psi(cx)| < 1$ and so $|psi(x)| < frac1c$. Since $c>0$ was arbitrary, $|psi(x)| = 0$ so $x in ker psi$.
answered Aug 1 at 12:12
Rhys Steele
5,5551827
add a comment |Â
up vote
1
down vote
Your argument is correct. But (perhaps) a better argument is the following: let $psi_k(x)=0$ for $1leq k leq n$. Let $n$ be any positive integer. Then $psi_k(Nx)=Npsi_k(x)=0$ for $1leq k leq n$. Hence $psi (Nx) <1$ and $psi (x) <frac 1 N$. Since this holds for all $N$ we get $psi (x) leq 0$. Changing $x$ to $-x$ we get $psi (x)geq 0$. Hence $psi (x)=0$.
answered Aug 1 at 12:13
Kavi Rama Murthy
19.5k2829
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You have the right kind of idea but I think that it is somewhat unclear from your argument why taking $c$ arbitrarily large will give the desired result (since this involves taking a limit of sets). One way to justify this is to conclude from what you've written that
beginalign*
bigcap_c in mathbbN bigg(bigcap_k=1^n x: psi_k(x) < fracepsiloncbigg) subset bigcap_c in mathbbN bigg(x: psi(x) < frac1c bigg)
endalign*
and then convince yourself that the left hand side is $bigcap_k=1^n ker psi_k$ and the right hand side is $ker psi$. If I'm honest, this argument feels unnecessarily complicated to me.
A much simpler way to do this kind of argument (that avoids set builder notations entirely) is to just take $x in bigcap_k=1^n ker psi_k$ and show it's in $ker psi$, using essentially the same idea as is used in your argument. Indeed, since $0 = |psi_k(cx)| < epsilon$ for all $c > 0$, $|psi(cx)| < 1$ and so $|psi(x)| < frac1c$. Since $c>0$ was arbitrary, $|psi(x)| = 0$ so $x in ker psi$.
answered Aug 1 at 12:12
Rhys Steele
5,5551827
add a comment |Â
up vote
1
down vote
accepted
You have the right kind of idea but I think that it is somewhat unclear from your argument why taking $c$ arbitrarily large will give the desired result (since this involves taking a limit of sets). One way to justify this is to conclude from what you've written that
beginalign*
bigcap_c in mathbbN bigg(bigcap_k=1^n x: psi_k(x) < fracepsiloncbigg) subset bigcap_c in mathbbN bigg(x: psi(x) < frac1c bigg)
endalign*
and then convince yourself that the left hand side is $bigcap_k=1^n ker psi_k$ and the right hand side is $ker psi$. If I'm honest, this argument feels unnecessarily complicated to me.
A much simpler way to do this kind of argument (that avoids set builder notations entirely) is to just take $x in bigcap_k=1^n ker psi_k$ and show it's in $ker psi$, using essentially the same idea as is used in your argument. Indeed, since $0 = |psi_k(cx)| < epsilon$ for all $c > 0$, $|psi(cx)| < 1$ and so $|psi(x)| < frac1c$. Since $c>0$ was arbitrary, $|psi(x)| = 0$ so $x in ker psi$.
answered Aug 1 at 12:12
Rhys Steele
5,5551827
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You have the right kind of idea but I think that it is somewhat unclear from your argument why taking $c$ arbitrarily large will give the desired result (since this involves taking a limit of sets). One way to justify this is to conclude from what you've written that
beginalign*
bigcap_c in mathbbN bigg(bigcap_k=1^n x: psi_k(x) < fracepsiloncbigg) subset bigcap_c in mathbbN bigg(x: psi(x) < frac1c bigg)
endalign*
and then convince yourself that the left hand side is $bigcap_k=1^n ker psi_k$ and the right hand side is $ker psi$. If I'm honest, this argument feels unnecessarily complicated to me.
A much simpler way to do this kind of argument (that avoids set builder notations entirely) is to just take $x in bigcap_k=1^n ker psi_k$ and show it's in $ker psi$, using essentially the same idea as is used in your argument. Indeed, since $0 = |psi_k(cx)| < epsilon$ for all $c > 0$, $|psi(cx)| < 1$ and so $|psi(x)| < frac1c$. Since $c>0$ was arbitrary, $|psi(x)| = 0$ so $x in ker psi$.
answered Aug 1 at 12:12
Rhys Steele
5,5551827
You have the right kind of idea but I think that it is somewhat unclear from your argument why taking $c$ arbitrarily large will give the desired result (since this involves taking a limit of sets). One way to justify this is to conclude from what you've written that
beginalign*
bigcap_c in mathbbN bigg(bigcap_k=1^n x: psi_k(x) < fracepsiloncbigg) subset bigcap_c in mathbbN bigg(x: psi(x) < frac1c bigg)
endalign*
and then convince yourself that the left hand side is $bigcap_k=1^n ker psi_k$ and the right hand side is $ker psi$. If I'm honest, this argument feels unnecessarily complicated to me.
A much simpler way to do this kind of argument (that avoids set builder notations entirely) is to just take $x in bigcap_k=1^n ker psi_k$ and show it's in $ker psi$, using essentially the same idea as is used in your argument. Indeed, since $0 = |psi_k(cx)| < epsilon$ for all $c > 0$, $|psi(cx)| < 1$ and so $|psi(x)| < frac1c$. Since $c>0$ was arbitrary, $|psi(x)| = 0$ so $x in ker psi$.
answered Aug 1 at 12:12
Rhys Steele
5,5551827
answered Aug 1 at 12:12
Rhys Steele
5,5551827
answered Aug 1 at 12:12
Rhys Steele
5,5551827
answered Aug 1 at 12:12
Rhys Steele
5,5551827
5,5551827
add a comment |Â
add a comment |Â
up vote
1
down vote
Your argument is correct. But (perhaps) a better argument is the following: let $psi_k(x)=0$ for $1leq k leq n$. Let $n$ be any positive integer. Then $psi_k(Nx)=Npsi_k(x)=0$ for $1leq k leq n$. Hence $psi (Nx) <1$ and $psi (x) <frac 1 N$. Since this holds for all $N$ we get $psi (x) leq 0$. Changing $x$ to $-x$ we get $psi (x)geq 0$. Hence $psi (x)=0$.
answered Aug 1 at 12:13
Kavi Rama Murthy
19.5k2829
add a comment |Â
up vote
1
down vote
Your argument is correct. But (perhaps) a better argument is the following: let $psi_k(x)=0$ for $1leq k leq n$. Let $n$ be any positive integer. Then $psi_k(Nx)=Npsi_k(x)=0$ for $1leq k leq n$. Hence $psi (Nx) <1$ and $psi (x) <frac 1 N$. Since this holds for all $N$ we get $psi (x) leq 0$. Changing $x$ to $-x$ we get $psi (x)geq 0$. Hence $psi (x)=0$.
answered Aug 1 at 12:13
Kavi Rama Murthy
19.5k2829
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Your argument is correct. But (perhaps) a better argument is the following: let $psi_k(x)=0$ for $1leq k leq n$. Let $n$ be any positive integer. Then $psi_k(Nx)=Npsi_k(x)=0$ for $1leq k leq n$. Hence $psi (Nx) <1$ and $psi (x) <frac 1 N$. Since this holds for all $N$ we get $psi (x) leq 0$. Changing $x$ to $-x$ we get $psi (x)geq 0$. Hence $psi (x)=0$.
answered Aug 1 at 12:13
Kavi Rama Murthy
19.5k2829
Your argument is correct. But (perhaps) a better argument is the following: let $psi_k(x)=0$ for $1leq k leq n$. Let $n$ be any positive integer. Then $psi_k(Nx)=Npsi_k(x)=0$ for $1leq k leq n$. Hence $psi (Nx) <1$ and $psi (x) <frac 1 N$. Since this holds for all $N$ we get $psi (x) leq 0$. Changing $x$ to $-x$ we get $psi (x)geq 0$. Hence $psi (x)=0$.
answered Aug 1 at 12:13
Kavi Rama Murthy
19.5k2829
answered Aug 1 at 12:13
Kavi Rama Murthy
19.5k2829
answered Aug 1 at 12:13
Kavi Rama Murthy
19.5k2829
answered Aug 1 at 12:13
Kavi Rama Murthy
19.5k2829
19.5k2829
add a comment |Â
add a comment |Â
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I've made some edits to improve formatting. In particular, you should be careful about writing lines that are too long in mathmode (especially with the way you had formatted Proposition 5). You might also want to define $mathcalN_epsilon,psi_1,...,psi_n$. I can guess what it should be but it would help other readers if the notation used is defined.
– Rhys Steele
Aug 1 at 11:45