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Giving a proof to the weak compactness of the unity ball in a reflexive normed space
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So, I am trying to prove that if $X$ is a normed and reflexive space, its unity ball, that is $B$=$lbrace xin X : parallel xparallel =1rbrace$, is weakly compact.
For that matter, I have already proven that:
$J:X longrightarrow X''$ the natural embedding is continuous provided that $X$ has the $tau_{d}$ topology (that is, the initial topology of $X'$) and $X''$ with $tau_{d}^{''}*$
If $X$ is reflexive, $J$ is an homeomorphism (with respect the previous topologies).
My idea is to use Alaoglu-Bourbaki's theorem,which stablishes that the unity ball in $X'$ is weakly* compact, and use that the image of a compact is a compact through a continuous function.
Therefore, my problem is: if $B$=$lbrace xin X : parallel xparallel =1rbrace$ is a subset of $X'$, how do I relate it? I mean, I just do not know how to finish the argument, which I feel to be very close to me.
Also, I have checked Using the Banach-Alaoglu Theorem to show that if $X$ is reflexive then $B_X$ is weakly compact. but I think it uses the algebraic dual, or I just don't fully understand his proof.
Thank you in advance.
general-topology functional-analysis normed-spaces reflexive-space
asked 2 days ago
Juan Zaragoza Chichell
11
New contributor
Juan Zaragoza Chichell is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
favorite
So, I am trying to prove that if $X$ is a normed and reflexive space, its unity ball, that is $B$=$lbrace xin X : parallel xparallel =1rbrace$, is weakly compact.
For that matter, I have already proven that:
$J:X longrightarrow X''$ the natural embedding is continuous provided that $X$ has the $tau_{d}$ topology (that is, the initial topology of $X'$) and $X''$ with $tau_{d}^{''}*$
If $X$ is reflexive, $J$ is an homeomorphism (with respect the previous topologies).
My idea is to use Alaoglu-Bourbaki's theorem,which stablishes that the unity ball in $X'$ is weakly* compact, and use that the image of a compact is a compact through a continuous function.
Therefore, my problem is: if $B$=$lbrace xin X : parallel xparallel =1rbrace$ is a subset of $X'$, how do I relate it? I mean, I just do not know how to finish the argument, which I feel to be very close to me.
Also, I have checked Using the Banach-Alaoglu Theorem to show that if $X$ is reflexive then $B_X$ is weakly compact. but I think it uses the algebraic dual, or I just don't fully understand his proof.
Thank you in advance.
general-topology functional-analysis normed-spaces reflexive-space
asked 2 days ago
Juan Zaragoza Chichell
11
New contributor
Juan Zaragoza Chichell is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
This is a theorem of Kakutani, proved in the answer. I have not seen a proof that does not use dual spaces.
– Matematleta
2 days ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
So, I am trying to prove that if $X$ is a normed and reflexive space, its unity ball, that is $B$=$lbrace xin X : parallel xparallel =1rbrace$, is weakly compact.
For that matter, I have already proven that:
$J:X longrightarrow X''$ the natural embedding is continuous provided that $X$ has the $tau_{d}$ topology (that is, the initial topology of $X'$) and $X''$ with $tau_{d}^{''}*$
If $X$ is reflexive, $J$ is an homeomorphism (with respect the previous topologies).
My idea is to use Alaoglu-Bourbaki's theorem,which stablishes that the unity ball in $X'$ is weakly* compact, and use that the image of a compact is a compact through a continuous function.
Therefore, my problem is: if $B$=$lbrace xin X : parallel xparallel =1rbrace$ is a subset of $X'$, how do I relate it? I mean, I just do not know how to finish the argument, which I feel to be very close to me.
Also, I have checked Using the Banach-Alaoglu Theorem to show that if $X$ is reflexive then $B_X$ is weakly compact. but I think it uses the algebraic dual, or I just don't fully understand his proof.
Thank you in advance.
general-topology functional-analysis normed-spaces reflexive-space
asked 2 days ago
Juan Zaragoza Chichell
11
New contributor
Juan Zaragoza Chichell is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
So, I am trying to prove that if $X$ is a normed and reflexive space, its unity ball, that is $B$=$lbrace xin X : parallel xparallel =1rbrace$, is weakly compact.
For that matter, I have already proven that:
$J:X longrightarrow X''$ the natural embedding is continuous provided that $X$ has the $tau_{d}$ topology (that is, the initial topology of $X'$) and $X''$ with $tau_{d}^{''}*$
If $X$ is reflexive, $J$ is an homeomorphism (with respect the previous topologies).
My idea is to use Alaoglu-Bourbaki's theorem,which stablishes that the unity ball in $X'$ is weakly* compact, and use that the image of a compact is a compact through a continuous function.
Therefore, my problem is: if $B$=$lbrace xin X : parallel xparallel =1rbrace$ is a subset of $X'$, how do I relate it? I mean, I just do not know how to finish the argument, which I feel to be very close to me.
Also, I have checked Using the Banach-Alaoglu Theorem to show that if $X$ is reflexive then $B_X$ is weakly compact. but I think it uses the algebraic dual, or I just don't fully understand his proof.
Thank you in advance.
general-topology functional-analysis normed-spaces reflexive-space
general-topology functional-analysis normed-spaces reflexive-space
asked 2 days ago
Juan Zaragoza Chichell
11
New contributor
Juan Zaragoza Chichell is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Juan Zaragoza Chichell
11
New contributor
Juan Zaragoza Chichell is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Juan Zaragoza Chichell
11
New contributor
Juan Zaragoza Chichell is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Juan Zaragoza Chichell
11
asked 2 days ago
Juan Zaragoza Chichell
11
11
New contributor
Juan Zaragoza Chichell is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Juan Zaragoza Chichell is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Juan Zaragoza Chichell is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
This is a theorem of Kakutani, proved in the answer. I have not seen a proof that does not use dual spaces.
– Matematleta
2 days ago
add a comment |
This is a theorem of Kakutani, proved in the answer. I have not seen a proof that does not use dual spaces.
– Matematleta
2 days ago
This is a theorem of Kakutani, proved in the answer. I have not seen a proof that does not use dual spaces.
– Matematleta
2 days ago
This is a theorem of Kakutani, proved in the answer. I have not seen a proof that does not use dual spaces.
– Matematleta
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
Here is a rather detailed proof.
If $X$ is reflexive, the natural embedding $J:Xto X''$
is an isometric isomorphism. In particular, the weak topology on $X$, that is, the topology $sigma(X,X')$, is identical to the weak topology on $X''$, that is, the topology $sigma(X'',X''')$. Let $B_X$ denote the unit ball of $X$. Then $J(B_X)$ is the unit ball of $X''$. By Alaoglu's theorem (Bourbaki had nothing to do with it), the unit ball of $X''$ is compact in the $sigma(X'',I(X'))$ topology, where $I$ is the natural embedding of $X'$ into $X'''$. But $I:X'to X'''$ is an injective isometry, so if a set is compact in the $sigma(X'',I(X'))$ topology then clearly it is also compact in the larger topology $sigma(X'',X''')$. Therefore, the unit ball of $X''$ is compact in the $sigma(X'',X''')$ topology, and therefore its inverse image under the homomorphism $J$, namely, $B_X$, is compact in the weak topology of $X$.
answered 2 days ago
uniquesolution
8,631823
I am sorry for asking this, but I'm definitely not proficient in topology or functional analysis: why is the weak topology on $X$ identical to the weak topology on $X''$ provided J is an homeomorphism? Also, where do you use the weak* topology? That last thing might be a notation issue.
– Juan Zaragoza Chichell
yesterday
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Here is a rather detailed proof.
If $X$ is reflexive, the natural embedding $J:Xto X''$
is an isometric isomorphism. In particular, the weak topology on $X$, that is, the topology $sigma(X,X')$, is identical to the weak topology on $X''$, that is, the topology $sigma(X'',X''')$. Let $B_X$ denote the unit ball of $X$. Then $J(B_X)$ is the unit ball of $X''$. By Alaoglu's theorem (Bourbaki had nothing to do with it), the unit ball of $X''$ is compact in the $sigma(X'',I(X'))$ topology, where $I$ is the natural embedding of $X'$ into $X'''$. But $I:X'to X'''$ is an injective isometry, so if a set is compact in the $sigma(X'',I(X'))$ topology then clearly it is also compact in the larger topology $sigma(X'',X''')$. Therefore, the unit ball of $X''$ is compact in the $sigma(X'',X''')$ topology, and therefore its inverse image under the homomorphism $J$, namely, $B_X$, is compact in the weak topology of $X$.
answered 2 days ago
uniquesolution
8,631823
I am sorry for asking this, but I'm definitely not proficient in topology or functional analysis: why is the weak topology on $X$ identical to the weak topology on $X''$ provided J is an homeomorphism? Also, where do you use the weak* topology? That last thing might be a notation issue.
– Juan Zaragoza Chichell
yesterday
add a comment |
up vote
1
down vote
Here is a rather detailed proof.
If $X$ is reflexive, the natural embedding $J:Xto X''$
is an isometric isomorphism. In particular, the weak topology on $X$, that is, the topology $sigma(X,X')$, is identical to the weak topology on $X''$, that is, the topology $sigma(X'',X''')$. Let $B_X$ denote the unit ball of $X$. Then $J(B_X)$ is the unit ball of $X''$. By Alaoglu's theorem (Bourbaki had nothing to do with it), the unit ball of $X''$ is compact in the $sigma(X'',I(X'))$ topology, where $I$ is the natural embedding of $X'$ into $X'''$. But $I:X'to X'''$ is an injective isometry, so if a set is compact in the $sigma(X'',I(X'))$ topology then clearly it is also compact in the larger topology $sigma(X'',X''')$. Therefore, the unit ball of $X''$ is compact in the $sigma(X'',X''')$ topology, and therefore its inverse image under the homomorphism $J$, namely, $B_X$, is compact in the weak topology of $X$.
answered 2 days ago
uniquesolution
8,631823
I am sorry for asking this, but I'm definitely not proficient in topology or functional analysis: why is the weak topology on $X$ identical to the weak topology on $X''$ provided J is an homeomorphism? Also, where do you use the weak* topology? That last thing might be a notation issue.
– Juan Zaragoza Chichell
yesterday
add a comment |
up vote
1
down vote
up vote
1
down vote
Here is a rather detailed proof.
If $X$ is reflexive, the natural embedding $J:Xto X''$
is an isometric isomorphism. In particular, the weak topology on $X$, that is, the topology $sigma(X,X')$, is identical to the weak topology on $X''$, that is, the topology $sigma(X'',X''')$. Let $B_X$ denote the unit ball of $X$. Then $J(B_X)$ is the unit ball of $X''$. By Alaoglu's theorem (Bourbaki had nothing to do with it), the unit ball of $X''$ is compact in the $sigma(X'',I(X'))$ topology, where $I$ is the natural embedding of $X'$ into $X'''$. But $I:X'to X'''$ is an injective isometry, so if a set is compact in the $sigma(X'',I(X'))$ topology then clearly it is also compact in the larger topology $sigma(X'',X''')$. Therefore, the unit ball of $X''$ is compact in the $sigma(X'',X''')$ topology, and therefore its inverse image under the homomorphism $J$, namely, $B_X$, is compact in the weak topology of $X$.
answered 2 days ago
uniquesolution
8,631823
Here is a rather detailed proof.
If $X$ is reflexive, the natural embedding $J:Xto X''$
is an isometric isomorphism. In particular, the weak topology on $X$, that is, the topology $sigma(X,X')$, is identical to the weak topology on $X''$, that is, the topology $sigma(X'',X''')$. Let $B_X$ denote the unit ball of $X$. Then $J(B_X)$ is the unit ball of $X''$. By Alaoglu's theorem (Bourbaki had nothing to do with it), the unit ball of $X''$ is compact in the $sigma(X'',I(X'))$ topology, where $I$ is the natural embedding of $X'$ into $X'''$. But $I:X'to X'''$ is an injective isometry, so if a set is compact in the $sigma(X'',I(X'))$ topology then clearly it is also compact in the larger topology $sigma(X'',X''')$. Therefore, the unit ball of $X''$ is compact in the $sigma(X'',X''')$ topology, and therefore its inverse image under the homomorphism $J$, namely, $B_X$, is compact in the weak topology of $X$.
answered 2 days ago
uniquesolution
8,631823
edited 2 days ago
answered 2 days ago
uniquesolution
8,631823
answered 2 days ago
uniquesolution
8,631823
answered 2 days ago
uniquesolution
8,631823
8,631823
I am sorry for asking this, but I'm definitely not proficient in topology or functional analysis: why is the weak topology on $X$ identical to the weak topology on $X''$ provided J is an homeomorphism? Also, where do you use the weak* topology? That last thing might be a notation issue.
– Juan Zaragoza Chichell
yesterday
add a comment |
I am sorry for asking this, but I'm definitely not proficient in topology or functional analysis: why is the weak topology on $X$ identical to the weak topology on $X''$ provided J is an homeomorphism? Also, where do you use the weak* topology? That last thing might be a notation issue.
– Juan Zaragoza Chichell
yesterday
I am sorry for asking this, but I'm definitely not proficient in topology or functional analysis: why is the weak topology on $X$ identical to the weak topology on $X''$ provided J is an homeomorphism? Also, where do you use the weak* topology? That last thing might be a notation issue.
– Juan Zaragoza Chichell
yesterday
I am sorry for asking this, but I'm definitely not proficient in topology or functional analysis: why is the weak topology on $X$ identical to the weak topology on $X''$ provided J is an homeomorphism? Also, where do you use the weak* topology? That last thing might be a notation issue.
– Juan Zaragoza Chichell
yesterday
add a comment |
Juan Zaragoza Chichell is a new contributor. Be nice, and check out our Code of Conduct.
Juan Zaragoza Chichell is a new contributor. Be nice, and check out our Code of Conduct.
Juan Zaragoza Chichell is a new contributor. Be nice, and check out our Code of Conduct.
Juan Zaragoza Chichell is a new contributor. Be nice, and check out our Code of Conduct.
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This is a theorem of Kakutani, proved in the answer. I have not seen a proof that does not use dual spaces.
– Matematleta
2 days ago