Why the reflexivity of a Banach set depend on the bijectivity of $Jx(f)=f(x)$?
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Let $E$ a Banach set. We say that $E$ is reflexive if $$J: Eto E^''$$
where $$Jx(f)=f(x)$$ is bijective (where $E''$ is the topological bidual). Now, I know if there is a bijective isometry $Eto F$, then $E$ and $F$ are the same (in a sense that is still not clear... but anyway). Now in Brezis functional analysis book (analyse fonctionnelle published by Dunod) page 43 remark 13, they say that : "We must use $J$ in the definition of reflexivity because there are spaces $E$ such that there is an onto isometry $Eto E''$ but $E$ is not reflexive".
I thought that if there is a bijective isometry between two spaces $A$ and $B$, then $A$ and $B$ are the same... So why do we must talk $J$ in the definition of reflexivity ? (since if there is a bijective isometry between two spaces they are the same).
general-topology functional-analysis
edited Aug 25 at 0:32
Bernard
111k635103
asked Aug 25 at 0:06
Henri
1176
add a comment |Â
up vote
1
down vote
favorite
Let $E$ a Banach set. We say that $E$ is reflexive if $$J: Eto E^''$$
where $$Jx(f)=f(x)$$ is bijective (where $E''$ is the topological bidual). Now, I know if there is a bijective isometry $Eto F$, then $E$ and $F$ are the same (in a sense that is still not clear... but anyway). Now in Brezis functional analysis book (analyse fonctionnelle published by Dunod) page 43 remark 13, they say that : "We must use $J$ in the definition of reflexivity because there are spaces $E$ such that there is an onto isometry $Eto E''$ but $E$ is not reflexive".
I thought that if there is a bijective isometry between two spaces $A$ and $B$, then $A$ and $B$ are the same... So why do we must talk $J$ in the definition of reflexivity ? (since if there is a bijective isometry between two spaces they are the same).
general-topology functional-analysis
edited Aug 25 at 0:32
Bernard
111k635103
asked Aug 25 at 0:06
Henri
1176
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $E$ a Banach set. We say that $E$ is reflexive if $$J: Eto E^''$$
where $$Jx(f)=f(x)$$ is bijective (where $E''$ is the topological bidual). Now, I know if there is a bijective isometry $Eto F$, then $E$ and $F$ are the same (in a sense that is still not clear... but anyway). Now in Brezis functional analysis book (analyse fonctionnelle published by Dunod) page 43 remark 13, they say that : "We must use $J$ in the definition of reflexivity because there are spaces $E$ such that there is an onto isometry $Eto E''$ but $E$ is not reflexive".
I thought that if there is a bijective isometry between two spaces $A$ and $B$, then $A$ and $B$ are the same... So why do we must talk $J$ in the definition of reflexivity ? (since if there is a bijective isometry between two spaces they are the same).
general-topology functional-analysis
edited Aug 25 at 0:32
Bernard
111k635103
asked Aug 25 at 0:06
Henri
1176
Let $E$ a Banach set. We say that $E$ is reflexive if $$J: Eto E^''$$
where $$Jx(f)=f(x)$$ is bijective (where $E''$ is the topological bidual). Now, I know if there is a bijective isometry $Eto F$, then $E$ and $F$ are the same (in a sense that is still not clear... but anyway). Now in Brezis functional analysis book (analyse fonctionnelle published by Dunod) page 43 remark 13, they say that : "We must use $J$ in the definition of reflexivity because there are spaces $E$ such that there is an onto isometry $Eto E''$ but $E$ is not reflexive".
I thought that if there is a bijective isometry between two spaces $A$ and $B$, then $A$ and $B$ are the same... So why do we must talk $J$ in the definition of reflexivity ? (since if there is a bijective isometry between two spaces they are the same).
general-topology functional-analysis
edited Aug 25 at 0:32
Bernard
111k635103
asked Aug 25 at 0:06
Henri
1176
edited Aug 25 at 0:32
Bernard
111k635103
edited Aug 25 at 0:32
Bernard
111k635103
edited Aug 25 at 0:32
Bernard
111k635103
111k635103
asked Aug 25 at 0:06
Henri
1176
asked Aug 25 at 0:06
Henri
1176
asked Aug 25 at 0:06
Henri
1176
1176
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
Reflexivity [in the $j:Xrightarrow X^**$ sense]is equivalent to weak compactness of the unit ball. The requirement that this map be onto is quite specific.
answered Aug 25 at 0:29
ncmathsadist
41.4k257100
Thank you for your answer. Do you agree that it's equivalent to a reflexivity in $k:xto X^**$ where $k(x)(f)=2f(x)$ for example ? i.e. we don't need an isometry, just a homeomorphism that preserve balls.
– Henri
Aug 31 at 21:29
I mean $k:Xto X^**$ (not $k:xto X^**$)
– Henri
Sep 1 at 8:46
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Reflexivity [in the $j:Xrightarrow X^**$ sense]is equivalent to weak compactness of the unit ball. The requirement that this map be onto is quite specific.
answered Aug 25 at 0:29
ncmathsadist
41.4k257100
Thank you for your answer. Do you agree that it's equivalent to a reflexivity in $k:xto X^**$ where $k(x)(f)=2f(x)$ for example ? i.e. we don't need an isometry, just a homeomorphism that preserve balls.
– Henri
Aug 31 at 21:29
I mean $k:Xto X^**$ (not $k:xto X^**$)
– Henri
Sep 1 at 8:46
add a comment |Â
up vote
1
down vote
Reflexivity [in the $j:Xrightarrow X^**$ sense]is equivalent to weak compactness of the unit ball. The requirement that this map be onto is quite specific.
answered Aug 25 at 0:29
ncmathsadist
41.4k257100
Thank you for your answer. Do you agree that it's equivalent to a reflexivity in $k:xto X^**$ where $k(x)(f)=2f(x)$ for example ? i.e. we don't need an isometry, just a homeomorphism that preserve balls.
– Henri
Aug 31 at 21:29
I mean $k:Xto X^**$ (not $k:xto X^**$)
– Henri
Sep 1 at 8:46
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Reflexivity [in the $j:Xrightarrow X^**$ sense]is equivalent to weak compactness of the unit ball. The requirement that this map be onto is quite specific.
answered Aug 25 at 0:29
ncmathsadist
41.4k257100
Reflexivity [in the $j:Xrightarrow X^**$ sense]is equivalent to weak compactness of the unit ball. The requirement that this map be onto is quite specific.
answered Aug 25 at 0:29
ncmathsadist
41.4k257100
answered Aug 25 at 0:29
ncmathsadist
41.4k257100
answered Aug 25 at 0:29
ncmathsadist
41.4k257100
answered Aug 25 at 0:29
ncmathsadist
41.4k257100
41.4k257100
Thank you for your answer. Do you agree that it's equivalent to a reflexivity in $k:xto X^**$ where $k(x)(f)=2f(x)$ for example ? i.e. we don't need an isometry, just a homeomorphism that preserve balls.
– Henri
Aug 31 at 21:29
I mean $k:Xto X^**$ (not $k:xto X^**$)
– Henri
Sep 1 at 8:46
add a comment |Â
Thank you for your answer. Do you agree that it's equivalent to a reflexivity in $k:xto X^**$ where $k(x)(f)=2f(x)$ for example ? i.e. we don't need an isometry, just a homeomorphism that preserve balls.
– Henri
Aug 31 at 21:29
I mean $k:Xto X^**$ (not $k:xto X^**$)
– Henri
Sep 1 at 8:46
Thank you for your answer. Do you agree that it's equivalent to a reflexivity in $k:xto X^**$ where $k(x)(f)=2f(x)$ for example ? i.e. we don't need an isometry, just a homeomorphism that preserve balls.
– Henri
Aug 31 at 21:29
Thank you for your answer. Do you agree that it's equivalent to a reflexivity in $k:xto X^**$ where $k(x)(f)=2f(x)$ for example ? i.e. we don't need an isometry, just a homeomorphism that preserve balls.
– Henri
Aug 31 at 21:29
I mean $k:Xto X^**$ (not $k:xto X^**$)
– Henri
Sep 1 at 8:46
I mean $k:Xto X^**$ (not $k:xto X^**$)
– Henri
Sep 1 at 8:46
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2893682%2fwhy-the-reflexivity-of-a-banach-set-depend-on-the-bijectivity-of-jxf-fx%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
