Mixing
Dualization map is surjective
1
$begingroup$
I am practicing for my exam and I want to solve the following problem.
Let $X,Y$ be normed reflexive spaces. Show that the "Dualization map" $':B(X,Y)to B(Y',X')$, $Tmapsto T'$ is surjective
I want to use reflexivity since $X''$ and $X$, likewise for $Y$, are isometrically isomorphic. But I get stuck. Has anybody an idea?
functional-analysis dual-spaces
asked Jan 21 at 17:03
JamesJames
936318
$endgroup$
add a comment |
1
$begingroup$
I am practicing for my exam and I want to solve the following problem.
Let $X,Y$ be normed reflexive spaces. Show that the "Dualization map" $':B(X,Y)to B(Y',X')$, $Tmapsto T'$ is surjective
I want to use reflexivity since $X''$ and $X$, likewise for $Y$, are isometrically isomorphic. But I get stuck. Has anybody an idea?
functional-analysis dual-spaces
asked Jan 21 at 17:03
JamesJames
936318
$endgroup$
$begingroup$
Maybe you should try and consider that $X’$ and $Y’$ are reflexive and consider two dualizations starting from $Y’ rightarrow X’$.
$endgroup$
– Mindlack
Jan 21 at 17:06
$begingroup$
But I don't know whether $X'$ is reflexive, since $X$ is not Banach
$endgroup$
– James
Jan 21 at 17:18
$begingroup$
If you are using the usual definition, if $X$ is reflexive, $X$ is a Banach, no? en.m.wikipedia.org/wiki/Reflexive_space
$endgroup$
– Mindlack
Jan 21 at 17:22
$begingroup$
You are right :)
$endgroup$
– James
Jan 21 at 17:23
add a comment |
1
1
1
$begingroup$
I am practicing for my exam and I want to solve the following problem.
Let $X,Y$ be normed reflexive spaces. Show that the "Dualization map" $':B(X,Y)to B(Y',X')$, $Tmapsto T'$ is surjective
I want to use reflexivity since $X''$ and $X$, likewise for $Y$, are isometrically isomorphic. But I get stuck. Has anybody an idea?
functional-analysis dual-spaces
asked Jan 21 at 17:03
JamesJames
936318
$endgroup$
I am practicing for my exam and I want to solve the following problem.
Let $X,Y$ be normed reflexive spaces. Show that the "Dualization map" $':B(X,Y)to B(Y',X')$, $Tmapsto T'$ is surjective
I want to use reflexivity since $X''$ and $X$, likewise for $Y$, are isometrically isomorphic. But I get stuck. Has anybody an idea?
functional-analysis dual-spaces
functional-analysis dual-spaces
asked Jan 21 at 17:03
JamesJames
936318
asked Jan 21 at 17:03
JamesJames
936318
asked Jan 21 at 17:03
JamesJames
936318
asked Jan 21 at 17:03
JamesJames
936318
asked Jan 21 at 17:03
JamesJames
936318
936318
$begingroup$
Maybe you should try and consider that $X’$ and $Y’$ are reflexive and consider two dualizations starting from $Y’ rightarrow X’$.
$endgroup$
– Mindlack
Jan 21 at 17:06
$begingroup$
But I don't know whether $X'$ is reflexive, since $X$ is not Banach
$endgroup$
– James
Jan 21 at 17:18
$begingroup$
If you are using the usual definition, if $X$ is reflexive, $X$ is a Banach, no? en.m.wikipedia.org/wiki/Reflexive_space
$endgroup$
– Mindlack
Jan 21 at 17:22
$begingroup$
You are right :)
$endgroup$
– James
Jan 21 at 17:23
add a comment |
$begingroup$
Maybe you should try and consider that $X’$ and $Y’$ are reflexive and consider two dualizations starting from $Y’ rightarrow X’$.
$endgroup$
– Mindlack
Jan 21 at 17:06
$begingroup$
But I don't know whether $X'$ is reflexive, since $X$ is not Banach
$endgroup$
– James
Jan 21 at 17:18
$begingroup$
If you are using the usual definition, if $X$ is reflexive, $X$ is a Banach, no? en.m.wikipedia.org/wiki/Reflexive_space
$endgroup$
– Mindlack
Jan 21 at 17:22
$begingroup$
You are right :)
$endgroup$
– James
Jan 21 at 17:23
$begingroup$
Maybe you should try and consider that $X’$ and $Y’$ are reflexive and consider two dualizations starting from $Y’ rightarrow X’$.
$endgroup$
– Mindlack
Jan 21 at 17:06
$begingroup$
Maybe you should try and consider that $X’$ and $Y’$ are reflexive and consider two dualizations starting from $Y’ rightarrow X’$.
$endgroup$
– Mindlack
Jan 21 at 17:06
$begingroup$
But I don't know whether $X'$ is reflexive, since $X$ is not Banach
$endgroup$
– James
Jan 21 at 17:18
$begingroup$
But I don't know whether $X'$ is reflexive, since $X$ is not Banach
$endgroup$
– James
Jan 21 at 17:18
$begingroup$
If you are using the usual definition, if $X$ is reflexive, $X$ is a Banach, no? en.m.wikipedia.org/wiki/Reflexive_space
$endgroup$
– Mindlack
Jan 21 at 17:22
$begingroup$
If you are using the usual definition, if $X$ is reflexive, $X$ is a Banach, no? en.m.wikipedia.org/wiki/Reflexive_space
$endgroup$
– Mindlack
Jan 21 at 17:22
$begingroup$
You are right :)
$endgroup$
– James
Jan 21 at 17:23
$begingroup$
You are right :)
$endgroup$
– James
Jan 21 at 17:23
add a comment |
1 Answer
1
active
oldest
votes
1
$begingroup$
Hint: for every $U'in B(Y',
X') $
Consider $U"=(U')':X"=Xrightarrow Y"=Y$.
answered Jan 21 at 17:21
Tsemo AristideTsemo Aristide
59.1k11445
$endgroup$
$begingroup$
Okay so I can show that $U''(x'')(y')=U''(iota_X(x))(y')=iota_X(x)circ U'(y')=U'(y')(x)=y'circ U(x)=i_Y(U(x))(y')$ but I don't really see how this helps
$endgroup$
– James
Jan 21 at 17:37
$begingroup$
You can compute $(U") '=U' $.
$endgroup$
– Tsemo Aristide
Jan 21 at 18:08
$begingroup$
I don't see what you mean
$endgroup$
– James
Jan 21 at 19:16
$begingroup$
$(U") '(y') (x) =y'(U"(x)) =(U"(x)) (y') $, identify $U"(x) $ with an element of the bidual. $(U"(x))(y') =x((U'(y')) =U'(y') (x) $.
$endgroup$
– Tsemo Aristide
Jan 21 at 19:39
add a comment |
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1 Answer
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1
$begingroup$
Hint: for every $U'in B(Y',
X') $
Consider $U"=(U')':X"=Xrightarrow Y"=Y$.
answered Jan 21 at 17:21
Tsemo AristideTsemo Aristide
59.1k11445
$endgroup$
$begingroup$
Okay so I can show that $U''(x'')(y')=U''(iota_X(x))(y')=iota_X(x)circ U'(y')=U'(y')(x)=y'circ U(x)=i_Y(U(x))(y')$ but I don't really see how this helps
$endgroup$
– James
Jan 21 at 17:37
$begingroup$
You can compute $(U") '=U' $.
$endgroup$
– Tsemo Aristide
Jan 21 at 18:08
$begingroup$
I don't see what you mean
$endgroup$
– James
Jan 21 at 19:16
$begingroup$
$(U") '(y') (x) =y'(U"(x)) =(U"(x)) (y') $, identify $U"(x) $ with an element of the bidual. $(U"(x))(y') =x((U'(y')) =U'(y') (x) $.
$endgroup$
– Tsemo Aristide
Jan 21 at 19:39
add a comment |
1
$begingroup$
Hint: for every $U'in B(Y',
X') $
Consider $U"=(U')':X"=Xrightarrow Y"=Y$.
answered Jan 21 at 17:21
Tsemo AristideTsemo Aristide
59.1k11445
$endgroup$
$begingroup$
Okay so I can show that $U''(x'')(y')=U''(iota_X(x))(y')=iota_X(x)circ U'(y')=U'(y')(x)=y'circ U(x)=i_Y(U(x))(y')$ but I don't really see how this helps
$endgroup$
– James
Jan 21 at 17:37
$begingroup$
You can compute $(U") '=U' $.
$endgroup$
– Tsemo Aristide
Jan 21 at 18:08
$begingroup$
I don't see what you mean
$endgroup$
– James
Jan 21 at 19:16
$begingroup$
$(U") '(y') (x) =y'(U"(x)) =(U"(x)) (y') $, identify $U"(x) $ with an element of the bidual. $(U"(x))(y') =x((U'(y')) =U'(y') (x) $.
$endgroup$
– Tsemo Aristide
Jan 21 at 19:39
add a comment |
1
1
1
$begingroup$
Hint: for every $U'in B(Y',
X') $
Consider $U"=(U')':X"=Xrightarrow Y"=Y$.
answered Jan 21 at 17:21
Tsemo AristideTsemo Aristide
59.1k11445
$endgroup$
Hint: for every $U'in B(Y',
X') $
Consider $U"=(U')':X"=Xrightarrow Y"=Y$.
answered Jan 21 at 17:21
Tsemo AristideTsemo Aristide
59.1k11445
answered Jan 21 at 17:21
Tsemo AristideTsemo Aristide
59.1k11445
answered Jan 21 at 17:21
Tsemo AristideTsemo Aristide
59.1k11445
answered Jan 21 at 17:21
Tsemo AristideTsemo Aristide
59.1k11445
59.1k11445
$begingroup$
Okay so I can show that $U''(x'')(y')=U''(iota_X(x))(y')=iota_X(x)circ U'(y')=U'(y')(x)=y'circ U(x)=i_Y(U(x))(y')$ but I don't really see how this helps
$endgroup$
– James
Jan 21 at 17:37
$begingroup$
You can compute $(U") '=U' $.
$endgroup$
– Tsemo Aristide
Jan 21 at 18:08
$begingroup$
I don't see what you mean
$endgroup$
– James
Jan 21 at 19:16
$begingroup$
$(U") '(y') (x) =y'(U"(x)) =(U"(x)) (y') $, identify $U"(x) $ with an element of the bidual. $(U"(x))(y') =x((U'(y')) =U'(y') (x) $.
$endgroup$
– Tsemo Aristide
Jan 21 at 19:39
add a comment |
$begingroup$
Okay so I can show that $U''(x'')(y')=U''(iota_X(x))(y')=iota_X(x)circ U'(y')=U'(y')(x)=y'circ U(x)=i_Y(U(x))(y')$ but I don't really see how this helps
$endgroup$
– James
Jan 21 at 17:37
$begingroup$
You can compute $(U") '=U' $.
$endgroup$
– Tsemo Aristide
Jan 21 at 18:08
$begingroup$
I don't see what you mean
$endgroup$
– James
Jan 21 at 19:16
$begingroup$
$(U") '(y') (x) =y'(U"(x)) =(U"(x)) (y') $, identify $U"(x) $ with an element of the bidual. $(U"(x))(y') =x((U'(y')) =U'(y') (x) $.
$endgroup$
– Tsemo Aristide
Jan 21 at 19:39
$begingroup$
Okay so I can show that $U''(x'')(y')=U''(iota_X(x))(y')=iota_X(x)circ U'(y')=U'(y')(x)=y'circ U(x)=i_Y(U(x))(y')$ but I don't really see how this helps
$endgroup$
– James
Jan 21 at 17:37
$begingroup$
Okay so I can show that $U''(x'')(y')=U''(iota_X(x))(y')=iota_X(x)circ U'(y')=U'(y')(x)=y'circ U(x)=i_Y(U(x))(y')$ but I don't really see how this helps
$endgroup$
– James
Jan 21 at 17:37
$begingroup$
You can compute $(U") '=U' $.
$endgroup$
– Tsemo Aristide
Jan 21 at 18:08
$begingroup$
You can compute $(U") '=U' $.
$endgroup$
– Tsemo Aristide
Jan 21 at 18:08
$begingroup$
I don't see what you mean
$endgroup$
– James
Jan 21 at 19:16
$begingroup$
I don't see what you mean
$endgroup$
– James
Jan 21 at 19:16
$begingroup$
$(U") '(y') (x) =y'(U"(x)) =(U"(x)) (y') $, identify $U"(x) $ with an element of the bidual. $(U"(x))(y') =x((U'(y')) =U'(y') (x) $.
$endgroup$
– Tsemo Aristide
Jan 21 at 19:39
$begingroup$
$(U") '(y') (x) =y'(U"(x)) =(U"(x)) (y') $, identify $U"(x) $ with an element of the bidual. $(U"(x))(y') =x((U'(y')) =U'(y') (x) $.
$endgroup$
– Tsemo Aristide
Jan 21 at 19:39
add a comment |
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$begingroup$
Maybe you should try and consider that $X’$ and $Y’$ are reflexive and consider two dualizations starting from $Y’ rightarrow X’$.
$endgroup$
– Mindlack
Jan 21 at 17:06
$begingroup$
But I don't know whether $X'$ is reflexive, since $X$ is not Banach
$endgroup$
– James
Jan 21 at 17:18
$begingroup$
If you are using the usual definition, if $X$ is reflexive, $X$ is a Banach, no? en.m.wikipedia.org/wiki/Reflexive_space
$endgroup$
– Mindlack
Jan 21 at 17:22
$begingroup$
You are right :)
$endgroup$
– James
Jan 21 at 17:23