Mixing
surjectivity of dual operator congruence implies reflexivity
$begingroup$
I tried to prove the following:
Let $X,Y$ be normed spaces and $Phi:mathcal{L}(X,Y)tomathcal{L}(Y^*,X^*)$, $Phi(A)=A^*$ is surjective. Then $Y$ is reflexive. Here $X^*$ $(A^*)$ denotes the dual space of $X$ (the dual operator of $A$).
My try is: Let $Fin Y^{**}setminus{0}$ and $xin Xsetminus{0}$. By Hahn-Banach theorem, there exists $varphiin X^*$ such that $varphi(x)=1$. Denote operator $A:Y^*to X^*$, $A(phi)y=varphi(y)F(phi)$; this operator is bounded and linear. Due to our assumption, there exists $Binmathcal{L}(X,Y)$ such that $B^*=A$. Now we choose $phiin Y^*$ and compute: $$F(phi)=varphi(x)F(phi)=A(phi)x=B^*(phi)x=phi(Bx).$$ Hence $Y$ is reflexive due to the definition of reflexivity.
Is this proof correct ? If yes, why $Bx$ does not depend on choice of $x$, only on choice of $F$ ?
Thank you for any comments.
dual-spaces reflexive-space
asked Jan 14 at 9:31
ellipticelliptic
1008
$endgroup$
add a comment |
$begingroup$
I tried to prove the following:
Let $X,Y$ be normed spaces and $Phi:mathcal{L}(X,Y)tomathcal{L}(Y^*,X^*)$, $Phi(A)=A^*$ is surjective. Then $Y$ is reflexive. Here $X^*$ $(A^*)$ denotes the dual space of $X$ (the dual operator of $A$).
My try is: Let $Fin Y^{**}setminus{0}$ and $xin Xsetminus{0}$. By Hahn-Banach theorem, there exists $varphiin X^*$ such that $varphi(x)=1$. Denote operator $A:Y^*to X^*$, $A(phi)y=varphi(y)F(phi)$; this operator is bounded and linear. Due to our assumption, there exists $Binmathcal{L}(X,Y)$ such that $B^*=A$. Now we choose $phiin Y^*$ and compute: $$F(phi)=varphi(x)F(phi)=A(phi)x=B^*(phi)x=phi(Bx).$$ Hence $Y$ is reflexive due to the definition of reflexivity.
Is this proof correct ? If yes, why $Bx$ does not depend on choice of $x$, only on choice of $F$ ?
Thank you for any comments.
dual-spaces reflexive-space
asked Jan 14 at 9:31
ellipticelliptic
1008
$endgroup$
$begingroup$
You are asked to show that $Y$ is reflexive, not $X$. You have to start with an element of $Y^{**}$.
$endgroup$
– Kavi Rama Murthy
Jan 14 at 9:38
$begingroup$
yes, of course. sorry.
$endgroup$
– elliptic
Jan 14 at 10:36
add a comment |
$begingroup$
I tried to prove the following:
Let $X,Y$ be normed spaces and $Phi:mathcal{L}(X,Y)tomathcal{L}(Y^*,X^*)$, $Phi(A)=A^*$ is surjective. Then $Y$ is reflexive. Here $X^*$ $(A^*)$ denotes the dual space of $X$ (the dual operator of $A$).
My try is: Let $Fin Y^{**}setminus{0}$ and $xin Xsetminus{0}$. By Hahn-Banach theorem, there exists $varphiin X^*$ such that $varphi(x)=1$. Denote operator $A:Y^*to X^*$, $A(phi)y=varphi(y)F(phi)$; this operator is bounded and linear. Due to our assumption, there exists $Binmathcal{L}(X,Y)$ such that $B^*=A$. Now we choose $phiin Y^*$ and compute: $$F(phi)=varphi(x)F(phi)=A(phi)x=B^*(phi)x=phi(Bx).$$ Hence $Y$ is reflexive due to the definition of reflexivity.
Is this proof correct ? If yes, why $Bx$ does not depend on choice of $x$, only on choice of $F$ ?
Thank you for any comments.
dual-spaces reflexive-space
asked Jan 14 at 9:31
ellipticelliptic
1008
$endgroup$
I tried to prove the following:
Let $X,Y$ be normed spaces and $Phi:mathcal{L}(X,Y)tomathcal{L}(Y^*,X^*)$, $Phi(A)=A^*$ is surjective. Then $Y$ is reflexive. Here $X^*$ $(A^*)$ denotes the dual space of $X$ (the dual operator of $A$).
My try is: Let $Fin Y^{**}setminus{0}$ and $xin Xsetminus{0}$. By Hahn-Banach theorem, there exists $varphiin X^*$ such that $varphi(x)=1$. Denote operator $A:Y^*to X^*$, $A(phi)y=varphi(y)F(phi)$; this operator is bounded and linear. Due to our assumption, there exists $Binmathcal{L}(X,Y)$ such that $B^*=A$. Now we choose $phiin Y^*$ and compute: $$F(phi)=varphi(x)F(phi)=A(phi)x=B^*(phi)x=phi(Bx).$$ Hence $Y$ is reflexive due to the definition of reflexivity.
Is this proof correct ? If yes, why $Bx$ does not depend on choice of $x$, only on choice of $F$ ?
Thank you for any comments.
dual-spaces reflexive-space
dual-spaces reflexive-space
asked Jan 14 at 9:31
ellipticelliptic
1008
asked Jan 14 at 9:31
ellipticelliptic
1008
edited Jan 14 at 10:35
elliptic
asked Jan 14 at 9:31
ellipticelliptic
1008
asked Jan 14 at 9:31
ellipticelliptic
1008
asked Jan 14 at 9:31
ellipticelliptic
1008
1008
$begingroup$
You are asked to show that $Y$ is reflexive, not $X$. You have to start with an element of $Y^{**}$.
$endgroup$
– Kavi Rama Murthy
Jan 14 at 9:38
$begingroup$
yes, of course. sorry.
$endgroup$
– elliptic
Jan 14 at 10:36
add a comment |
$begingroup$
You are asked to show that $Y$ is reflexive, not $X$. You have to start with an element of $Y^{**}$.
$endgroup$
– Kavi Rama Murthy
Jan 14 at 9:38
$begingroup$
yes, of course. sorry.
$endgroup$
– elliptic
Jan 14 at 10:36
$begingroup$
You are asked to show that $Y$ is reflexive, not $X$. You have to start with an element of $Y^{**}$.
$endgroup$
– Kavi Rama Murthy
Jan 14 at 9:38
$begingroup$
You are asked to show that $Y$ is reflexive, not $X$. You have to start with an element of $Y^{**}$.
$endgroup$
– Kavi Rama Murthy
Jan 14 at 9:38
$begingroup$
yes, of course. sorry.
$endgroup$
– elliptic
Jan 14 at 10:36
$begingroup$
yes, of course. sorry.
$endgroup$
– elliptic
Jan 14 at 10:36
add a comment |
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$begingroup$
You are asked to show that $Y$ is reflexive, not $X$. You have to start with an element of $Y^{**}$.
$endgroup$
– Kavi Rama Murthy
Jan 14 at 9:38
$begingroup$
yes, of course. sorry.
$endgroup$
– elliptic
Jan 14 at 10:36