Show that a Banach space $E$ is reflexive $iff$ for all closed subspace $F$ of $E$ we have $F^{perp perp} =...
$begingroup$
Problem: Let $E$ be a normed space over $mathbb{C}$.
- Show that we are able to embed $E$ into the second dual space (bidual) $E''$ of $E$ by a linear isometry i.e we can consider $E$ as a subspace of $E''$.
- For each subset $A$ of $E$, let $A^{perp} = {f in E' : f|_A = 0}$. Show that a Banach space $E$ is reflexive $iff$ for all closed subspace $F$ of $E$ we have $F^{perp perp} = J(F)$ in which $J$ is canonical embedding $E$ into $E'$.
My attempt:
- Let $E$ be a normed space. For each $x in E$, let $(Jx)(u) = u(x)$ for all $u in E'$.
We show that $J: E rightarrow E''$ is a linear isometry. Clearly, $Jx$ is a linear form on $E'$. Since $(Jx)(u) = |u(x)| le ||u||$ $||x|| $, $Jx$ is continuous on $E'$and $||Jx|| le ||x||$. Therefore, $Jx in E''$. It is routine to show that $J$ is a linear map from $E$ into $E''$. Take any $x in E$. There is $v in E'$ such that $||v|| = 1$ and $v(x) = ||x||$. Hence
$$||Jx|| = sup{|(Jx)(u)|:||u|| le 1}$$
$$= sup{|u(x)|: u in E', ||u|| le 1} ge v(x) = ||x||.$$
Therefore $J$ is a linear isometry.
Anybody can help me to solve question number $2$? Thank all!
functional-analysis proof-verification hilbert-spaces
asked Jan 15 at 18:31
MinhMinh
31119
$endgroup$
add a comment |
$begingroup$
Problem: Let $E$ be a normed space over $mathbb{C}$.
- Show that we are able to embed $E$ into the second dual space (bidual) $E''$ of $E$ by a linear isometry i.e we can consider $E$ as a subspace of $E''$.
- For each subset $A$ of $E$, let $A^{perp} = {f in E' : f|_A = 0}$. Show that a Banach space $E$ is reflexive $iff$ for all closed subspace $F$ of $E$ we have $F^{perp perp} = J(F)$ in which $J$ is canonical embedding $E$ into $E'$.
My attempt:
- Let $E$ be a normed space. For each $x in E$, let $(Jx)(u) = u(x)$ for all $u in E'$.
We show that $J: E rightarrow E''$ is a linear isometry. Clearly, $Jx$ is a linear form on $E'$. Since $(Jx)(u) = |u(x)| le ||u||$ $||x|| $, $Jx$ is continuous on $E'$and $||Jx|| le ||x||$. Therefore, $Jx in E''$. It is routine to show that $J$ is a linear map from $E$ into $E''$. Take any $x in E$. There is $v in E'$ such that $||v|| = 1$ and $v(x) = ||x||$. Hence
$$||Jx|| = sup{|(Jx)(u)|:||u|| le 1}$$
$$= sup{|u(x)|: u in E', ||u|| le 1} ge v(x) = ||x||.$$
Therefore $J$ is a linear isometry.
Anybody can help me to solve question number $2$? Thank all!
functional-analysis proof-verification hilbert-spaces
asked Jan 15 at 18:31
MinhMinh
31119
$endgroup$
add a comment |
$begingroup$
Problem: Let $E$ be a normed space over $mathbb{C}$.
- Show that we are able to embed $E$ into the second dual space (bidual) $E''$ of $E$ by a linear isometry i.e we can consider $E$ as a subspace of $E''$.
- For each subset $A$ of $E$, let $A^{perp} = {f in E' : f|_A = 0}$. Show that a Banach space $E$ is reflexive $iff$ for all closed subspace $F$ of $E$ we have $F^{perp perp} = J(F)$ in which $J$ is canonical embedding $E$ into $E'$.
My attempt:
- Let $E$ be a normed space. For each $x in E$, let $(Jx)(u) = u(x)$ for all $u in E'$.
We show that $J: E rightarrow E''$ is a linear isometry. Clearly, $Jx$ is a linear form on $E'$. Since $(Jx)(u) = |u(x)| le ||u||$ $||x|| $, $Jx$ is continuous on $E'$and $||Jx|| le ||x||$. Therefore, $Jx in E''$. It is routine to show that $J$ is a linear map from $E$ into $E''$. Take any $x in E$. There is $v in E'$ such that $||v|| = 1$ and $v(x) = ||x||$. Hence
$$||Jx|| = sup{|(Jx)(u)|:||u|| le 1}$$
$$= sup{|u(x)|: u in E', ||u|| le 1} ge v(x) = ||x||.$$
Therefore $J$ is a linear isometry.
Anybody can help me to solve question number $2$? Thank all!
functional-analysis proof-verification hilbert-spaces
asked Jan 15 at 18:31
MinhMinh
31119
$endgroup$
Problem: Let $E$ be a normed space over $mathbb{C}$.
- Show that we are able to embed $E$ into the second dual space (bidual) $E''$ of $E$ by a linear isometry i.e we can consider $E$ as a subspace of $E''$.
- For each subset $A$ of $E$, let $A^{perp} = {f in E' : f|_A = 0}$. Show that a Banach space $E$ is reflexive $iff$ for all closed subspace $F$ of $E$ we have $F^{perp perp} = J(F)$ in which $J$ is canonical embedding $E$ into $E'$.
My attempt:
- Let $E$ be a normed space. For each $x in E$, let $(Jx)(u) = u(x)$ for all $u in E'$.
We show that $J: E rightarrow E''$ is a linear isometry. Clearly, $Jx$ is a linear form on $E'$. Since $(Jx)(u) = |u(x)| le ||u||$ $||x|| $, $Jx$ is continuous on $E'$and $||Jx|| le ||x||$. Therefore, $Jx in E''$. It is routine to show that $J$ is a linear map from $E$ into $E''$. Take any $x in E$. There is $v in E'$ such that $||v|| = 1$ and $v(x) = ||x||$. Hence
$$||Jx|| = sup{|(Jx)(u)|:||u|| le 1}$$
$$= sup{|u(x)|: u in E', ||u|| le 1} ge v(x) = ||x||.$$
Therefore $J$ is a linear isometry.
Anybody can help me to solve question number $2$? Thank all!
functional-analysis proof-verification hilbert-spaces
functional-analysis proof-verification hilbert-spaces
asked Jan 15 at 18:31
MinhMinh
31119
asked Jan 15 at 18:31
MinhMinh
31119
edited Jan 16 at 10:24
Minh
asked Jan 15 at 18:31
MinhMinh
31119
asked Jan 15 at 18:31
MinhMinh
31119
asked Jan 15 at 18:31
MinhMinh
31119
31119
add a comment |
add a comment |
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$begingroup$
The implication $Leftarrow$ direction is very simple. Since $E^{perpperp}=E^{primeprime}$, it follows from $E^{perpperp}=J(E)$ that $J$ is surjective.
For the converse implication let $F$ be a closed subspace of $E$. If $phiin E^perp$, then $0=phi(x)=Jx(phi)$ for all $xin F$. Thus $J(F)subset F^{perpperp}$ (this inclusion does not require reflexivity or closedness). On the other hand let $psiin F^{perpperp}$. By assumption there exists $yin E$ such that $Jy=psi$. Then $0=psi(u)=u(y)$ for all $uin F^perp$. It follows from the Hahn-Banach theorem that $yin F$ (here one needs the closedness of $F$).
answered Jan 16 at 10:38
MaoWaoMaoWao
3,958618
$endgroup$
$begingroup$
Thanks. This implies from the definition.
$endgroup$
– Minh
Jan 16 at 11:48
add a comment |
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1 Answer
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$begingroup$
The implication $Leftarrow$ direction is very simple. Since $E^{perpperp}=E^{primeprime}$, it follows from $E^{perpperp}=J(E)$ that $J$ is surjective.
For the converse implication let $F$ be a closed subspace of $E$. If $phiin E^perp$, then $0=phi(x)=Jx(phi)$ for all $xin F$. Thus $J(F)subset F^{perpperp}$ (this inclusion does not require reflexivity or closedness). On the other hand let $psiin F^{perpperp}$. By assumption there exists $yin E$ such that $Jy=psi$. Then $0=psi(u)=u(y)$ for all $uin F^perp$. It follows from the Hahn-Banach theorem that $yin F$ (here one needs the closedness of $F$).
answered Jan 16 at 10:38
MaoWaoMaoWao
3,958618
$endgroup$
$begingroup$
Thanks. This implies from the definition.
$endgroup$
– Minh
Jan 16 at 11:48
add a comment |
$begingroup$
The implication $Leftarrow$ direction is very simple. Since $E^{perpperp}=E^{primeprime}$, it follows from $E^{perpperp}=J(E)$ that $J$ is surjective.
For the converse implication let $F$ be a closed subspace of $E$. If $phiin E^perp$, then $0=phi(x)=Jx(phi)$ for all $xin F$. Thus $J(F)subset F^{perpperp}$ (this inclusion does not require reflexivity or closedness). On the other hand let $psiin F^{perpperp}$. By assumption there exists $yin E$ such that $Jy=psi$. Then $0=psi(u)=u(y)$ for all $uin F^perp$. It follows from the Hahn-Banach theorem that $yin F$ (here one needs the closedness of $F$).
answered Jan 16 at 10:38
MaoWaoMaoWao
3,958618
$endgroup$
$begingroup$
Thanks. This implies from the definition.
$endgroup$
– Minh
Jan 16 at 11:48
add a comment |
$begingroup$
The implication $Leftarrow$ direction is very simple. Since $E^{perpperp}=E^{primeprime}$, it follows from $E^{perpperp}=J(E)$ that $J$ is surjective.
For the converse implication let $F$ be a closed subspace of $E$. If $phiin E^perp$, then $0=phi(x)=Jx(phi)$ for all $xin F$. Thus $J(F)subset F^{perpperp}$ (this inclusion does not require reflexivity or closedness). On the other hand let $psiin F^{perpperp}$. By assumption there exists $yin E$ such that $Jy=psi$. Then $0=psi(u)=u(y)$ for all $uin F^perp$. It follows from the Hahn-Banach theorem that $yin F$ (here one needs the closedness of $F$).
answered Jan 16 at 10:38
MaoWaoMaoWao
3,958618
$endgroup$
The implication $Leftarrow$ direction is very simple. Since $E^{perpperp}=E^{primeprime}$, it follows from $E^{perpperp}=J(E)$ that $J$ is surjective.
For the converse implication let $F$ be a closed subspace of $E$. If $phiin E^perp$, then $0=phi(x)=Jx(phi)$ for all $xin F$. Thus $J(F)subset F^{perpperp}$ (this inclusion does not require reflexivity or closedness). On the other hand let $psiin F^{perpperp}$. By assumption there exists $yin E$ such that $Jy=psi$. Then $0=psi(u)=u(y)$ for all $uin F^perp$. It follows from the Hahn-Banach theorem that $yin F$ (here one needs the closedness of $F$).
answered Jan 16 at 10:38
MaoWaoMaoWao
3,958618
answered Jan 16 at 10:38
MaoWaoMaoWao
3,958618
answered Jan 16 at 10:38
MaoWaoMaoWao
3,958618
answered Jan 16 at 10:38
MaoWaoMaoWao
3,958618
3,958618
$begingroup$
Thanks. This implies from the definition.
$endgroup$
– Minh
Jan 16 at 11:48
add a comment |
$begingroup$
Thanks. This implies from the definition.
$endgroup$
– Minh
Jan 16 at 11:48
$begingroup$
Thanks. This implies from the definition.
$endgroup$
– Minh
Jan 16 at 11:48
$begingroup$
Thanks. This implies from the definition.
$endgroup$
– Minh
Jan 16 at 11:48
add a comment |
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