Mixing
Show that a Banach space $X$ is not reflexive if a closed subspace of its dual separates the points of $X$
$begingroup$
Given a Banach space $X$ and a closed subspace $Z$ of $X^*$ such that $Z neq X^*$, suppose that $Z$ separates the points of $X$, I mean:
$x in X, , , , x^*(x) = 0 , , forall x^* in Z implies x=0$.
Show that $X$ is not reflexive.
With $X^*$ I mean the dual of $X$.
Any suggestion? This is a functional analysis exercise about duality and reflexivity but I don’t know how to show that the given Banach space is not reflexive
functional-analysis banach-spaces separation-axioms dual-spaces reflexive-space
edited Jan 3 at 9:45
Song
8,311625
asked Jan 3 at 9:30
Maggie94Maggie94
876
$endgroup$
add a comment |
$begingroup$
Given a Banach space $X$ and a closed subspace $Z$ of $X^*$ such that $Z neq X^*$, suppose that $Z$ separates the points of $X$, I mean:
$x in X, , , , x^*(x) = 0 , , forall x^* in Z implies x=0$.
Show that $X$ is not reflexive.
With $X^*$ I mean the dual of $X$.
Any suggestion? This is a functional analysis exercise about duality and reflexivity but I don’t know how to show that the given Banach space is not reflexive
functional-analysis banach-spaces separation-axioms dual-spaces reflexive-space
edited Jan 3 at 9:45
Song
8,311625
asked Jan 3 at 9:30
Maggie94Maggie94
876
$endgroup$
1
$begingroup$
In one place you say 'show that $X$ is not reflexive' and in two other places you ask us to show that $X$ is reflexive.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 9:33
$begingroup$
You still have a mistake in the last part.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 9:42
add a comment |
$begingroup$
Given a Banach space $X$ and a closed subspace $Z$ of $X^*$ such that $Z neq X^*$, suppose that $Z$ separates the points of $X$, I mean:
$x in X, , , , x^*(x) = 0 , , forall x^* in Z implies x=0$.
Show that $X$ is not reflexive.
With $X^*$ I mean the dual of $X$.
Any suggestion? This is a functional analysis exercise about duality and reflexivity but I don’t know how to show that the given Banach space is not reflexive
functional-analysis banach-spaces separation-axioms dual-spaces reflexive-space
edited Jan 3 at 9:45
Song
8,311625
asked Jan 3 at 9:30
Maggie94Maggie94
876
$endgroup$
Given a Banach space $X$ and a closed subspace $Z$ of $X^*$ such that $Z neq X^*$, suppose that $Z$ separates the points of $X$, I mean:
$x in X, , , , x^*(x) = 0 , , forall x^* in Z implies x=0$.
Show that $X$ is not reflexive.
With $X^*$ I mean the dual of $X$.
Any suggestion? This is a functional analysis exercise about duality and reflexivity but I don’t know how to show that the given Banach space is not reflexive
functional-analysis banach-spaces separation-axioms dual-spaces reflexive-space
functional-analysis banach-spaces separation-axioms dual-spaces reflexive-space
edited Jan 3 at 9:45
Song
8,311625
asked Jan 3 at 9:30
Maggie94Maggie94
876
edited Jan 3 at 9:45
Song
8,311625
asked Jan 3 at 9:30
Maggie94Maggie94
876
edited Jan 3 at 9:45
Song
8,311625
edited Jan 3 at 9:45
Song
8,311625
edited Jan 3 at 9:45
Song
8,311625
8,311625
asked Jan 3 at 9:30
Maggie94Maggie94
876
asked Jan 3 at 9:30
Maggie94Maggie94
876
asked Jan 3 at 9:30
Maggie94Maggie94
876
876
1
$begingroup$
In one place you say 'show that $X$ is not reflexive' and in two other places you ask us to show that $X$ is reflexive.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 9:33
$begingroup$
You still have a mistake in the last part.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 9:42
add a comment |
1
$begingroup$
In one place you say 'show that $X$ is not reflexive' and in two other places you ask us to show that $X$ is reflexive.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 9:33
$begingroup$
You still have a mistake in the last part.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 9:42
1
1
$begingroup$
In one place you say 'show that $X$ is not reflexive' and in two other places you ask us to show that $X$ is reflexive.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 9:33
$begingroup$
In one place you say 'show that $X$ is not reflexive' and in two other places you ask us to show that $X$ is reflexive.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 9:33
$begingroup$
You still have a mistake in the last part.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 9:42
$begingroup$
You still have a mistake in the last part.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 9:42
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is in fact quite straightforward by the definition of a reflexive space and Hahn-Banach theorem. If $Zneq X^*$, then there exists $phi in X^{**}setminus{{0}}$ such that $phi$ vanishes on $Z$. If $X=X^{**}$, then $phi = x$ for some $xin X$ and this implies
$$
f(x) = 0
$$ for all $fin Z$. Since $Z$ separates points in $X$, it follows $x = 0 =phi$ contradicting $phi neq 0$.
answered Jan 3 at 9:45
SongSong
8,311625
$endgroup$
add a comment |
$begingroup$
Consider $y$ which is not in $Z$ by Hahn Banach there exists $fin (X^*)^*$ such that $f(Z)=0, f(y)=1$, $f$ is not in $Xrightarrow (X^*)^*$ since if there exists $xin X$ such that $f(z)=z(x), z(x)=0$ for every $zin Z$ implies that $x=0$. Contradiction.
answered Jan 3 at 9:45
Tsemo AristideTsemo Aristide
56.9k11444
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
$begingroup$
It is in fact quite straightforward by the definition of a reflexive space and Hahn-Banach theorem. If $Zneq X^*$, then there exists $phi in X^{**}setminus{{0}}$ such that $phi$ vanishes on $Z$. If $X=X^{**}$, then $phi = x$ for some $xin X$ and this implies
$$
f(x) = 0
$$ for all $fin Z$. Since $Z$ separates points in $X$, it follows $x = 0 =phi$ contradicting $phi neq 0$.
answered Jan 3 at 9:45
SongSong
8,311625
$endgroup$
add a comment |
$begingroup$
It is in fact quite straightforward by the definition of a reflexive space and Hahn-Banach theorem. If $Zneq X^*$, then there exists $phi in X^{**}setminus{{0}}$ such that $phi$ vanishes on $Z$. If $X=X^{**}$, then $phi = x$ for some $xin X$ and this implies
$$
f(x) = 0
$$ for all $fin Z$. Since $Z$ separates points in $X$, it follows $x = 0 =phi$ contradicting $phi neq 0$.
answered Jan 3 at 9:45
SongSong
8,311625
$endgroup$
add a comment |
$begingroup$
It is in fact quite straightforward by the definition of a reflexive space and Hahn-Banach theorem. If $Zneq X^*$, then there exists $phi in X^{**}setminus{{0}}$ such that $phi$ vanishes on $Z$. If $X=X^{**}$, then $phi = x$ for some $xin X$ and this implies
$$
f(x) = 0
$$ for all $fin Z$. Since $Z$ separates points in $X$, it follows $x = 0 =phi$ contradicting $phi neq 0$.
answered Jan 3 at 9:45
SongSong
8,311625
$endgroup$
It is in fact quite straightforward by the definition of a reflexive space and Hahn-Banach theorem. If $Zneq X^*$, then there exists $phi in X^{**}setminus{{0}}$ such that $phi$ vanishes on $Z$. If $X=X^{**}$, then $phi = x$ for some $xin X$ and this implies
$$
f(x) = 0
$$ for all $fin Z$. Since $Z$ separates points in $X$, it follows $x = 0 =phi$ contradicting $phi neq 0$.
answered Jan 3 at 9:45
SongSong
8,311625
answered Jan 3 at 9:45
SongSong
8,311625
answered Jan 3 at 9:45
SongSong
8,311625
answered Jan 3 at 9:45
SongSong
8,311625
8,311625
add a comment |
add a comment |
$begingroup$
Consider $y$ which is not in $Z$ by Hahn Banach there exists $fin (X^*)^*$ such that $f(Z)=0, f(y)=1$, $f$ is not in $Xrightarrow (X^*)^*$ since if there exists $xin X$ such that $f(z)=z(x), z(x)=0$ for every $zin Z$ implies that $x=0$. Contradiction.
answered Jan 3 at 9:45
Tsemo AristideTsemo Aristide
56.9k11444
$endgroup$
add a comment |
$begingroup$
Consider $y$ which is not in $Z$ by Hahn Banach there exists $fin (X^*)^*$ such that $f(Z)=0, f(y)=1$, $f$ is not in $Xrightarrow (X^*)^*$ since if there exists $xin X$ such that $f(z)=z(x), z(x)=0$ for every $zin Z$ implies that $x=0$. Contradiction.
answered Jan 3 at 9:45
Tsemo AristideTsemo Aristide
56.9k11444
$endgroup$
add a comment |
$begingroup$
Consider $y$ which is not in $Z$ by Hahn Banach there exists $fin (X^*)^*$ such that $f(Z)=0, f(y)=1$, $f$ is not in $Xrightarrow (X^*)^*$ since if there exists $xin X$ such that $f(z)=z(x), z(x)=0$ for every $zin Z$ implies that $x=0$. Contradiction.
answered Jan 3 at 9:45
Tsemo AristideTsemo Aristide
56.9k11444
$endgroup$
Consider $y$ which is not in $Z$ by Hahn Banach there exists $fin (X^*)^*$ such that $f(Z)=0, f(y)=1$, $f$ is not in $Xrightarrow (X^*)^*$ since if there exists $xin X$ such that $f(z)=z(x), z(x)=0$ for every $zin Z$ implies that $x=0$. Contradiction.
answered Jan 3 at 9:45
Tsemo AristideTsemo Aristide
56.9k11444
answered Jan 3 at 9:45
Tsemo AristideTsemo Aristide
56.9k11444
answered Jan 3 at 9:45
Tsemo AristideTsemo Aristide
56.9k11444
answered Jan 3 at 9:45
Tsemo AristideTsemo Aristide
56.9k11444
56.9k11444
add a comment |
add a comment |
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1
$begingroup$
In one place you say 'show that $X$ is not reflexive' and in two other places you ask us to show that $X$ is reflexive.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 9:33
$begingroup$
You still have a mistake in the last part.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 9:42