Rudin functional analysis, theorem 4.9 part (b)Question about a proof in Rudin's book - annihilatorsIs it necessary to use the Hahn-Banach theorem to show that $(X/M)^*simeq M^perp$?Question on Rudin's functional analysis Theorem 4.9A question in Functional AnalysisTheorem 2.9 rudin functional analysis (application of Baire's theorem)Rudin functional analysis, theorem 2.11 (Open mapping theorem)Theorem 3.18, Rudin's functional analysisRudin's functional analysis Theorem 3.18, second part.Rudin functional analysis, theorem 3.27Rudin functional analysis theorem 3.28, application of Reisz representation theorem.
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Rudin functional analysis, theorem 4.9 part (b)
Question about a proof in Rudin's book - annihilatorsIs it necessary to use the Hahn-Banach theorem to show that $(X/M)^*simeq M^perp$?Question on Rudin's functional analysis Theorem 4.9A question in Functional AnalysisTheorem 2.9 rudin functional analysis (application of Baire's theorem)Rudin functional analysis, theorem 2.11 (Open mapping theorem)Theorem 3.18, Rudin's functional analysisRudin's functional analysis Theorem 3.18, second part.Rudin functional analysis, theorem 3.27Rudin functional analysis theorem 3.28, application of Reisz representation theorem.
$begingroup$
Let $M$ a closed subspace of a Banach space $X$.
b) Let $pi:X to X/M$ be the quotient map. Put $Y = X/M$, For each $y^* in Y^*$, define
$$
tau y^* = y^* pi
$$
Then $tau$ is an isometric isomorphism of $Y^*$.
Partial proof below:
If $x in X, y^* in Y^*$, then $pi x in Y$; hence $x to y^*pi x$ is a continuous linear functional on $X$ which vanishes for $x in M$. Thus $tau y^* in M^perp$. The linearity of $tau$ is obvious. Fix $x^* in M^perp$. Let $N$ be the null space of $x^*$. Since $M subset N$ there's a linear functional $Lambda$ on $Y$ such that $Lambda pi = x^*$.
Here's the question, which there's such a functional?
functional-analysis proof-explanation topological-vector-spaces quotient-spaces
asked yesterday
user8469759user8469759
1,5541618
$endgroup$
add a comment |
$begingroup$
Let $M$ a closed subspace of a Banach space $X$.
b) Let $pi:X to X/M$ be the quotient map. Put $Y = X/M$, For each $y^* in Y^*$, define
$$
tau y^* = y^* pi
$$
Then $tau$ is an isometric isomorphism of $Y^*$.
Partial proof below:
If $x in X, y^* in Y^*$, then $pi x in Y$; hence $x to y^*pi x$ is a continuous linear functional on $X$ which vanishes for $x in M$. Thus $tau y^* in M^perp$. The linearity of $tau$ is obvious. Fix $x^* in M^perp$. Let $N$ be the null space of $x^*$. Since $M subset N$ there's a linear functional $Lambda$ on $Y$ such that $Lambda pi = x^*$.
Here's the question, which there's such a functional?
functional-analysis proof-explanation topological-vector-spaces quotient-spaces
asked yesterday
user8469759user8469759
1,5541618
$endgroup$
add a comment |
$begingroup$
Let $M$ a closed subspace of a Banach space $X$.
b) Let $pi:X to X/M$ be the quotient map. Put $Y = X/M$, For each $y^* in Y^*$, define
$$
tau y^* = y^* pi
$$
Then $tau$ is an isometric isomorphism of $Y^*$.
Partial proof below:
If $x in X, y^* in Y^*$, then $pi x in Y$; hence $x to y^*pi x$ is a continuous linear functional on $X$ which vanishes for $x in M$. Thus $tau y^* in M^perp$. The linearity of $tau$ is obvious. Fix $x^* in M^perp$. Let $N$ be the null space of $x^*$. Since $M subset N$ there's a linear functional $Lambda$ on $Y$ such that $Lambda pi = x^*$.
Here's the question, which there's such a functional?
functional-analysis proof-explanation topological-vector-spaces quotient-spaces
asked yesterday
user8469759user8469759
1,5541618
$endgroup$
Let $M$ a closed subspace of a Banach space $X$.
b) Let $pi:X to X/M$ be the quotient map. Put $Y = X/M$, For each $y^* in Y^*$, define
$$
tau y^* = y^* pi
$$
Then $tau$ is an isometric isomorphism of $Y^*$.
Partial proof below:
If $x in X, y^* in Y^*$, then $pi x in Y$; hence $x to y^*pi x$ is a continuous linear functional on $X$ which vanishes for $x in M$. Thus $tau y^* in M^perp$. The linearity of $tau$ is obvious. Fix $x^* in M^perp$. Let $N$ be the null space of $x^*$. Since $M subset N$ there's a linear functional $Lambda$ on $Y$ such that $Lambda pi = x^*$.
Here's the question, which there's such a functional?
functional-analysis proof-explanation topological-vector-spaces quotient-spaces
functional-analysis proof-explanation topological-vector-spaces quotient-spaces
asked yesterday
user8469759user8469759
1,5541618
asked yesterday
user8469759user8469759
1,5541618
asked yesterday
user8469759user8469759
1,5541618
asked yesterday
user8469759user8469759
1,5541618
asked yesterday
user8469759user8469759
1,5541618
1,5541618
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Take $yin Y$, there exists $xin X$ such that $pi(x_y)=y$, write $Lambda(y)=x^*(x_y)$, if $pi(x'_y)=y$, $x_y-x'_yin M$, this implies that $x^*(x_y-x'_y)=0$ and $Lambda$ is well defined.
answered yesterday
Tsemo AristideTsemo Aristide
59.1k11445
$endgroup$
$begingroup$
How do you assert the existance of the quotient map for given y?
$endgroup$
– user8469759
yesterday
$begingroup$
Does it just follow from the definition of coset? Or is it more subtle?
$endgroup$
– user8469759
yesterday
$begingroup$
It just follows from the definition of cosets, but you have to show that $Lambda$ is bounded.
$endgroup$
– Tsemo Aristide
yesterday
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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active
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$begingroup$
Take $yin Y$, there exists $xin X$ such that $pi(x_y)=y$, write $Lambda(y)=x^*(x_y)$, if $pi(x'_y)=y$, $x_y-x'_yin M$, this implies that $x^*(x_y-x'_y)=0$ and $Lambda$ is well defined.
answered yesterday
Tsemo AristideTsemo Aristide
59.1k11445
$endgroup$
$begingroup$
How do you assert the existance of the quotient map for given y?
$endgroup$
– user8469759
yesterday
$begingroup$
Does it just follow from the definition of coset? Or is it more subtle?
$endgroup$
– user8469759
yesterday
$begingroup$
It just follows from the definition of cosets, but you have to show that $Lambda$ is bounded.
$endgroup$
– Tsemo Aristide
yesterday
add a comment |
$begingroup$
Take $yin Y$, there exists $xin X$ such that $pi(x_y)=y$, write $Lambda(y)=x^*(x_y)$, if $pi(x'_y)=y$, $x_y-x'_yin M$, this implies that $x^*(x_y-x'_y)=0$ and $Lambda$ is well defined.
answered yesterday
Tsemo AristideTsemo Aristide
59.1k11445
$endgroup$
$begingroup$
How do you assert the existance of the quotient map for given y?
$endgroup$
– user8469759
yesterday
$begingroup$
Does it just follow from the definition of coset? Or is it more subtle?
$endgroup$
– user8469759
yesterday
$begingroup$
It just follows from the definition of cosets, but you have to show that $Lambda$ is bounded.
$endgroup$
– Tsemo Aristide
yesterday
add a comment |
$begingroup$
Take $yin Y$, there exists $xin X$ such that $pi(x_y)=y$, write $Lambda(y)=x^*(x_y)$, if $pi(x'_y)=y$, $x_y-x'_yin M$, this implies that $x^*(x_y-x'_y)=0$ and $Lambda$ is well defined.
answered yesterday
Tsemo AristideTsemo Aristide
59.1k11445
$endgroup$
Take $yin Y$, there exists $xin X$ such that $pi(x_y)=y$, write $Lambda(y)=x^*(x_y)$, if $pi(x'_y)=y$, $x_y-x'_yin M$, this implies that $x^*(x_y-x'_y)=0$ and $Lambda$ is well defined.
answered yesterday
Tsemo AristideTsemo Aristide
59.1k11445
answered yesterday
Tsemo AristideTsemo Aristide
59.1k11445
answered yesterday
Tsemo AristideTsemo Aristide
59.1k11445
answered yesterday
Tsemo AristideTsemo Aristide
59.1k11445
59.1k11445
$begingroup$
How do you assert the existance of the quotient map for given y?
$endgroup$
– user8469759
yesterday
$begingroup$
Does it just follow from the definition of coset? Or is it more subtle?
$endgroup$
– user8469759
yesterday
$begingroup$
It just follows from the definition of cosets, but you have to show that $Lambda$ is bounded.
$endgroup$
– Tsemo Aristide
yesterday
add a comment |
$begingroup$
How do you assert the existance of the quotient map for given y?
$endgroup$
– user8469759
yesterday
$begingroup$
Does it just follow from the definition of coset? Or is it more subtle?
$endgroup$
– user8469759
yesterday
$begingroup$
It just follows from the definition of cosets, but you have to show that $Lambda$ is bounded.
$endgroup$
– Tsemo Aristide
yesterday
$begingroup$
How do you assert the existance of the quotient map for given y?
$endgroup$
– user8469759
yesterday
$begingroup$
How do you assert the existance of the quotient map for given y?
$endgroup$
– user8469759
yesterday
$begingroup$
Does it just follow from the definition of coset? Or is it more subtle?
$endgroup$
– user8469759
yesterday
$begingroup$
Does it just follow from the definition of coset? Or is it more subtle?
$endgroup$
– user8469759
yesterday
$begingroup$
It just follows from the definition of cosets, but you have to show that $Lambda$ is bounded.
$endgroup$
– Tsemo Aristide
yesterday
$begingroup$
It just follows from the definition of cosets, but you have to show that $Lambda$ is bounded.
$endgroup$
– Tsemo Aristide
yesterday
add a comment |
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