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Hahn-Banch theorem in Rudin functional analysis
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On page 59, in the first part of the geometric form of Hahn-Banach theorem, it only assumed that $X$ is a topological space and it says
there exists $Lambdain X^*$ and ...
Here $X^*$ is the continuous dual.
But we know when the topological vector space is not locally convex, the continuous dual space might be trivial, $L^p[0,1]$ where $0<p<1$ is such example. How is this possible?
In the second part of the theorem, we have the additional assumption that $X$ is locally convex.
real-analysis functional-analysis dual-spaces
edited 4 hours ago
Clayton
18.5k22983
asked 5 hours ago
Xiao
4,53711334
add a comment |Â
up vote
1
down vote
favorite
On page 59, in the first part of the geometric form of Hahn-Banach theorem, it only assumed that $X$ is a topological space and it says
there exists $Lambdain X^*$ and ...
Here $X^*$ is the continuous dual.
But we know when the topological vector space is not locally convex, the continuous dual space might be trivial, $L^p[0,1]$ where $0<p<1$ is such example. How is this possible?
In the second part of the theorem, we have the additional assumption that $X$ is locally convex.
real-analysis functional-analysis dual-spaces
edited 4 hours ago
Clayton
18.5k22983
asked 5 hours ago
Xiao
4,53711334
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
On page 59, in the first part of the geometric form of Hahn-Banach theorem, it only assumed that $X$ is a topological space and it says
there exists $Lambdain X^*$ and ...
Here $X^*$ is the continuous dual.
But we know when the topological vector space is not locally convex, the continuous dual space might be trivial, $L^p[0,1]$ where $0<p<1$ is such example. How is this possible?
In the second part of the theorem, we have the additional assumption that $X$ is locally convex.
real-analysis functional-analysis dual-spaces
edited 4 hours ago
Clayton
18.5k22983
asked 5 hours ago
Xiao
4,53711334
On page 59, in the first part of the geometric form of Hahn-Banach theorem, it only assumed that $X$ is a topological space and it says
there exists $Lambdain X^*$ and ...
Here $X^*$ is the continuous dual.
But we know when the topological vector space is not locally convex, the continuous dual space might be trivial, $L^p[0,1]$ where $0<p<1$ is such example. How is this possible?
In the second part of the theorem, we have the additional assumption that $X$ is locally convex.
real-analysis functional-analysis dual-spaces
real-analysis functional-analysis dual-spaces
edited 4 hours ago
Clayton
18.5k22983
asked 5 hours ago
Xiao
4,53711334
edited 4 hours ago
Clayton
18.5k22983
asked 5 hours ago
Xiao
4,53711334
edited 4 hours ago
Clayton
18.5k22983
edited 4 hours ago
Clayton
18.5k22983
edited 4 hours ago
Clayton
18.5k22983
18.5k22983
asked 5 hours ago
Xiao
4,53711334
asked 5 hours ago
Xiao
4,53711334
asked 5 hours ago
Xiao
4,53711334
4,53711334
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
The theorem never in any way asserts that a nontrivial element of $X^*$ must exist for arbitrary $X$. It only asserts that given disjoint, nonempty convex sets $A,Bsubset X$ with $A$ open, then there exists a $Lambdain X^*$ satisfying the stated inequalities. It's entirely possible that for some $X$, no such sets $A$ and $B$ exist.
answered 4 hours ago
Eric Wofsey
169k12196313
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
The theorem never in any way asserts that a nontrivial element of $X^*$ must exist for arbitrary $X$. It only asserts that given disjoint, nonempty convex sets $A,Bsubset X$ with $A$ open, then there exists a $Lambdain X^*$ satisfying the stated inequalities. It's entirely possible that for some $X$, no such sets $A$ and $B$ exist.
answered 4 hours ago
Eric Wofsey
169k12196313
add a comment |Â
up vote
4
down vote
accepted
The theorem never in any way asserts that a nontrivial element of $X^*$ must exist for arbitrary $X$. It only asserts that given disjoint, nonempty convex sets $A,Bsubset X$ with $A$ open, then there exists a $Lambdain X^*$ satisfying the stated inequalities. It's entirely possible that for some $X$, no such sets $A$ and $B$ exist.
answered 4 hours ago
Eric Wofsey
169k12196313
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
The theorem never in any way asserts that a nontrivial element of $X^*$ must exist for arbitrary $X$. It only asserts that given disjoint, nonempty convex sets $A,Bsubset X$ with $A$ open, then there exists a $Lambdain X^*$ satisfying the stated inequalities. It's entirely possible that for some $X$, no such sets $A$ and $B$ exist.
answered 4 hours ago
Eric Wofsey
169k12196313
The theorem never in any way asserts that a nontrivial element of $X^*$ must exist for arbitrary $X$. It only asserts that given disjoint, nonempty convex sets $A,Bsubset X$ with $A$ open, then there exists a $Lambdain X^*$ satisfying the stated inequalities. It's entirely possible that for some $X$, no such sets $A$ and $B$ exist.
answered 4 hours ago
Eric Wofsey
169k12196313
answered 4 hours ago
Eric Wofsey
169k12196313
answered 4 hours ago
Eric Wofsey
169k12196313
answered 4 hours ago
Eric Wofsey
169k12196313
169k12196313
add a comment |Â
add a comment |Â
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