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







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    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







    share|cite|improve this question

























      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







      share|cite|improve this question















      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






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      share|cite|improve this question













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      edited 4 hours ago









      Clayton

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      Xiao

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          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.






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            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.






            share|cite|improve this answer
























              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.






              share|cite|improve this answer






















                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.






                share|cite|improve this answer












                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.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 4 hours ago









                Eric Wofsey

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