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Is a linearly independent set whose span is dense a Schauder basis?



The Next CEO of Stack OverflowCoordinate functions of Schauder basisLinearly independentSchauder basis for a separable Banach spaceWhat is the difference between a Hamel basis and a Schauder basis?Hamel basis for subspacesExistence of weak Schauder-basis for concrete example.Isomorphisms with invariant linearly independent dense subset.Linear independence and Schauder basisWhy isn't every Hamel basis a Schauder basis?Schauder basis that is not Hilbert basis










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$begingroup$


If $X$ is a Banach space, then a Schauder basis of $X$ is a subset $B$ of $X$ such that every element of $X$ can be written uniquely as an infinite linear combination of elements of $B$. My question is, if $A$ is a linearly independent subset of $X$ such that the closure of the span of $A$ equals $X$, then is $A$ necessarily a Schauder basis of $X$?



If not, does anyone know of any counterexamples?










share|cite|improve this question









$endgroup$
















    3












    $begingroup$


    If $X$ is a Banach space, then a Schauder basis of $X$ is a subset $B$ of $X$ such that every element of $X$ can be written uniquely as an infinite linear combination of elements of $B$. My question is, if $A$ is a linearly independent subset of $X$ such that the closure of the span of $A$ equals $X$, then is $A$ necessarily a Schauder basis of $X$?



    If not, does anyone know of any counterexamples?










    share|cite|improve this question









    $endgroup$














      3












      3








      3


      1



      $begingroup$


      If $X$ is a Banach space, then a Schauder basis of $X$ is a subset $B$ of $X$ such that every element of $X$ can be written uniquely as an infinite linear combination of elements of $B$. My question is, if $A$ is a linearly independent subset of $X$ such that the closure of the span of $A$ equals $X$, then is $A$ necessarily a Schauder basis of $X$?



      If not, does anyone know of any counterexamples?










      share|cite|improve this question









      $endgroup$




      If $X$ is a Banach space, then a Schauder basis of $X$ is a subset $B$ of $X$ such that every element of $X$ can be written uniquely as an infinite linear combination of elements of $B$. My question is, if $A$ is a linearly independent subset of $X$ such that the closure of the span of $A$ equals $X$, then is $A$ necessarily a Schauder basis of $X$?



      If not, does anyone know of any counterexamples?







      linear-algebra functional-analysis banach-spaces normed-spaces schauder-basis






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      share|cite|improve this question











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      share|cite|improve this question










      asked 2 hours ago









      Keshav SrinivasanKeshav Srinivasan

      2,39121446




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          $begingroup$

          No, certainly not. The linearly independent set $1, x, x^2, x^3, dots$ has span dense in $C[0,1]$, but is not a Schauder basis of that space. (Not every continuous function is given by a power series.)



          A Schauder basis is, in general, much harder to construct than a set with dense span.



          Since Enflo we know that there are separable Banach spaces (hence they have countable, dense, linearly independent set) that have no Schauder basis at all.






          share|cite|improve this answer









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            8












            $begingroup$

            No, certainly not. The linearly independent set $1, x, x^2, x^3, dots$ has span dense in $C[0,1]$, but is not a Schauder basis of that space. (Not every continuous function is given by a power series.)



            A Schauder basis is, in general, much harder to construct than a set with dense span.



            Since Enflo we know that there are separable Banach spaces (hence they have countable, dense, linearly independent set) that have no Schauder basis at all.






            share|cite|improve this answer









            $endgroup$

















              8












              $begingroup$

              No, certainly not. The linearly independent set $1, x, x^2, x^3, dots$ has span dense in $C[0,1]$, but is not a Schauder basis of that space. (Not every continuous function is given by a power series.)



              A Schauder basis is, in general, much harder to construct than a set with dense span.



              Since Enflo we know that there are separable Banach spaces (hence they have countable, dense, linearly independent set) that have no Schauder basis at all.






              share|cite|improve this answer









              $endgroup$















                8












                8








                8





                $begingroup$

                No, certainly not. The linearly independent set $1, x, x^2, x^3, dots$ has span dense in $C[0,1]$, but is not a Schauder basis of that space. (Not every continuous function is given by a power series.)



                A Schauder basis is, in general, much harder to construct than a set with dense span.



                Since Enflo we know that there are separable Banach spaces (hence they have countable, dense, linearly independent set) that have no Schauder basis at all.






                share|cite|improve this answer









                $endgroup$



                No, certainly not. The linearly independent set $1, x, x^2, x^3, dots$ has span dense in $C[0,1]$, but is not a Schauder basis of that space. (Not every continuous function is given by a power series.)



                A Schauder basis is, in general, much harder to construct than a set with dense span.



                Since Enflo we know that there are separable Banach spaces (hence they have countable, dense, linearly independent set) that have no Schauder basis at all.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 hours ago









                GEdgarGEdgar

                63.3k268172




                63.3k268172



























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