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






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







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    5












    $begingroup$


    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?






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







    share|cite|improve this question









    $endgroup$















      5












      5








      5


      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?






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







      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




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






      share|cite|improve this question













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      asked 7 hours ago









      Keshav SrinivasanKeshav Srinivasan

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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 subset set) that have no Schauder basis at all.






          share|cite|improve this answer











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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 subset set) that have no Schauder basis at all.






            share|cite|improve this answer











            $endgroup$


















              12












              $begingroup$

              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 subset set) that have no Schauder basis at all.






              share|cite|improve this answer











              $endgroup$
















                12












                12








                12





                $begingroup$

                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 subset 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 subset set) that have no Schauder basis at all.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 4 hours ago

























                answered 7 hours ago









                GEdgarGEdgar

                63.3k268173




                63.3k268173






























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