Prove that $S_{n} : X to X$ is continuous mapping.












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$begingroup$


Consider $X$ - Banach space with Schauder basis : $forall x in X$ $exists!$ $x = sum_{i=1}^{infty} x_{i}e_{i}$.



Let's consider $S_{n} = sum_{i = 1}^{n} x_{i}e_{i}$. We want to show that it is a continous mapping (i.e. $exists C > 0: $ $|S_n| le C |x|$, for all $x in X$).



How should I start with? Should I use Cauchy-sequences?










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$endgroup$

















    0












    $begingroup$


    Consider $X$ - Banach space with Schauder basis : $forall x in X$ $exists!$ $x = sum_{i=1}^{infty} x_{i}e_{i}$.



    Let's consider $S_{n} = sum_{i = 1}^{n} x_{i}e_{i}$. We want to show that it is a continous mapping (i.e. $exists C > 0: $ $|S_n| le C |x|$, for all $x in X$).



    How should I start with? Should I use Cauchy-sequences?










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Consider $X$ - Banach space with Schauder basis : $forall x in X$ $exists!$ $x = sum_{i=1}^{infty} x_{i}e_{i}$.



      Let's consider $S_{n} = sum_{i = 1}^{n} x_{i}e_{i}$. We want to show that it is a continous mapping (i.e. $exists C > 0: $ $|S_n| le C |x|$, for all $x in X$).



      How should I start with? Should I use Cauchy-sequences?










      share|cite|improve this question









      $endgroup$




      Consider $X$ - Banach space with Schauder basis : $forall x in X$ $exists!$ $x = sum_{i=1}^{infty} x_{i}e_{i}$.



      Let's consider $S_{n} = sum_{i = 1}^{n} x_{i}e_{i}$. We want to show that it is a continous mapping (i.e. $exists C > 0: $ $|S_n| le C |x|$, for all $x in X$).



      How should I start with? Should I use Cauchy-sequences?







      real-analysis functional-analysis banach-spaces






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      asked Dec 20 '18 at 17:45









      openspaceopenspace

      3,4512822




      3,4512822






















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          $begingroup$

          $x = sum_{i=1}^infty x_i e_i$ means that the partial sums converge to $x$ in norm. So, if you add and subtract $x$ inside $|S_n|$, then apply the triangle inequality, you should be able to finish the proof.






          share|cite|improve this answer









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            1 Answer
            1






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            active

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            0












            $begingroup$

            $x = sum_{i=1}^infty x_i e_i$ means that the partial sums converge to $x$ in norm. So, if you add and subtract $x$ inside $|S_n|$, then apply the triangle inequality, you should be able to finish the proof.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              $x = sum_{i=1}^infty x_i e_i$ means that the partial sums converge to $x$ in norm. So, if you add and subtract $x$ inside $|S_n|$, then apply the triangle inequality, you should be able to finish the proof.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                $x = sum_{i=1}^infty x_i e_i$ means that the partial sums converge to $x$ in norm. So, if you add and subtract $x$ inside $|S_n|$, then apply the triangle inequality, you should be able to finish the proof.






                share|cite|improve this answer









                $endgroup$



                $x = sum_{i=1}^infty x_i e_i$ means that the partial sums converge to $x$ in norm. So, if you add and subtract $x$ inside $|S_n|$, then apply the triangle inequality, you should be able to finish the proof.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 20 '18 at 19:05









                DunhamDunham

                2,074614




                2,074614






























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