Show that $f$ is a bounded linear functional and find its norm












0












$begingroup$


Let $f : l^{1} to {mathbb R}$ and



$f(x) = sum (1-1/n) x_n$



Where $x = (x_1, x_2 , ldots)$



Show that $f$ is a bounded linear functional and find its norm.



My work :
$| f | = | sum (1-1/n) x_n |$



$leq sum | (1-1/n) | cdot | x_n |$



$<leq | x | sum (1-1/n)$



But what I can do after that?






functional-analysis norm







share|cite|improve this question











$endgroup$












  • $begingroup$
    Since the series $sum_{n=1}^inftyleft(1-frac1nright)$ diverges, you did nothing.
    $endgroup$
    – José Carlos Santos
    Dec 19 '18 at 13:42










  • $begingroup$
    use the triangle inequality
    $endgroup$
    – tonydo
    Dec 19 '18 at 13:45
















0












$begingroup$


Let $f : l^{1} to {mathbb R}$ and



$f(x) = sum (1-1/n) x_n$



Where $x = (x_1, x_2 , ldots)$



Show that $f$ is a bounded linear functional and find its norm.



My work :
$| f | = | sum (1-1/n) x_n |$



$leq sum | (1-1/n) | cdot | x_n |$



$<leq | x | sum (1-1/n)$



But what I can do after that?






functional-analysis norm







share|cite|improve this question











$endgroup$












  • $begingroup$
    Since the series $sum_{n=1}^inftyleft(1-frac1nright)$ diverges, you did nothing.
    $endgroup$
    – José Carlos Santos
    Dec 19 '18 at 13:42










  • $begingroup$
    use the triangle inequality
    $endgroup$
    – tonydo
    Dec 19 '18 at 13:45














0












0








0





$begingroup$


Let $f : l^{1} to {mathbb R}$ and



$f(x) = sum (1-1/n) x_n$



Where $x = (x_1, x_2 , ldots)$



Show that $f$ is a bounded linear functional and find its norm.



My work :
$| f | = | sum (1-1/n) x_n |$



$leq sum | (1-1/n) | cdot | x_n |$



$<leq | x | sum (1-1/n)$



But what I can do after that?






functional-analysis norm







share|cite|improve this question











$endgroup$




Let $f : l^{1} to {mathbb R}$ and



$f(x) = sum (1-1/n) x_n$



Where $x = (x_1, x_2 , ldots)$



Show that $f$ is a bounded linear functional and find its norm.



My work :
$| f | = | sum (1-1/n) x_n |$



$leq sum | (1-1/n) | cdot | x_n |$



$<leq | x | sum (1-1/n)$



But what I can do after that?






functional-analysis norm




functional-analysis norm






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 14:54









Namaste

1




1










asked Dec 19 '18 at 13:39









Duaa HamzehDuaa Hamzeh

524




524












  • $begingroup$
    Since the series $sum_{n=1}^inftyleft(1-frac1nright)$ diverges, you did nothing.
    $endgroup$
    – José Carlos Santos
    Dec 19 '18 at 13:42










  • $begingroup$
    use the triangle inequality
    $endgroup$
    – tonydo
    Dec 19 '18 at 13:45


















  • $begingroup$
    Since the series $sum_{n=1}^inftyleft(1-frac1nright)$ diverges, you did nothing.
    $endgroup$
    – José Carlos Santos
    Dec 19 '18 at 13:42










  • $begingroup$
    use the triangle inequality
    $endgroup$
    – tonydo
    Dec 19 '18 at 13:45
















$begingroup$
Since the series $sum_{n=1}^inftyleft(1-frac1nright)$ diverges, you did nothing.
$endgroup$
– José Carlos Santos
Dec 19 '18 at 13:42




$begingroup$
Since the series $sum_{n=1}^inftyleft(1-frac1nright)$ diverges, you did nothing.
$endgroup$
– José Carlos Santos
Dec 19 '18 at 13:42












$begingroup$
use the triangle inequality
$endgroup$
– tonydo
Dec 19 '18 at 13:45




$begingroup$
use the triangle inequality
$endgroup$
– tonydo
Dec 19 '18 at 13:45










2 Answers
2






active

oldest

votes


















1












$begingroup$

We have $|f(x)| le sum_{n=1}^{infty}(1- frac{1}{n})|x_n| le sum_{n=1}^{infty}|x_n| = ||x||_1$.



Hence $f$ is bounded and $||f|| le 1$. It is your turn to determine $||f||$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    $xin l^1Rightarrow |x|=sum |x_n|<infty.$ $|f(x)|leqsum (1-1/n)|x_n|leq sum |x_n|=|x|.$ So, $|f|leq 1.$ For $|f|=sup_{|x|leq 1}|f(x)|.$ Taking $x_i=0$ for $i=1(1)n$ with $|x|=1$ and $x_igeq 0$ we get $|f|geq |f(x)|=sum (1-1/i)x_igeq (1-1/n)|x|=1-1/n.$ So, $|f|=1.$






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Space is free :) Your answer would be a lot more readable if you moved distinct steps onto separate lines.
      $endgroup$
      – postmortes
      Dec 19 '18 at 14:16











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    We have $|f(x)| le sum_{n=1}^{infty}(1- frac{1}{n})|x_n| le sum_{n=1}^{infty}|x_n| = ||x||_1$.



    Hence $f$ is bounded and $||f|| le 1$. It is your turn to determine $||f||$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      We have $|f(x)| le sum_{n=1}^{infty}(1- frac{1}{n})|x_n| le sum_{n=1}^{infty}|x_n| = ||x||_1$.



      Hence $f$ is bounded and $||f|| le 1$. It is your turn to determine $||f||$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        We have $|f(x)| le sum_{n=1}^{infty}(1- frac{1}{n})|x_n| le sum_{n=1}^{infty}|x_n| = ||x||_1$.



        Hence $f$ is bounded and $||f|| le 1$. It is your turn to determine $||f||$.






        share|cite|improve this answer









        $endgroup$



        We have $|f(x)| le sum_{n=1}^{infty}(1- frac{1}{n})|x_n| le sum_{n=1}^{infty}|x_n| = ||x||_1$.



        Hence $f$ is bounded and $||f|| le 1$. It is your turn to determine $||f||$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 19 '18 at 13:46









        FredFred

        48.6k11849




        48.6k11849























            0












            $begingroup$

            $xin l^1Rightarrow |x|=sum |x_n|<infty.$ $|f(x)|leqsum (1-1/n)|x_n|leq sum |x_n|=|x|.$ So, $|f|leq 1.$ For $|f|=sup_{|x|leq 1}|f(x)|.$ Taking $x_i=0$ for $i=1(1)n$ with $|x|=1$ and $x_igeq 0$ we get $|f|geq |f(x)|=sum (1-1/i)x_igeq (1-1/n)|x|=1-1/n.$ So, $|f|=1.$






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              Space is free :) Your answer would be a lot more readable if you moved distinct steps onto separate lines.
              $endgroup$
              – postmortes
              Dec 19 '18 at 14:16
















            0












            $begingroup$

            $xin l^1Rightarrow |x|=sum |x_n|<infty.$ $|f(x)|leqsum (1-1/n)|x_n|leq sum |x_n|=|x|.$ So, $|f|leq 1.$ For $|f|=sup_{|x|leq 1}|f(x)|.$ Taking $x_i=0$ for $i=1(1)n$ with $|x|=1$ and $x_igeq 0$ we get $|f|geq |f(x)|=sum (1-1/i)x_igeq (1-1/n)|x|=1-1/n.$ So, $|f|=1.$






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              Space is free :) Your answer would be a lot more readable if you moved distinct steps onto separate lines.
              $endgroup$
              – postmortes
              Dec 19 '18 at 14:16














            0












            0








            0





            $begingroup$

            $xin l^1Rightarrow |x|=sum |x_n|<infty.$ $|f(x)|leqsum (1-1/n)|x_n|leq sum |x_n|=|x|.$ So, $|f|leq 1.$ For $|f|=sup_{|x|leq 1}|f(x)|.$ Taking $x_i=0$ for $i=1(1)n$ with $|x|=1$ and $x_igeq 0$ we get $|f|geq |f(x)|=sum (1-1/i)x_igeq (1-1/n)|x|=1-1/n.$ So, $|f|=1.$






            share|cite|improve this answer









            $endgroup$



            $xin l^1Rightarrow |x|=sum |x_n|<infty.$ $|f(x)|leqsum (1-1/n)|x_n|leq sum |x_n|=|x|.$ So, $|f|leq 1.$ For $|f|=sup_{|x|leq 1}|f(x)|.$ Taking $x_i=0$ for $i=1(1)n$ with $|x|=1$ and $x_igeq 0$ we get $|f|geq |f(x)|=sum (1-1/i)x_igeq (1-1/n)|x|=1-1/n.$ So, $|f|=1.$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 19 '18 at 13:50









            John_WickJohn_Wick

            1,616111




            1,616111








            • 1




              $begingroup$
              Space is free :) Your answer would be a lot more readable if you moved distinct steps onto separate lines.
              $endgroup$
              – postmortes
              Dec 19 '18 at 14:16














            • 1




              $begingroup$
              Space is free :) Your answer would be a lot more readable if you moved distinct steps onto separate lines.
              $endgroup$
              – postmortes
              Dec 19 '18 at 14:16








            1




            1




            $begingroup$
            Space is free :) Your answer would be a lot more readable if you moved distinct steps onto separate lines.
            $endgroup$
            – postmortes
            Dec 19 '18 at 14:16




            $begingroup$
            Space is free :) Your answer would be a lot more readable if you moved distinct steps onto separate lines.
            $endgroup$
            – postmortes
            Dec 19 '18 at 14:16


















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