Proof that $l^p$ with $1 leq p < infty$ is dense in $c_0$











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Proof that $l^p$ with $1 leq p < infty$ is dense in $c_0$
My definition of $c_0$ is the following:



$$ c_0 = left{ x vert lim_{kto infty} x_k = 0right}.$$



I want to prove it like this: every element of $c_0$ can be written as limit of $l^p$.
First question, should I use the $p$ norm or the sup norm?
Second question, how should I start given that I think this statement is true?










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    up vote
    1
    down vote

    favorite












    Proof that $l^p$ with $1 leq p < infty$ is dense in $c_0$
    My definition of $c_0$ is the following:



    $$ c_0 = left{ x vert lim_{kto infty} x_k = 0right}.$$



    I want to prove it like this: every element of $c_0$ can be written as limit of $l^p$.
    First question, should I use the $p$ norm or the sup norm?
    Second question, how should I start given that I think this statement is true?










    share|cite|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Proof that $l^p$ with $1 leq p < infty$ is dense in $c_0$
      My definition of $c_0$ is the following:



      $$ c_0 = left{ x vert lim_{kto infty} x_k = 0right}.$$



      I want to prove it like this: every element of $c_0$ can be written as limit of $l^p$.
      First question, should I use the $p$ norm or the sup norm?
      Second question, how should I start given that I think this statement is true?










      share|cite|improve this question















      Proof that $l^p$ with $1 leq p < infty$ is dense in $c_0$
      My definition of $c_0$ is the following:



      $$ c_0 = left{ x vert lim_{kto infty} x_k = 0right}.$$



      I want to prove it like this: every element of $c_0$ can be written as limit of $l^p$.
      First question, should I use the $p$ norm or the sup norm?
      Second question, how should I start given that I think this statement is true?







      general-topology functional-analysis lp-spaces






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      edited Nov 20 at 21:38









      Davide Giraudo

      124k16150257




      124k16150257










      asked Nov 20 at 21:24









      qcc101

      454112




      454112






















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          The space $c_0$ is usually considered as a subspace of $ell^infty$, and hence automatically inherits the sup-norm.
          For any given $x = (x_1, x_2, ldots) in c_0$, you can truncate it to obtain an $ell^p$ sequence, i.e. you can consider
          $$ hat x^{(n)} = (x_1, x_2, ldots, x_n , 0, 0 ,ldots) in ell^p. $$
          Then
          $$ Vert hat x^{(n)} - xVert_{infty} = sup_{j geq n+1} |x_j|, $$
          which can evidently be made arbitrarily small by taking $n$ to be large, since $x_j to 0$ as $j to infty$.






          share|cite|improve this answer





















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            1 Answer
            1






            active

            oldest

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            active

            oldest

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            active

            oldest

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            up vote
            2
            down vote



            accepted










            The space $c_0$ is usually considered as a subspace of $ell^infty$, and hence automatically inherits the sup-norm.
            For any given $x = (x_1, x_2, ldots) in c_0$, you can truncate it to obtain an $ell^p$ sequence, i.e. you can consider
            $$ hat x^{(n)} = (x_1, x_2, ldots, x_n , 0, 0 ,ldots) in ell^p. $$
            Then
            $$ Vert hat x^{(n)} - xVert_{infty} = sup_{j geq n+1} |x_j|, $$
            which can evidently be made arbitrarily small by taking $n$ to be large, since $x_j to 0$ as $j to infty$.






            share|cite|improve this answer

























              up vote
              2
              down vote



              accepted










              The space $c_0$ is usually considered as a subspace of $ell^infty$, and hence automatically inherits the sup-norm.
              For any given $x = (x_1, x_2, ldots) in c_0$, you can truncate it to obtain an $ell^p$ sequence, i.e. you can consider
              $$ hat x^{(n)} = (x_1, x_2, ldots, x_n , 0, 0 ,ldots) in ell^p. $$
              Then
              $$ Vert hat x^{(n)} - xVert_{infty} = sup_{j geq n+1} |x_j|, $$
              which can evidently be made arbitrarily small by taking $n$ to be large, since $x_j to 0$ as $j to infty$.






              share|cite|improve this answer























                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                The space $c_0$ is usually considered as a subspace of $ell^infty$, and hence automatically inherits the sup-norm.
                For any given $x = (x_1, x_2, ldots) in c_0$, you can truncate it to obtain an $ell^p$ sequence, i.e. you can consider
                $$ hat x^{(n)} = (x_1, x_2, ldots, x_n , 0, 0 ,ldots) in ell^p. $$
                Then
                $$ Vert hat x^{(n)} - xVert_{infty} = sup_{j geq n+1} |x_j|, $$
                which can evidently be made arbitrarily small by taking $n$ to be large, since $x_j to 0$ as $j to infty$.






                share|cite|improve this answer












                The space $c_0$ is usually considered as a subspace of $ell^infty$, and hence automatically inherits the sup-norm.
                For any given $x = (x_1, x_2, ldots) in c_0$, you can truncate it to obtain an $ell^p$ sequence, i.e. you can consider
                $$ hat x^{(n)} = (x_1, x_2, ldots, x_n , 0, 0 ,ldots) in ell^p. $$
                Then
                $$ Vert hat x^{(n)} - xVert_{infty} = sup_{j geq n+1} |x_j|, $$
                which can evidently be made arbitrarily small by taking $n$ to be large, since $x_j to 0$ as $j to infty$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 20 at 21:32









                MisterRiemann

                5,4471623




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