Continuity of a function between $l^p$ spaces












2












$begingroup$


Let us consider the function defined as
$$
F: l^4 rightarrow l^6 \
(x_1, dots,x_n, dots) mapsto (x_1^{20}, dots, x_n^{20}, dots)
$$
I am asked to prove whether this function is continuous or not.
Let us try proving the continuity, then using the $epsilon - delta$ definition I need to find, given $epsilon > 0$ a suitable $delta$ such that
$$
lvert lvert (x_1, dots,x_n, dots) - (y_1, dots,y_n, dots) rvert rvert_{l^4} = left(sum_{n in mathbb{N}}left| x_n-y_n right|^{4}right)^frac{1}{4} < delta
$$
Implies
$$
lvertlvert F(x_1, dots,x_n, dots) - F(y_1, dots,y_n, dots)rvertrvert_{l^{6}} = left(sum_{n in mathbb{N}}left| x_n^{20}-y_n^{20} right|^{6}right)^frac{1}{6} < epsilon
$$
Anyway I am stuck here since I cannot find any good inequalities, so I am starting having doubts about its continuity.






real-analysis functional-analysis lp-spaces







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$endgroup$

















    2












    $begingroup$


    Let us consider the function defined as
    $$
    F: l^4 rightarrow l^6 \
    (x_1, dots,x_n, dots) mapsto (x_1^{20}, dots, x_n^{20}, dots)
    $$
    I am asked to prove whether this function is continuous or not.
    Let us try proving the continuity, then using the $epsilon - delta$ definition I need to find, given $epsilon > 0$ a suitable $delta$ such that
    $$
    lvert lvert (x_1, dots,x_n, dots) - (y_1, dots,y_n, dots) rvert rvert_{l^4} = left(sum_{n in mathbb{N}}left| x_n-y_n right|^{4}right)^frac{1}{4} < delta
    $$
    Implies
    $$
    lvertlvert F(x_1, dots,x_n, dots) - F(y_1, dots,y_n, dots)rvertrvert_{l^{6}} = left(sum_{n in mathbb{N}}left| x_n^{20}-y_n^{20} right|^{6}right)^frac{1}{6} < epsilon
    $$
    Anyway I am stuck here since I cannot find any good inequalities, so I am starting having doubts about its continuity.






    real-analysis functional-analysis lp-spaces







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      2



      $begingroup$


      Let us consider the function defined as
      $$
      F: l^4 rightarrow l^6 \
      (x_1, dots,x_n, dots) mapsto (x_1^{20}, dots, x_n^{20}, dots)
      $$
      I am asked to prove whether this function is continuous or not.
      Let us try proving the continuity, then using the $epsilon - delta$ definition I need to find, given $epsilon > 0$ a suitable $delta$ such that
      $$
      lvert lvert (x_1, dots,x_n, dots) - (y_1, dots,y_n, dots) rvert rvert_{l^4} = left(sum_{n in mathbb{N}}left| x_n-y_n right|^{4}right)^frac{1}{4} < delta
      $$
      Implies
      $$
      lvertlvert F(x_1, dots,x_n, dots) - F(y_1, dots,y_n, dots)rvertrvert_{l^{6}} = left(sum_{n in mathbb{N}}left| x_n^{20}-y_n^{20} right|^{6}right)^frac{1}{6} < epsilon
      $$
      Anyway I am stuck here since I cannot find any good inequalities, so I am starting having doubts about its continuity.






      real-analysis functional-analysis lp-spaces







      share|cite|improve this question









      $endgroup$




      Let us consider the function defined as
      $$
      F: l^4 rightarrow l^6 \
      (x_1, dots,x_n, dots) mapsto (x_1^{20}, dots, x_n^{20}, dots)
      $$
      I am asked to prove whether this function is continuous or not.
      Let us try proving the continuity, then using the $epsilon - delta$ definition I need to find, given $epsilon > 0$ a suitable $delta$ such that
      $$
      lvert lvert (x_1, dots,x_n, dots) - (y_1, dots,y_n, dots) rvert rvert_{l^4} = left(sum_{n in mathbb{N}}left| x_n-y_n right|^{4}right)^frac{1}{4} < delta
      $$
      Implies
      $$
      lvertlvert F(x_1, dots,x_n, dots) - F(y_1, dots,y_n, dots)rvertrvert_{l^{6}} = left(sum_{n in mathbb{N}}left| x_n^{20}-y_n^{20} right|^{6}right)^frac{1}{6} < epsilon
      $$
      Anyway I am stuck here since I cannot find any good inequalities, so I am starting having doubts about its continuity.






      real-analysis functional-analysis lp-spaces




      real-analysis functional-analysis lp-spaces






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 27 at 17:40









      JCFJCF

      354112




      354112






















          1 Answer
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          $begingroup$

          Probably not the best proof, but the best I can think of right now.



          For sequences, $|boldsymbol{x}|_qle|boldsymbol{x}|_p$ whenever $ple q$. Also, $$|x^{20}-y^{20}|=|x-y||x^{19}+cdots+y^{19}|le c|x-y|.$$ Hence $$|(x_n^{20})-(y_n^{20})|_6le c|x-y|_6le c|x-y|_4<cdelta$$



          To justify the constant $c$, note that begin{align*}left|sum_{i=0}^{19}x_n^iy_n^{19-i}right|&lesum_{i=0}^{19}|x_n|^i|y_n|^{19-i}\
          &lesum_{i=0}^{19}frac{i|x_n|^{19}+(19-i)|y_n|^{19}}{19}\
          &= 20(|x_n|^{19}+|y_n|^{19})\
          &le 20(|boldsymbol{x}|_infty^{19}+|boldsymbol{y}|_infty^{19})=c
          end{align*}






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            What about the constant c: How do you control it? I think that it depends on $x,y$ hence it shouldn't be uniform for all $x_n, y_n$. Maybe I am missing something.
            $endgroup$
            – JCF
            Jan 31 at 10:52











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          1 Answer
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          1 Answer
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          active

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          1












          $begingroup$

          Probably not the best proof, but the best I can think of right now.



          For sequences, $|boldsymbol{x}|_qle|boldsymbol{x}|_p$ whenever $ple q$. Also, $$|x^{20}-y^{20}|=|x-y||x^{19}+cdots+y^{19}|le c|x-y|.$$ Hence $$|(x_n^{20})-(y_n^{20})|_6le c|x-y|_6le c|x-y|_4<cdelta$$



          To justify the constant $c$, note that begin{align*}left|sum_{i=0}^{19}x_n^iy_n^{19-i}right|&lesum_{i=0}^{19}|x_n|^i|y_n|^{19-i}\
          &lesum_{i=0}^{19}frac{i|x_n|^{19}+(19-i)|y_n|^{19}}{19}\
          &= 20(|x_n|^{19}+|y_n|^{19})\
          &le 20(|boldsymbol{x}|_infty^{19}+|boldsymbol{y}|_infty^{19})=c
          end{align*}






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            What about the constant c: How do you control it? I think that it depends on $x,y$ hence it shouldn't be uniform for all $x_n, y_n$. Maybe I am missing something.
            $endgroup$
            – JCF
            Jan 31 at 10:52
















          1












          $begingroup$

          Probably not the best proof, but the best I can think of right now.



          For sequences, $|boldsymbol{x}|_qle|boldsymbol{x}|_p$ whenever $ple q$. Also, $$|x^{20}-y^{20}|=|x-y||x^{19}+cdots+y^{19}|le c|x-y|.$$ Hence $$|(x_n^{20})-(y_n^{20})|_6le c|x-y|_6le c|x-y|_4<cdelta$$



          To justify the constant $c$, note that begin{align*}left|sum_{i=0}^{19}x_n^iy_n^{19-i}right|&lesum_{i=0}^{19}|x_n|^i|y_n|^{19-i}\
          &lesum_{i=0}^{19}frac{i|x_n|^{19}+(19-i)|y_n|^{19}}{19}\
          &= 20(|x_n|^{19}+|y_n|^{19})\
          &le 20(|boldsymbol{x}|_infty^{19}+|boldsymbol{y}|_infty^{19})=c
          end{align*}






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            What about the constant c: How do you control it? I think that it depends on $x,y$ hence it shouldn't be uniform for all $x_n, y_n$. Maybe I am missing something.
            $endgroup$
            – JCF
            Jan 31 at 10:52














          1












          1








          1





          $begingroup$

          Probably not the best proof, but the best I can think of right now.



          For sequences, $|boldsymbol{x}|_qle|boldsymbol{x}|_p$ whenever $ple q$. Also, $$|x^{20}-y^{20}|=|x-y||x^{19}+cdots+y^{19}|le c|x-y|.$$ Hence $$|(x_n^{20})-(y_n^{20})|_6le c|x-y|_6le c|x-y|_4<cdelta$$



          To justify the constant $c$, note that begin{align*}left|sum_{i=0}^{19}x_n^iy_n^{19-i}right|&lesum_{i=0}^{19}|x_n|^i|y_n|^{19-i}\
          &lesum_{i=0}^{19}frac{i|x_n|^{19}+(19-i)|y_n|^{19}}{19}\
          &= 20(|x_n|^{19}+|y_n|^{19})\
          &le 20(|boldsymbol{x}|_infty^{19}+|boldsymbol{y}|_infty^{19})=c
          end{align*}






          share|cite|improve this answer











          $endgroup$



          Probably not the best proof, but the best I can think of right now.



          For sequences, $|boldsymbol{x}|_qle|boldsymbol{x}|_p$ whenever $ple q$. Also, $$|x^{20}-y^{20}|=|x-y||x^{19}+cdots+y^{19}|le c|x-y|.$$ Hence $$|(x_n^{20})-(y_n^{20})|_6le c|x-y|_6le c|x-y|_4<cdelta$$



          To justify the constant $c$, note that begin{align*}left|sum_{i=0}^{19}x_n^iy_n^{19-i}right|&lesum_{i=0}^{19}|x_n|^i|y_n|^{19-i}\
          &lesum_{i=0}^{19}frac{i|x_n|^{19}+(19-i)|y_n|^{19}}{19}\
          &= 20(|x_n|^{19}+|y_n|^{19})\
          &le 20(|boldsymbol{x}|_infty^{19}+|boldsymbol{y}|_infty^{19})=c
          end{align*}







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Feb 1 at 9:13

























          answered Jan 29 at 12:11









          ChrystomathChrystomath

          1,838513




          1,838513












          • $begingroup$
            What about the constant c: How do you control it? I think that it depends on $x,y$ hence it shouldn't be uniform for all $x_n, y_n$. Maybe I am missing something.
            $endgroup$
            – JCF
            Jan 31 at 10:52


















          • $begingroup$
            What about the constant c: How do you control it? I think that it depends on $x,y$ hence it shouldn't be uniform for all $x_n, y_n$. Maybe I am missing something.
            $endgroup$
            – JCF
            Jan 31 at 10:52
















          $begingroup$
          What about the constant c: How do you control it? I think that it depends on $x,y$ hence it shouldn't be uniform for all $x_n, y_n$. Maybe I am missing something.
          $endgroup$
          – JCF
          Jan 31 at 10:52




          $begingroup$
          What about the constant c: How do you control it? I think that it depends on $x,y$ hence it shouldn't be uniform for all $x_n, y_n$. Maybe I am missing something.
          $endgroup$
          – JCF
          Jan 31 at 10:52


















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