Continuous function and Lp norm












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On the vector space $C([0,1])$ of all continuous functions $f : [0,1] to K$ consider the $p$-norm $$|f|_p =left(int_0^1 |f(t)|^p dtright)^{frac{1}{p}},$$ $f in C([0,1])$, where $1 le p < infty$, as well as the uniform norm $|f|_{infty} = sup_{tin[0,1]}|f(t)|$.



I try to show $|f|_p leq |f|_infty$ for all $f in C([0,1])$ and that $(C([0,1]),|·|_p)$ is not complete.



Can please someone help? I am thinking the first part could be releated to the Minkowski inequality.






functional-analysis lp-spaces







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  • $begingroup$
    What is $K$? Presumably, it is either $mathbb{C}$ or $mathbb{R}$, but it would be good to specify this.
    $endgroup$
    – Xander Henderson
    Dec 16 '18 at 20:52


















0












$begingroup$


On the vector space $C([0,1])$ of all continuous functions $f : [0,1] to K$ consider the $p$-norm $$|f|_p =left(int_0^1 |f(t)|^p dtright)^{frac{1}{p}},$$ $f in C([0,1])$, where $1 le p < infty$, as well as the uniform norm $|f|_{infty} = sup_{tin[0,1]}|f(t)|$.



I try to show $|f|_p leq |f|_infty$ for all $f in C([0,1])$ and that $(C([0,1]),|·|_p)$ is not complete.



Can please someone help? I am thinking the first part could be releated to the Minkowski inequality.






functional-analysis lp-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is $K$? Presumably, it is either $mathbb{C}$ or $mathbb{R}$, but it would be good to specify this.
    $endgroup$
    – Xander Henderson
    Dec 16 '18 at 20:52
















0












0








0





$begingroup$


On the vector space $C([0,1])$ of all continuous functions $f : [0,1] to K$ consider the $p$-norm $$|f|_p =left(int_0^1 |f(t)|^p dtright)^{frac{1}{p}},$$ $f in C([0,1])$, where $1 le p < infty$, as well as the uniform norm $|f|_{infty} = sup_{tin[0,1]}|f(t)|$.



I try to show $|f|_p leq |f|_infty$ for all $f in C([0,1])$ and that $(C([0,1]),|·|_p)$ is not complete.



Can please someone help? I am thinking the first part could be releated to the Minkowski inequality.






functional-analysis lp-spaces







share|cite|improve this question











$endgroup$




On the vector space $C([0,1])$ of all continuous functions $f : [0,1] to K$ consider the $p$-norm $$|f|_p =left(int_0^1 |f(t)|^p dtright)^{frac{1}{p}},$$ $f in C([0,1])$, where $1 le p < infty$, as well as the uniform norm $|f|_{infty} = sup_{tin[0,1]}|f(t)|$.



I try to show $|f|_p leq |f|_infty$ for all $f in C([0,1])$ and that $(C([0,1]),|·|_p)$ is not complete.



Can please someone help? I am thinking the first part could be releated to the Minkowski inequality.






functional-analysis lp-spaces




functional-analysis lp-spaces






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edited Dec 16 '18 at 20:51









Xander Henderson

14.6k103555




14.6k103555










asked Dec 16 '18 at 20:49









Anna SchmitzAnna Schmitz

917




917












  • $begingroup$
    What is $K$? Presumably, it is either $mathbb{C}$ or $mathbb{R}$, but it would be good to specify this.
    $endgroup$
    – Xander Henderson
    Dec 16 '18 at 20:52




















  • $begingroup$
    What is $K$? Presumably, it is either $mathbb{C}$ or $mathbb{R}$, but it would be good to specify this.
    $endgroup$
    – Xander Henderson
    Dec 16 '18 at 20:52


















$begingroup$
What is $K$? Presumably, it is either $mathbb{C}$ or $mathbb{R}$, but it would be good to specify this.
$endgroup$
– Xander Henderson
Dec 16 '18 at 20:52






$begingroup$
What is $K$? Presumably, it is either $mathbb{C}$ or $mathbb{R}$, but it would be good to specify this.
$endgroup$
– Xander Henderson
Dec 16 '18 at 20:52












2 Answers
2






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$begingroup$

Since $|f(x)|leq sup_{[0,1]}|f|$ for all $xin [0,1]$,
$$int_0^1|f(x)|^pdxleq (sup_{[0,1]}|f|)^pint_0^1dx=|f|_infty ^p.$$



Therefore $$|f|_{p}=sqrt[p]{int_0^1|f(x)|^pdx}leq |f|_{infty }$$






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    For the second part, consider for example
    $$ f_n(x) = tanh left(n(x - 1/2)right).$$
    As $n to infty$, $f_n to g$ in $L^p$ if $p < infty$, where
    $$ g(x) = begin{cases} & -1 quad text{if} , x leq 1/2 \
    & 1 quad text{if} , x > 1/2 end{cases},$$
    but $g notin C([0,1])$.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

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      active

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      active

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      3












      $begingroup$

      Since $|f(x)|leq sup_{[0,1]}|f|$ for all $xin [0,1]$,
      $$int_0^1|f(x)|^pdxleq (sup_{[0,1]}|f|)^pint_0^1dx=|f|_infty ^p.$$



      Therefore $$|f|_{p}=sqrt[p]{int_0^1|f(x)|^pdx}leq |f|_{infty }$$






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        Since $|f(x)|leq sup_{[0,1]}|f|$ for all $xin [0,1]$,
        $$int_0^1|f(x)|^pdxleq (sup_{[0,1]}|f|)^pint_0^1dx=|f|_infty ^p.$$



        Therefore $$|f|_{p}=sqrt[p]{int_0^1|f(x)|^pdx}leq |f|_{infty }$$






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          Since $|f(x)|leq sup_{[0,1]}|f|$ for all $xin [0,1]$,
          $$int_0^1|f(x)|^pdxleq (sup_{[0,1]}|f|)^pint_0^1dx=|f|_infty ^p.$$



          Therefore $$|f|_{p}=sqrt[p]{int_0^1|f(x)|^pdx}leq |f|_{infty }$$






          share|cite|improve this answer









          $endgroup$



          Since $|f(x)|leq sup_{[0,1]}|f|$ for all $xin [0,1]$,
          $$int_0^1|f(x)|^pdxleq (sup_{[0,1]}|f|)^pint_0^1dx=|f|_infty ^p.$$



          Therefore $$|f|_{p}=sqrt[p]{int_0^1|f(x)|^pdx}leq |f|_{infty }$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 16 '18 at 20:51









          NewMathNewMath

          4059




          4059























              2












              $begingroup$

              For the second part, consider for example
              $$ f_n(x) = tanh left(n(x - 1/2)right).$$
              As $n to infty$, $f_n to g$ in $L^p$ if $p < infty$, where
              $$ g(x) = begin{cases} & -1 quad text{if} , x leq 1/2 \
              & 1 quad text{if} , x > 1/2 end{cases},$$
              but $g notin C([0,1])$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                For the second part, consider for example
                $$ f_n(x) = tanh left(n(x - 1/2)right).$$
                As $n to infty$, $f_n to g$ in $L^p$ if $p < infty$, where
                $$ g(x) = begin{cases} & -1 quad text{if} , x leq 1/2 \
                & 1 quad text{if} , x > 1/2 end{cases},$$
                but $g notin C([0,1])$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  For the second part, consider for example
                  $$ f_n(x) = tanh left(n(x - 1/2)right).$$
                  As $n to infty$, $f_n to g$ in $L^p$ if $p < infty$, where
                  $$ g(x) = begin{cases} & -1 quad text{if} , x leq 1/2 \
                  & 1 quad text{if} , x > 1/2 end{cases},$$
                  but $g notin C([0,1])$.






                  share|cite|improve this answer









                  $endgroup$



                  For the second part, consider for example
                  $$ f_n(x) = tanh left(n(x - 1/2)right).$$
                  As $n to infty$, $f_n to g$ in $L^p$ if $p < infty$, where
                  $$ g(x) = begin{cases} & -1 quad text{if} , x leq 1/2 \
                  & 1 quad text{if} , x > 1/2 end{cases},$$
                  but $g notin C([0,1])$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 16 '18 at 20:57









                  Roberto RastapopoulosRoberto Rastapopoulos

                  928425




                  928425






























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