Operator norm on Lebesgue integrable functions












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Let $L_1([0,1],m)$ be the Banach space of $mathbb{K}$-valued integrable functions with respect to Lebesgue measure $m$, where $mathbb{K}$ is either $mathbb{R}$ or $mathbb{C}$. The norm on this space is defined like this: $||f||_1=int_{[0,1]}|f| dm$. I have to show that:
$a)$ For $n geq 2$ the operator $varphi_n(f)=int_{[0,1]} f g_n dm$, where $g_n(x)=n sin(n^2x)$ for $x in [0,1]$ is bounded with $||varphi_n||=n$.
$b)$ Show that there exists $f in L_1([0,1],m)$ such that $lim_{n to infty} |varphi_n( f)|=infty$.

MY ATTEMPT:

$g_n$ is Lebesgue integrable on $[0,1]$ since it's Riemann integrable. Hence $fg_n in L_1([0,1],m)$ and $|int_{[0,1]}fg_n dm| leq int_{[0,1]}|fg_n| dm$. We also have that $||fg_n||_1 leq ||f||_1||g_n||_infty$.
Thus $||varphi_n(f)||=|int_{[0,1]}fg_n dm| leq int_{[0,1]}|fg_n| dm=||fg_n||_1 leq ||f||_1 ||g_n||_infty$, i.e. $varphi_n$ is bounded. Now $||g_n||_infty=n$ since it's continuous on a bounded interval and the $essential$ $supremum$ is the same as the $max$. Now I would like to attain the equality with some function, and once that I find it I can use in part $b)$. Any ideas on the function?






functional-analysis lebesgue-integral lebesgue-measure norm







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    6












    $begingroup$


    Let $L_1([0,1],m)$ be the Banach space of $mathbb{K}$-valued integrable functions with respect to Lebesgue measure $m$, where $mathbb{K}$ is either $mathbb{R}$ or $mathbb{C}$. The norm on this space is defined like this: $||f||_1=int_{[0,1]}|f| dm$. I have to show that:
    $a)$ For $n geq 2$ the operator $varphi_n(f)=int_{[0,1]} f g_n dm$, where $g_n(x)=n sin(n^2x)$ for $x in [0,1]$ is bounded with $||varphi_n||=n$.
    $b)$ Show that there exists $f in L_1([0,1],m)$ such that $lim_{n to infty} |varphi_n( f)|=infty$.

    MY ATTEMPT:

    $g_n$ is Lebesgue integrable on $[0,1]$ since it's Riemann integrable. Hence $fg_n in L_1([0,1],m)$ and $|int_{[0,1]}fg_n dm| leq int_{[0,1]}|fg_n| dm$. We also have that $||fg_n||_1 leq ||f||_1||g_n||_infty$.
    Thus $||varphi_n(f)||=|int_{[0,1]}fg_n dm| leq int_{[0,1]}|fg_n| dm=||fg_n||_1 leq ||f||_1 ||g_n||_infty$, i.e. $varphi_n$ is bounded. Now $||g_n||_infty=n$ since it's continuous on a bounded interval and the $essential$ $supremum$ is the same as the $max$. Now I would like to attain the equality with some function, and once that I find it I can use in part $b)$. Any ideas on the function?






    functional-analysis lebesgue-integral lebesgue-measure norm







    share|cite|improve this question









    $endgroup$















      6












      6








      6


      1



      $begingroup$


      Let $L_1([0,1],m)$ be the Banach space of $mathbb{K}$-valued integrable functions with respect to Lebesgue measure $m$, where $mathbb{K}$ is either $mathbb{R}$ or $mathbb{C}$. The norm on this space is defined like this: $||f||_1=int_{[0,1]}|f| dm$. I have to show that:
      $a)$ For $n geq 2$ the operator $varphi_n(f)=int_{[0,1]} f g_n dm$, where $g_n(x)=n sin(n^2x)$ for $x in [0,1]$ is bounded with $||varphi_n||=n$.
      $b)$ Show that there exists $f in L_1([0,1],m)$ such that $lim_{n to infty} |varphi_n( f)|=infty$.

      MY ATTEMPT:

      $g_n$ is Lebesgue integrable on $[0,1]$ since it's Riemann integrable. Hence $fg_n in L_1([0,1],m)$ and $|int_{[0,1]}fg_n dm| leq int_{[0,1]}|fg_n| dm$. We also have that $||fg_n||_1 leq ||f||_1||g_n||_infty$.
      Thus $||varphi_n(f)||=|int_{[0,1]}fg_n dm| leq int_{[0,1]}|fg_n| dm=||fg_n||_1 leq ||f||_1 ||g_n||_infty$, i.e. $varphi_n$ is bounded. Now $||g_n||_infty=n$ since it's continuous on a bounded interval and the $essential$ $supremum$ is the same as the $max$. Now I would like to attain the equality with some function, and once that I find it I can use in part $b)$. Any ideas on the function?






      functional-analysis lebesgue-integral lebesgue-measure norm







      share|cite|improve this question









      $endgroup$




      Let $L_1([0,1],m)$ be the Banach space of $mathbb{K}$-valued integrable functions with respect to Lebesgue measure $m$, where $mathbb{K}$ is either $mathbb{R}$ or $mathbb{C}$. The norm on this space is defined like this: $||f||_1=int_{[0,1]}|f| dm$. I have to show that:
      $a)$ For $n geq 2$ the operator $varphi_n(f)=int_{[0,1]} f g_n dm$, where $g_n(x)=n sin(n^2x)$ for $x in [0,1]$ is bounded with $||varphi_n||=n$.
      $b)$ Show that there exists $f in L_1([0,1],m)$ such that $lim_{n to infty} |varphi_n( f)|=infty$.

      MY ATTEMPT:

      $g_n$ is Lebesgue integrable on $[0,1]$ since it's Riemann integrable. Hence $fg_n in L_1([0,1],m)$ and $|int_{[0,1]}fg_n dm| leq int_{[0,1]}|fg_n| dm$. We also have that $||fg_n||_1 leq ||f||_1||g_n||_infty$.
      Thus $||varphi_n(f)||=|int_{[0,1]}fg_n dm| leq int_{[0,1]}|fg_n| dm=||fg_n||_1 leq ||f||_1 ||g_n||_infty$, i.e. $varphi_n$ is bounded. Now $||g_n||_infty=n$ since it's continuous on a bounded interval and the $essential$ $supremum$ is the same as the $max$. Now I would like to attain the equality with some function, and once that I find it I can use in part $b)$. Any ideas on the function?






      functional-analysis lebesgue-integral lebesgue-measure norm




      functional-analysis lebesgue-integral lebesgue-measure norm






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      asked Dec 16 '18 at 22:00









      user289143user289143

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          Answer to part a): it is not necessary to get equality in $|int_{[0,1]} f(x)nsin(n^{2}x),dx | leq |f|_1|g|_{infty}$. Instead, we get an 'approximate equality' as follows: let $epsilon >0$. Choose $delta >0$ such that $sin, x>1-epsilon$ for $frac {pi} 2 -delta <x <frac {pi} 2 +delta $. Let $f=frac {n^{2}} {2delta} I_A$ where $A=(frac {pi} {2n^{2}} -frac {delta} {n^{2}}, frac {pi} {2n^{2}} +frac {delta} {n^{2}})$. Simple calculations show that $|f|_1=1$ and $phi_n(f) >n(1-epsilon)$. Hence $|phi_n| geq n(1-epsilon)$ for all $epsilon >0$. Hence $|phi_n|geq n$.






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            $begingroup$

            Answer to part a): it is not necessary to get equality in $|int_{[0,1]} f(x)nsin(n^{2}x),dx | leq |f|_1|g|_{infty}$. Instead, we get an 'approximate equality' as follows: let $epsilon >0$. Choose $delta >0$ such that $sin, x>1-epsilon$ for $frac {pi} 2 -delta <x <frac {pi} 2 +delta $. Let $f=frac {n^{2}} {2delta} I_A$ where $A=(frac {pi} {2n^{2}} -frac {delta} {n^{2}}, frac {pi} {2n^{2}} +frac {delta} {n^{2}})$. Simple calculations show that $|f|_1=1$ and $phi_n(f) >n(1-epsilon)$. Hence $|phi_n| geq n(1-epsilon)$ for all $epsilon >0$. Hence $|phi_n|geq n$.






            share|cite|improve this answer











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              1












              $begingroup$

              Answer to part a): it is not necessary to get equality in $|int_{[0,1]} f(x)nsin(n^{2}x),dx | leq |f|_1|g|_{infty}$. Instead, we get an 'approximate equality' as follows: let $epsilon >0$. Choose $delta >0$ such that $sin, x>1-epsilon$ for $frac {pi} 2 -delta <x <frac {pi} 2 +delta $. Let $f=frac {n^{2}} {2delta} I_A$ where $A=(frac {pi} {2n^{2}} -frac {delta} {n^{2}}, frac {pi} {2n^{2}} +frac {delta} {n^{2}})$. Simple calculations show that $|f|_1=1$ and $phi_n(f) >n(1-epsilon)$. Hence $|phi_n| geq n(1-epsilon)$ for all $epsilon >0$. Hence $|phi_n|geq n$.






              share|cite|improve this answer











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                $begingroup$

                Answer to part a): it is not necessary to get equality in $|int_{[0,1]} f(x)nsin(n^{2}x),dx | leq |f|_1|g|_{infty}$. Instead, we get an 'approximate equality' as follows: let $epsilon >0$. Choose $delta >0$ such that $sin, x>1-epsilon$ for $frac {pi} 2 -delta <x <frac {pi} 2 +delta $. Let $f=frac {n^{2}} {2delta} I_A$ where $A=(frac {pi} {2n^{2}} -frac {delta} {n^{2}}, frac {pi} {2n^{2}} +frac {delta} {n^{2}})$. Simple calculations show that $|f|_1=1$ and $phi_n(f) >n(1-epsilon)$. Hence $|phi_n| geq n(1-epsilon)$ for all $epsilon >0$. Hence $|phi_n|geq n$.






                share|cite|improve this answer











                $endgroup$



                Answer to part a): it is not necessary to get equality in $|int_{[0,1]} f(x)nsin(n^{2}x),dx | leq |f|_1|g|_{infty}$. Instead, we get an 'approximate equality' as follows: let $epsilon >0$. Choose $delta >0$ such that $sin, x>1-epsilon$ for $frac {pi} 2 -delta <x <frac {pi} 2 +delta $. Let $f=frac {n^{2}} {2delta} I_A$ where $A=(frac {pi} {2n^{2}} -frac {delta} {n^{2}}, frac {pi} {2n^{2}} +frac {delta} {n^{2}})$. Simple calculations show that $|f|_1=1$ and $phi_n(f) >n(1-epsilon)$. Hence $|phi_n| geq n(1-epsilon)$ for all $epsilon >0$. Hence $|phi_n|geq n$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 17 '18 at 6:41

























                answered Dec 17 '18 at 4:57









                Kavi Rama MurthyKavi Rama Murthy

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