Calculating the norm of a specific bounded linear functional












3












$begingroup$


Let $L_1([0,1], m)$ be the Banach space of $mathbb{K}$-valued (i.e. $mathbb{C}$ or $mathbb{R}$-valued) Lebesgue-integrable functions, where $m$ is the Lebesgue measure, be equipped with the norm $‖f‖_1=int_{[0,1]}|f|,dm$ for $f in L_1([0,1], m)$. For $n geq 1$ define $g_n(x) =nsin(n^2x)$, for $x in [0,1]$.
Furthermore, let $phi_n(f)=int_{[0,1]} f g_n , dm$.



I'm then asked to show that $phi_n(f)$ defines a bounded linear map (which I've done) and to show $‖phi_n‖=n$ for $n geq 2$. What I've done is that I have shown $‖phi_n‖leq n$. If I could find a function $f in L_1([0,1], m)$ such that $|phi_n(f)|=n$ and $‖f‖_1=1$, then I would have shown equality. How can I find such a function?



The problem also asks whether I can find a function $f in L_1([0,1], m)$ such that $lim_{n rightarrow infty} |phi_n(f)|=infty$ (perhaps I could use the same function as above?)










share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    Let $L_1([0,1], m)$ be the Banach space of $mathbb{K}$-valued (i.e. $mathbb{C}$ or $mathbb{R}$-valued) Lebesgue-integrable functions, where $m$ is the Lebesgue measure, be equipped with the norm $‖f‖_1=int_{[0,1]}|f|,dm$ for $f in L_1([0,1], m)$. For $n geq 1$ define $g_n(x) =nsin(n^2x)$, for $x in [0,1]$.
    Furthermore, let $phi_n(f)=int_{[0,1]} f g_n , dm$.



    I'm then asked to show that $phi_n(f)$ defines a bounded linear map (which I've done) and to show $‖phi_n‖=n$ for $n geq 2$. What I've done is that I have shown $‖phi_n‖leq n$. If I could find a function $f in L_1([0,1], m)$ such that $|phi_n(f)|=n$ and $‖f‖_1=1$, then I would have shown equality. How can I find such a function?



    The problem also asks whether I can find a function $f in L_1([0,1], m)$ such that $lim_{n rightarrow infty} |phi_n(f)|=infty$ (perhaps I could use the same function as above?)










    share|cite|improve this question









    $endgroup$















      3












      3








      3


      1



      $begingroup$


      Let $L_1([0,1], m)$ be the Banach space of $mathbb{K}$-valued (i.e. $mathbb{C}$ or $mathbb{R}$-valued) Lebesgue-integrable functions, where $m$ is the Lebesgue measure, be equipped with the norm $‖f‖_1=int_{[0,1]}|f|,dm$ for $f in L_1([0,1], m)$. For $n geq 1$ define $g_n(x) =nsin(n^2x)$, for $x in [0,1]$.
      Furthermore, let $phi_n(f)=int_{[0,1]} f g_n , dm$.



      I'm then asked to show that $phi_n(f)$ defines a bounded linear map (which I've done) and to show $‖phi_n‖=n$ for $n geq 2$. What I've done is that I have shown $‖phi_n‖leq n$. If I could find a function $f in L_1([0,1], m)$ such that $|phi_n(f)|=n$ and $‖f‖_1=1$, then I would have shown equality. How can I find such a function?



      The problem also asks whether I can find a function $f in L_1([0,1], m)$ such that $lim_{n rightarrow infty} |phi_n(f)|=infty$ (perhaps I could use the same function as above?)










      share|cite|improve this question









      $endgroup$




      Let $L_1([0,1], m)$ be the Banach space of $mathbb{K}$-valued (i.e. $mathbb{C}$ or $mathbb{R}$-valued) Lebesgue-integrable functions, where $m$ is the Lebesgue measure, be equipped with the norm $‖f‖_1=int_{[0,1]}|f|,dm$ for $f in L_1([0,1], m)$. For $n geq 1$ define $g_n(x) =nsin(n^2x)$, for $x in [0,1]$.
      Furthermore, let $phi_n(f)=int_{[0,1]} f g_n , dm$.



      I'm then asked to show that $phi_n(f)$ defines a bounded linear map (which I've done) and to show $‖phi_n‖=n$ for $n geq 2$. What I've done is that I have shown $‖phi_n‖leq n$. If I could find a function $f in L_1([0,1], m)$ such that $|phi_n(f)|=n$ and $‖f‖_1=1$, then I would have shown equality. How can I find such a function?



      The problem also asks whether I can find a function $f in L_1([0,1], m)$ such that $lim_{n rightarrow infty} |phi_n(f)|=infty$ (perhaps I could use the same function as above?)







      functional-analysis lebesgue-measure lp-spaces






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Dec 6 '18 at 22:01









      VercingetorixVercingetorix

      285




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          1 Answer
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          1












          $begingroup$

          It is a general fact that if $g$ is a bounded function, then the norm of the linear functional $Lcolon mathbb L^1tomathbb R$ defined by $L(f)=int_{(0,1)}g(x)f(x)mathrm dlambda(x)$ is $leftlVert grightrVert_infty$. Indeed, fix a positive $varepsilon<leftlVert grightrVert_infty$ and define the sets $A^+:=left{xin (0,1), g(x)gt leftlVert grightrVert_infty-varepsilonright}$ and $A^-:=left{xin (0,1), -g(x)gt leftlVert grightrVert_infty-varepsilonright}$. Then
          $$
          maxleft{Lleft(frac 1{lambdaleft(A^+right)}mathbf 1_{A^+}right),Lleft(frac 1{lambdaleft(A^-right)}mathbf 1_{A^-}right) right}geqslant leftlVert grightrVert_infty-varepsilon
          $$

          and (at least one of) the involved function are well-defined and their $L^1$-norm is $1$.



          For the other question, look at one of the functions $xmapsto pm x^{-alpha}$ for $1/2ltalphalt 1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I can see that the first one is correct, but if you calculate $phi_n(x^{-alpha})$ I get $n^{2alpha-1} int_0^{n^2} frac{sin(x)}{x^{alpha}} , dm$ (but it's not that easy to verify that this diverges to infinity}. I also know that we can show the last part without explicitly having to find $f$ (using methods from functional analysis)
            $endgroup$
            – Vercingetorix
            Dec 11 '18 at 18:20












          • $begingroup$
            but I don't know how to do it without explicitly finding a function
            $endgroup$
            – Vercingetorix
            Dec 11 '18 at 18:27










          • $begingroup$
            The integral $int_0^{infty}sin x/x^{alpha}$ is convergent. Which methods of functional analysis would you have at your disposal for showing the existence of $f$ without giving a function.
            $endgroup$
            – Davide Giraudo
            Dec 11 '18 at 19:05










          • $begingroup$
            Perhaps the Banach-Steinhaus Theorem? Using the contraposed of Banach-Steinhaus gives that there exist $f in L_1([0,1],m)$ such that $sup_{n in mathbb{N}} |phi_n(f)|=infty$, but I'm not sure where I can go from there to conclude that $|phi_n(f)| rightarrow infty$.
            $endgroup$
            – Vercingetorix
            Dec 11 '18 at 19:49













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          active

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          1












          $begingroup$

          It is a general fact that if $g$ is a bounded function, then the norm of the linear functional $Lcolon mathbb L^1tomathbb R$ defined by $L(f)=int_{(0,1)}g(x)f(x)mathrm dlambda(x)$ is $leftlVert grightrVert_infty$. Indeed, fix a positive $varepsilon<leftlVert grightrVert_infty$ and define the sets $A^+:=left{xin (0,1), g(x)gt leftlVert grightrVert_infty-varepsilonright}$ and $A^-:=left{xin (0,1), -g(x)gt leftlVert grightrVert_infty-varepsilonright}$. Then
          $$
          maxleft{Lleft(frac 1{lambdaleft(A^+right)}mathbf 1_{A^+}right),Lleft(frac 1{lambdaleft(A^-right)}mathbf 1_{A^-}right) right}geqslant leftlVert grightrVert_infty-varepsilon
          $$

          and (at least one of) the involved function are well-defined and their $L^1$-norm is $1$.



          For the other question, look at one of the functions $xmapsto pm x^{-alpha}$ for $1/2ltalphalt 1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I can see that the first one is correct, but if you calculate $phi_n(x^{-alpha})$ I get $n^{2alpha-1} int_0^{n^2} frac{sin(x)}{x^{alpha}} , dm$ (but it's not that easy to verify that this diverges to infinity}. I also know that we can show the last part without explicitly having to find $f$ (using methods from functional analysis)
            $endgroup$
            – Vercingetorix
            Dec 11 '18 at 18:20












          • $begingroup$
            but I don't know how to do it without explicitly finding a function
            $endgroup$
            – Vercingetorix
            Dec 11 '18 at 18:27










          • $begingroup$
            The integral $int_0^{infty}sin x/x^{alpha}$ is convergent. Which methods of functional analysis would you have at your disposal for showing the existence of $f$ without giving a function.
            $endgroup$
            – Davide Giraudo
            Dec 11 '18 at 19:05










          • $begingroup$
            Perhaps the Banach-Steinhaus Theorem? Using the contraposed of Banach-Steinhaus gives that there exist $f in L_1([0,1],m)$ such that $sup_{n in mathbb{N}} |phi_n(f)|=infty$, but I'm not sure where I can go from there to conclude that $|phi_n(f)| rightarrow infty$.
            $endgroup$
            – Vercingetorix
            Dec 11 '18 at 19:49


















          1












          $begingroup$

          It is a general fact that if $g$ is a bounded function, then the norm of the linear functional $Lcolon mathbb L^1tomathbb R$ defined by $L(f)=int_{(0,1)}g(x)f(x)mathrm dlambda(x)$ is $leftlVert grightrVert_infty$. Indeed, fix a positive $varepsilon<leftlVert grightrVert_infty$ and define the sets $A^+:=left{xin (0,1), g(x)gt leftlVert grightrVert_infty-varepsilonright}$ and $A^-:=left{xin (0,1), -g(x)gt leftlVert grightrVert_infty-varepsilonright}$. Then
          $$
          maxleft{Lleft(frac 1{lambdaleft(A^+right)}mathbf 1_{A^+}right),Lleft(frac 1{lambdaleft(A^-right)}mathbf 1_{A^-}right) right}geqslant leftlVert grightrVert_infty-varepsilon
          $$

          and (at least one of) the involved function are well-defined and their $L^1$-norm is $1$.



          For the other question, look at one of the functions $xmapsto pm x^{-alpha}$ for $1/2ltalphalt 1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I can see that the first one is correct, but if you calculate $phi_n(x^{-alpha})$ I get $n^{2alpha-1} int_0^{n^2} frac{sin(x)}{x^{alpha}} , dm$ (but it's not that easy to verify that this diverges to infinity}. I also know that we can show the last part without explicitly having to find $f$ (using methods from functional analysis)
            $endgroup$
            – Vercingetorix
            Dec 11 '18 at 18:20












          • $begingroup$
            but I don't know how to do it without explicitly finding a function
            $endgroup$
            – Vercingetorix
            Dec 11 '18 at 18:27










          • $begingroup$
            The integral $int_0^{infty}sin x/x^{alpha}$ is convergent. Which methods of functional analysis would you have at your disposal for showing the existence of $f$ without giving a function.
            $endgroup$
            – Davide Giraudo
            Dec 11 '18 at 19:05










          • $begingroup$
            Perhaps the Banach-Steinhaus Theorem? Using the contraposed of Banach-Steinhaus gives that there exist $f in L_1([0,1],m)$ such that $sup_{n in mathbb{N}} |phi_n(f)|=infty$, but I'm not sure where I can go from there to conclude that $|phi_n(f)| rightarrow infty$.
            $endgroup$
            – Vercingetorix
            Dec 11 '18 at 19:49
















          1












          1








          1





          $begingroup$

          It is a general fact that if $g$ is a bounded function, then the norm of the linear functional $Lcolon mathbb L^1tomathbb R$ defined by $L(f)=int_{(0,1)}g(x)f(x)mathrm dlambda(x)$ is $leftlVert grightrVert_infty$. Indeed, fix a positive $varepsilon<leftlVert grightrVert_infty$ and define the sets $A^+:=left{xin (0,1), g(x)gt leftlVert grightrVert_infty-varepsilonright}$ and $A^-:=left{xin (0,1), -g(x)gt leftlVert grightrVert_infty-varepsilonright}$. Then
          $$
          maxleft{Lleft(frac 1{lambdaleft(A^+right)}mathbf 1_{A^+}right),Lleft(frac 1{lambdaleft(A^-right)}mathbf 1_{A^-}right) right}geqslant leftlVert grightrVert_infty-varepsilon
          $$

          and (at least one of) the involved function are well-defined and their $L^1$-norm is $1$.



          For the other question, look at one of the functions $xmapsto pm x^{-alpha}$ for $1/2ltalphalt 1$.






          share|cite|improve this answer









          $endgroup$



          It is a general fact that if $g$ is a bounded function, then the norm of the linear functional $Lcolon mathbb L^1tomathbb R$ defined by $L(f)=int_{(0,1)}g(x)f(x)mathrm dlambda(x)$ is $leftlVert grightrVert_infty$. Indeed, fix a positive $varepsilon<leftlVert grightrVert_infty$ and define the sets $A^+:=left{xin (0,1), g(x)gt leftlVert grightrVert_infty-varepsilonright}$ and $A^-:=left{xin (0,1), -g(x)gt leftlVert grightrVert_infty-varepsilonright}$. Then
          $$
          maxleft{Lleft(frac 1{lambdaleft(A^+right)}mathbf 1_{A^+}right),Lleft(frac 1{lambdaleft(A^-right)}mathbf 1_{A^-}right) right}geqslant leftlVert grightrVert_infty-varepsilon
          $$

          and (at least one of) the involved function are well-defined and their $L^1$-norm is $1$.



          For the other question, look at one of the functions $xmapsto pm x^{-alpha}$ for $1/2ltalphalt 1$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 10 '18 at 23:04









          Davide GiraudoDavide Giraudo

          127k17154268




          127k17154268












          • $begingroup$
            I can see that the first one is correct, but if you calculate $phi_n(x^{-alpha})$ I get $n^{2alpha-1} int_0^{n^2} frac{sin(x)}{x^{alpha}} , dm$ (but it's not that easy to verify that this diverges to infinity}. I also know that we can show the last part without explicitly having to find $f$ (using methods from functional analysis)
            $endgroup$
            – Vercingetorix
            Dec 11 '18 at 18:20












          • $begingroup$
            but I don't know how to do it without explicitly finding a function
            $endgroup$
            – Vercingetorix
            Dec 11 '18 at 18:27










          • $begingroup$
            The integral $int_0^{infty}sin x/x^{alpha}$ is convergent. Which methods of functional analysis would you have at your disposal for showing the existence of $f$ without giving a function.
            $endgroup$
            – Davide Giraudo
            Dec 11 '18 at 19:05










          • $begingroup$
            Perhaps the Banach-Steinhaus Theorem? Using the contraposed of Banach-Steinhaus gives that there exist $f in L_1([0,1],m)$ such that $sup_{n in mathbb{N}} |phi_n(f)|=infty$, but I'm not sure where I can go from there to conclude that $|phi_n(f)| rightarrow infty$.
            $endgroup$
            – Vercingetorix
            Dec 11 '18 at 19:49




















          • $begingroup$
            I can see that the first one is correct, but if you calculate $phi_n(x^{-alpha})$ I get $n^{2alpha-1} int_0^{n^2} frac{sin(x)}{x^{alpha}} , dm$ (but it's not that easy to verify that this diverges to infinity}. I also know that we can show the last part without explicitly having to find $f$ (using methods from functional analysis)
            $endgroup$
            – Vercingetorix
            Dec 11 '18 at 18:20












          • $begingroup$
            but I don't know how to do it without explicitly finding a function
            $endgroup$
            – Vercingetorix
            Dec 11 '18 at 18:27










          • $begingroup$
            The integral $int_0^{infty}sin x/x^{alpha}$ is convergent. Which methods of functional analysis would you have at your disposal for showing the existence of $f$ without giving a function.
            $endgroup$
            – Davide Giraudo
            Dec 11 '18 at 19:05










          • $begingroup$
            Perhaps the Banach-Steinhaus Theorem? Using the contraposed of Banach-Steinhaus gives that there exist $f in L_1([0,1],m)$ such that $sup_{n in mathbb{N}} |phi_n(f)|=infty$, but I'm not sure where I can go from there to conclude that $|phi_n(f)| rightarrow infty$.
            $endgroup$
            – Vercingetorix
            Dec 11 '18 at 19:49


















          $begingroup$
          I can see that the first one is correct, but if you calculate $phi_n(x^{-alpha})$ I get $n^{2alpha-1} int_0^{n^2} frac{sin(x)}{x^{alpha}} , dm$ (but it's not that easy to verify that this diverges to infinity}. I also know that we can show the last part without explicitly having to find $f$ (using methods from functional analysis)
          $endgroup$
          – Vercingetorix
          Dec 11 '18 at 18:20






          $begingroup$
          I can see that the first one is correct, but if you calculate $phi_n(x^{-alpha})$ I get $n^{2alpha-1} int_0^{n^2} frac{sin(x)}{x^{alpha}} , dm$ (but it's not that easy to verify that this diverges to infinity}. I also know that we can show the last part without explicitly having to find $f$ (using methods from functional analysis)
          $endgroup$
          – Vercingetorix
          Dec 11 '18 at 18:20














          $begingroup$
          but I don't know how to do it without explicitly finding a function
          $endgroup$
          – Vercingetorix
          Dec 11 '18 at 18:27




          $begingroup$
          but I don't know how to do it without explicitly finding a function
          $endgroup$
          – Vercingetorix
          Dec 11 '18 at 18:27












          $begingroup$
          The integral $int_0^{infty}sin x/x^{alpha}$ is convergent. Which methods of functional analysis would you have at your disposal for showing the existence of $f$ without giving a function.
          $endgroup$
          – Davide Giraudo
          Dec 11 '18 at 19:05




          $begingroup$
          The integral $int_0^{infty}sin x/x^{alpha}$ is convergent. Which methods of functional analysis would you have at your disposal for showing the existence of $f$ without giving a function.
          $endgroup$
          – Davide Giraudo
          Dec 11 '18 at 19:05












          $begingroup$
          Perhaps the Banach-Steinhaus Theorem? Using the contraposed of Banach-Steinhaus gives that there exist $f in L_1([0,1],m)$ such that $sup_{n in mathbb{N}} |phi_n(f)|=infty$, but I'm not sure where I can go from there to conclude that $|phi_n(f)| rightarrow infty$.
          $endgroup$
          – Vercingetorix
          Dec 11 '18 at 19:49






          $begingroup$
          Perhaps the Banach-Steinhaus Theorem? Using the contraposed of Banach-Steinhaus gives that there exist $f in L_1([0,1],m)$ such that $sup_{n in mathbb{N}} |phi_n(f)|=infty$, but I'm not sure where I can go from there to conclude that $|phi_n(f)| rightarrow infty$.
          $endgroup$
          – Vercingetorix
          Dec 11 '18 at 19:49




















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