Properties of Norm spaces












1












$begingroup$


Suppose $m(E)<infty$ and $finmathcal{L}^{infty}(E)$. The goal of this problem is to show
begin{align*}
lim_{pto infty}|f|_p=|f|_{infty}.
end{align*}
First, prove that
begin{align*}
lim_{ptoinfty}|f|_pleq |f|_{infty}.
end{align*}Proof:
Assume $m(E)<infty$ and $finmathcal{L}^{infty}(E)$. By definition,
begin{align*}
|f|_p=left(int_E|f|^pdmright)^{1/p}leq left(int_E|f|_{infty}^pright)^{1/p}=|f|_{infty}cdot m(E)^{1/p}<infty.
end{align*}
Letting $pto infty$, we have that $m(E)^{1/p}to 1$. Therefore,
begin{align*}
lim_{ptoinfty}|f|_pleq |f|_{infty}.
end{align*}



I'm not sure if I am missing anything here or need to justify being able to bring the limit inside and applying to $m(E)$.



Next, I need to prove that
begin{align*}
lim_{ptoinfty}|f|_pgeq |f|_{infty}-epsilon
end{align*}
for any $epsilon>0$. Hint: Look at the set
begin{align*}
F={xin E:|f|>|f|_{infty}-epsilon}.
end{align*}



I'm stuck on this part and would appreciate any help, thanks.






functional-analysis







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Suppose $m(E)<infty$ and $finmathcal{L}^{infty}(E)$. The goal of this problem is to show
    begin{align*}
    lim_{pto infty}|f|_p=|f|_{infty}.
    end{align*}
    First, prove that
    begin{align*}
    lim_{ptoinfty}|f|_pleq |f|_{infty}.
    end{align*}    Proof:
    Assume $m(E)<infty$ and $finmathcal{L}^{infty}(E)$. By definition,
    begin{align*}
    |f|_p=left(int_E|f|^pdmright)^{1/p}leq left(int_E|f|_{infty}^pright)^{1/p}=|f|_{infty}cdot m(E)^{1/p}<infty.
    end{align*}
    Letting $pto infty$, we have that $m(E)^{1/p}to 1$. Therefore,
    begin{align*}
    lim_{ptoinfty}|f|_pleq |f|_{infty}.
    end{align*}



    I'm not sure if I am missing anything here or need to justify being able to bring the limit inside and applying to $m(E)$.



    Next, I need to prove that
    begin{align*}
    lim_{ptoinfty}|f|_pgeq |f|_{infty}-epsilon
    end{align*}
    for any $epsilon>0$. Hint: Look at the set
    begin{align*}
    F={xin E:|f|>|f|_{infty}-epsilon}.
    end{align*}



    I'm stuck on this part and would appreciate any help, thanks.






    functional-analysis







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Suppose $m(E)<infty$ and $finmathcal{L}^{infty}(E)$. The goal of this problem is to show
      begin{align*}
      lim_{pto infty}|f|_p=|f|_{infty}.
      end{align*}
      First, prove that
      begin{align*}
      lim_{ptoinfty}|f|_pleq |f|_{infty}.
      end{align*}      Proof:
      Assume $m(E)<infty$ and $finmathcal{L}^{infty}(E)$. By definition,
      begin{align*}
      |f|_p=left(int_E|f|^pdmright)^{1/p}leq left(int_E|f|_{infty}^pright)^{1/p}=|f|_{infty}cdot m(E)^{1/p}<infty.
      end{align*}
      Letting $pto infty$, we have that $m(E)^{1/p}to 1$. Therefore,
      begin{align*}
      lim_{ptoinfty}|f|_pleq |f|_{infty}.
      end{align*}



      I'm not sure if I am missing anything here or need to justify being able to bring the limit inside and applying to $m(E)$.



      Next, I need to prove that
      begin{align*}
      lim_{ptoinfty}|f|_pgeq |f|_{infty}-epsilon
      end{align*}
      for any $epsilon>0$. Hint: Look at the set
      begin{align*}
      F={xin E:|f|>|f|_{infty}-epsilon}.
      end{align*}



      I'm stuck on this part and would appreciate any help, thanks.






      functional-analysis







      share|cite|improve this question









      $endgroup$




      Suppose $m(E)<infty$ and $finmathcal{L}^{infty}(E)$. The goal of this problem is to show
      begin{align*}
      lim_{pto infty}|f|_p=|f|_{infty}.
      end{align*}
      First, prove that
      begin{align*}
      lim_{ptoinfty}|f|_pleq |f|_{infty}.
      end{align*}      Proof:
      Assume $m(E)<infty$ and $finmathcal{L}^{infty}(E)$. By definition,
      begin{align*}
      |f|_p=left(int_E|f|^pdmright)^{1/p}leq left(int_E|f|_{infty}^pright)^{1/p}=|f|_{infty}cdot m(E)^{1/p}<infty.
      end{align*}
      Letting $pto infty$, we have that $m(E)^{1/p}to 1$. Therefore,
      begin{align*}
      lim_{ptoinfty}|f|_pleq |f|_{infty}.
      end{align*}



      I'm not sure if I am missing anything here or need to justify being able to bring the limit inside and applying to $m(E)$.



      Next, I need to prove that
      begin{align*}
      lim_{ptoinfty}|f|_pgeq |f|_{infty}-epsilon
      end{align*}
      for any $epsilon>0$. Hint: Look at the set
      begin{align*}
      F={xin E:|f|>|f|_{infty}-epsilon}.
      end{align*}



      I'm stuck on this part and would appreciate any help, thanks.






      functional-analysis




      functional-analysis






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 4 '18 at 19:18









      TNTTNT

      596




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          $begingroup$

          You have that



          $||f||_p^p=int_F|f|^p+int_{F^c}|f|^p>int_F|f|^p >$



          $>int_F (||f||_infty -epsilon)^p=(||f||_infty -epsilon)^pm(F) $



          Then



          $||f||_p> (||f||_infty -epsilon)(m(F))^frac{1}{p}$



          So if you fixed $epsilon$ (and $ F=F_{epsilon}$ ) you have that for $pto infty$



          $lim_p||f||_p>||f||_infty -epsilon$



          for every $epsilon>0$ so



          $lim_p ||f||_p>||f||_infty$






          share|cite|improve this answer











          $endgroup$













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            $begingroup$

            You have that



            $||f||_p^p=int_F|f|^p+int_{F^c}|f|^p>int_F|f|^p >$



            $>int_F (||f||_infty -epsilon)^p=(||f||_infty -epsilon)^pm(F) $



            Then



            $||f||_p> (||f||_infty -epsilon)(m(F))^frac{1}{p}$



            So if you fixed $epsilon$ (and $ F=F_{epsilon}$ ) you have that for $pto infty$



            $lim_p||f||_p>||f||_infty -epsilon$



            for every $epsilon>0$ so



            $lim_p ||f||_p>||f||_infty$






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              You have that



              $||f||_p^p=int_F|f|^p+int_{F^c}|f|^p>int_F|f|^p >$



              $>int_F (||f||_infty -epsilon)^p=(||f||_infty -epsilon)^pm(F) $



              Then



              $||f||_p> (||f||_infty -epsilon)(m(F))^frac{1}{p}$



              So if you fixed $epsilon$ (and $ F=F_{epsilon}$ ) you have that for $pto infty$



              $lim_p||f||_p>||f||_infty -epsilon$



              for every $epsilon>0$ so



              $lim_p ||f||_p>||f||_infty$






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                You have that



                $||f||_p^p=int_F|f|^p+int_{F^c}|f|^p>int_F|f|^p >$



                $>int_F (||f||_infty -epsilon)^p=(||f||_infty -epsilon)^pm(F) $



                Then



                $||f||_p> (||f||_infty -epsilon)(m(F))^frac{1}{p}$



                So if you fixed $epsilon$ (and $ F=F_{epsilon}$ ) you have that for $pto infty$



                $lim_p||f||_p>||f||_infty -epsilon$



                for every $epsilon>0$ so



                $lim_p ||f||_p>||f||_infty$






                share|cite|improve this answer











                $endgroup$



                You have that



                $||f||_p^p=int_F|f|^p+int_{F^c}|f|^p>int_F|f|^p >$



                $>int_F (||f||_infty -epsilon)^p=(||f||_infty -epsilon)^pm(F) $



                Then



                $||f||_p> (||f||_infty -epsilon)(m(F))^frac{1}{p}$



                So if you fixed $epsilon$ (and $ F=F_{epsilon}$ ) you have that for $pto infty$



                $lim_p||f||_p>||f||_infty -epsilon$



                for every $epsilon>0$ so



                $lim_p ||f||_p>||f||_infty$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 4 '18 at 19:38

























                answered Dec 4 '18 at 19:27









                Federico FalluccaFederico Fallucca

                2,185210




                2,185210






























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