Show that T is continuos linear functional and find the norm of T












0












$begingroup$


Let T : $L^{3}$ [0,1] $to$ R



T(f) = $int_{0}^1$ $t^{2}$ f(t) dt



1- Show that T is continuos linear functional



2- Find the norm of T



My solution :



1- first I proved that $t^{2}$ $in$ $L^{3/2}$



Now we have
f $in$ $L^{3}$ and $t^{2}$ $in$ $L^{3/2}$



And since p, q are conjugate then T is bounded linear functional
Then so it's continuous



Is it true?



2- I know the norm of T is :
|| T || = sup || Tt ||
where || t || = 1



But how can I apply on this?



Thanks a lot.






functional-analysis measure-theory norm







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Let T : $L^{3}$ [0,1] $to$ R



    T(f) = $int_{0}^1$ $t^{2}$ f(t) dt



    1- Show that T is continuos linear functional



    2- Find the norm of T



    My solution :



    1- first I proved that $t^{2}$ $in$ $L^{3/2}$



    Now we have
    f $in$ $L^{3}$ and $t^{2}$ $in$ $L^{3/2}$



    And since p, q are conjugate then T is bounded linear functional
    Then so it's continuous



    Is it true?



    2- I know the norm of T is :
    || T || = sup || Tt ||
    where || t || = 1



    But how can I apply on this?



    Thanks a lot.






    functional-analysis measure-theory norm







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let T : $L^{3}$ [0,1] $to$ R



      T(f) = $int_{0}^1$ $t^{2}$ f(t) dt



      1- Show that T is continuos linear functional



      2- Find the norm of T



      My solution :



      1- first I proved that $t^{2}$ $in$ $L^{3/2}$



      Now we have
      f $in$ $L^{3}$ and $t^{2}$ $in$ $L^{3/2}$



      And since p, q are conjugate then T is bounded linear functional
      Then so it's continuous



      Is it true?



      2- I know the norm of T is :
      || T || = sup || Tt ||
      where || t || = 1



      But how can I apply on this?



      Thanks a lot.






      functional-analysis measure-theory norm







      share|cite|improve this question









      $endgroup$




      Let T : $L^{3}$ [0,1] $to$ R



      T(f) = $int_{0}^1$ $t^{2}$ f(t) dt



      1- Show that T is continuos linear functional



      2- Find the norm of T



      My solution :



      1- first I proved that $t^{2}$ $in$ $L^{3/2}$



      Now we have
      f $in$ $L^{3}$ and $t^{2}$ $in$ $L^{3/2}$



      And since p, q are conjugate then T is bounded linear functional
      Then so it's continuous



      Is it true?



      2- I know the norm of T is :
      || T || = sup || Tt ||
      where || t || = 1



      But how can I apply on this?



      Thanks a lot.






      functional-analysis measure-theory norm




      functional-analysis measure-theory norm






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 18 '18 at 10:01









      Duaa HamzehDuaa Hamzeh

      524




      524






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          $|T| leq 4^{-2/3}$ because $|int f(t)t^{2}, dt| leq (int |f|^{3})^{1/3} (int t^{3})^{2/3}=4^{-2/3} |f|_3$. Now take $f(t)=4^{1/3}t$ and verify that $|f|_3=1$ and that $Tf=4^{-2/3}$. Hence $|T|=4^{-2/3}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much.. But how the " <= " became “ = “ ?
            $endgroup$
            – Duaa Hamzeh
            Dec 18 '18 at 10:30






          • 1




            $begingroup$
            By definition of operator norm, $|T|| geq |f|$ if $|f|=1$. I have given a specific $f$ such that $|f|=1$ and $Tf=4^{-2/3}$.
            $endgroup$
            – Kavi Rama Murthy
            Dec 18 '18 at 10:33










          • $begingroup$
            For the record I meant $|T| geq |Tf|$ if $|f|=1$.
            $endgroup$
            – Kavi Rama Murthy
            Dec 18 '18 at 12:02










          • $begingroup$
            It's help me very much.. Grateful 🌸
            $endgroup$
            – Duaa Hamzeh
            Dec 18 '18 at 12:15










          • $begingroup$
            Also.. I have seen a theorem now from Royden say that || T || = ||$t^{2}$ || on $L^{3/2}$ space since it is bounded
            $endgroup$
            – Duaa Hamzeh
            Dec 18 '18 at 12:19



















          1












          $begingroup$

          With Hölder we get



          $|T(f)| le (int_0^1t^3 dt)^{2/3} (int_0^1 |f(t)|^3)^{1/3} = c ||f||_3$,



          where $c:= (int_0^1t^3 dt)^{2/3}$.



          This shows that $T$ is continuous.



          It is your turn to determine $||T||$.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            $|T| leq 4^{-2/3}$ because $|int f(t)t^{2}, dt| leq (int |f|^{3})^{1/3} (int t^{3})^{2/3}=4^{-2/3} |f|_3$. Now take $f(t)=4^{1/3}t$ and verify that $|f|_3=1$ and that $Tf=4^{-2/3}$. Hence $|T|=4^{-2/3}$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you very much.. But how the " <= " became “ = “ ?
              $endgroup$
              – Duaa Hamzeh
              Dec 18 '18 at 10:30






            • 1




              $begingroup$
              By definition of operator norm, $|T|| geq |f|$ if $|f|=1$. I have given a specific $f$ such that $|f|=1$ and $Tf=4^{-2/3}$.
              $endgroup$
              – Kavi Rama Murthy
              Dec 18 '18 at 10:33










            • $begingroup$
              For the record I meant $|T| geq |Tf|$ if $|f|=1$.
              $endgroup$
              – Kavi Rama Murthy
              Dec 18 '18 at 12:02










            • $begingroup$
              It's help me very much.. Grateful 🌸
              $endgroup$
              – Duaa Hamzeh
              Dec 18 '18 at 12:15










            • $begingroup$
              Also.. I have seen a theorem now from Royden say that || T || = ||$t^{2}$ || on $L^{3/2}$ space since it is bounded
              $endgroup$
              – Duaa Hamzeh
              Dec 18 '18 at 12:19
















            1












            $begingroup$

            $|T| leq 4^{-2/3}$ because $|int f(t)t^{2}, dt| leq (int |f|^{3})^{1/3} (int t^{3})^{2/3}=4^{-2/3} |f|_3$. Now take $f(t)=4^{1/3}t$ and verify that $|f|_3=1$ and that $Tf=4^{-2/3}$. Hence $|T|=4^{-2/3}$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you very much.. But how the " <= " became “ = “ ?
              $endgroup$
              – Duaa Hamzeh
              Dec 18 '18 at 10:30






            • 1




              $begingroup$
              By definition of operator norm, $|T|| geq |f|$ if $|f|=1$. I have given a specific $f$ such that $|f|=1$ and $Tf=4^{-2/3}$.
              $endgroup$
              – Kavi Rama Murthy
              Dec 18 '18 at 10:33










            • $begingroup$
              For the record I meant $|T| geq |Tf|$ if $|f|=1$.
              $endgroup$
              – Kavi Rama Murthy
              Dec 18 '18 at 12:02










            • $begingroup$
              It's help me very much.. Grateful 🌸
              $endgroup$
              – Duaa Hamzeh
              Dec 18 '18 at 12:15










            • $begingroup$
              Also.. I have seen a theorem now from Royden say that || T || = ||$t^{2}$ || on $L^{3/2}$ space since it is bounded
              $endgroup$
              – Duaa Hamzeh
              Dec 18 '18 at 12:19














            1












            1








            1





            $begingroup$

            $|T| leq 4^{-2/3}$ because $|int f(t)t^{2}, dt| leq (int |f|^{3})^{1/3} (int t^{3})^{2/3}=4^{-2/3} |f|_3$. Now take $f(t)=4^{1/3}t$ and verify that $|f|_3=1$ and that $Tf=4^{-2/3}$. Hence $|T|=4^{-2/3}$.






            share|cite|improve this answer









            $endgroup$



            $|T| leq 4^{-2/3}$ because $|int f(t)t^{2}, dt| leq (int |f|^{3})^{1/3} (int t^{3})^{2/3}=4^{-2/3} |f|_3$. Now take $f(t)=4^{1/3}t$ and verify that $|f|_3=1$ and that $Tf=4^{-2/3}$. Hence $|T|=4^{-2/3}$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 18 '18 at 10:11









            Kavi Rama MurthyKavi Rama Murthy

            60k42161




            60k42161












            • $begingroup$
              Thank you very much.. But how the " <= " became “ = “ ?
              $endgroup$
              – Duaa Hamzeh
              Dec 18 '18 at 10:30






            • 1




              $begingroup$
              By definition of operator norm, $|T|| geq |f|$ if $|f|=1$. I have given a specific $f$ such that $|f|=1$ and $Tf=4^{-2/3}$.
              $endgroup$
              – Kavi Rama Murthy
              Dec 18 '18 at 10:33










            • $begingroup$
              For the record I meant $|T| geq |Tf|$ if $|f|=1$.
              $endgroup$
              – Kavi Rama Murthy
              Dec 18 '18 at 12:02










            • $begingroup$
              It's help me very much.. Grateful 🌸
              $endgroup$
              – Duaa Hamzeh
              Dec 18 '18 at 12:15










            • $begingroup$
              Also.. I have seen a theorem now from Royden say that || T || = ||$t^{2}$ || on $L^{3/2}$ space since it is bounded
              $endgroup$
              – Duaa Hamzeh
              Dec 18 '18 at 12:19


















            • $begingroup$
              Thank you very much.. But how the " <= " became “ = “ ?
              $endgroup$
              – Duaa Hamzeh
              Dec 18 '18 at 10:30






            • 1




              $begingroup$
              By definition of operator norm, $|T|| geq |f|$ if $|f|=1$. I have given a specific $f$ such that $|f|=1$ and $Tf=4^{-2/3}$.
              $endgroup$
              – Kavi Rama Murthy
              Dec 18 '18 at 10:33










            • $begingroup$
              For the record I meant $|T| geq |Tf|$ if $|f|=1$.
              $endgroup$
              – Kavi Rama Murthy
              Dec 18 '18 at 12:02










            • $begingroup$
              It's help me very much.. Grateful 🌸
              $endgroup$
              – Duaa Hamzeh
              Dec 18 '18 at 12:15










            • $begingroup$
              Also.. I have seen a theorem now from Royden say that || T || = ||$t^{2}$ || on $L^{3/2}$ space since it is bounded
              $endgroup$
              – Duaa Hamzeh
              Dec 18 '18 at 12:19
















            $begingroup$
            Thank you very much.. But how the " <= " became “ = “ ?
            $endgroup$
            – Duaa Hamzeh
            Dec 18 '18 at 10:30




            $begingroup$
            Thank you very much.. But how the " <= " became “ = “ ?
            $endgroup$
            – Duaa Hamzeh
            Dec 18 '18 at 10:30




            1




            1




            $begingroup$
            By definition of operator norm, $|T|| geq |f|$ if $|f|=1$. I have given a specific $f$ such that $|f|=1$ and $Tf=4^{-2/3}$.
            $endgroup$
            – Kavi Rama Murthy
            Dec 18 '18 at 10:33




            $begingroup$
            By definition of operator norm, $|T|| geq |f|$ if $|f|=1$. I have given a specific $f$ such that $|f|=1$ and $Tf=4^{-2/3}$.
            $endgroup$
            – Kavi Rama Murthy
            Dec 18 '18 at 10:33












            $begingroup$
            For the record I meant $|T| geq |Tf|$ if $|f|=1$.
            $endgroup$
            – Kavi Rama Murthy
            Dec 18 '18 at 12:02




            $begingroup$
            For the record I meant $|T| geq |Tf|$ if $|f|=1$.
            $endgroup$
            – Kavi Rama Murthy
            Dec 18 '18 at 12:02












            $begingroup$
            It's help me very much.. Grateful 🌸
            $endgroup$
            – Duaa Hamzeh
            Dec 18 '18 at 12:15




            $begingroup$
            It's help me very much.. Grateful 🌸
            $endgroup$
            – Duaa Hamzeh
            Dec 18 '18 at 12:15












            $begingroup$
            Also.. I have seen a theorem now from Royden say that || T || = ||$t^{2}$ || on $L^{3/2}$ space since it is bounded
            $endgroup$
            – Duaa Hamzeh
            Dec 18 '18 at 12:19




            $begingroup$
            Also.. I have seen a theorem now from Royden say that || T || = ||$t^{2}$ || on $L^{3/2}$ space since it is bounded
            $endgroup$
            – Duaa Hamzeh
            Dec 18 '18 at 12:19











            1












            $begingroup$

            With Hölder we get



            $|T(f)| le (int_0^1t^3 dt)^{2/3} (int_0^1 |f(t)|^3)^{1/3} = c ||f||_3$,



            where $c:= (int_0^1t^3 dt)^{2/3}$.



            This shows that $T$ is continuous.



            It is your turn to determine $||T||$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              With Hölder we get



              $|T(f)| le (int_0^1t^3 dt)^{2/3} (int_0^1 |f(t)|^3)^{1/3} = c ||f||_3$,



              where $c:= (int_0^1t^3 dt)^{2/3}$.



              This shows that $T$ is continuous.



              It is your turn to determine $||T||$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                With Hölder we get



                $|T(f)| le (int_0^1t^3 dt)^{2/3} (int_0^1 |f(t)|^3)^{1/3} = c ||f||_3$,



                where $c:= (int_0^1t^3 dt)^{2/3}$.



                This shows that $T$ is continuous.



                It is your turn to determine $||T||$.






                share|cite|improve this answer









                $endgroup$



                With Hölder we get



                $|T(f)| le (int_0^1t^3 dt)^{2/3} (int_0^1 |f(t)|^3)^{1/3} = c ||f||_3$,



                where $c:= (int_0^1t^3 dt)^{2/3}$.



                This shows that $T$ is continuous.



                It is your turn to determine $||T||$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 18 '18 at 10:09









                FredFred

                46.7k1848




                46.7k1848






























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