Convergence in $L^{3/2}$ and in $L^2$












3












$begingroup$


Let $(f_n)_n$ be a bounded sequence in $L^3(mathbb R)$, such that $f_nrightarrow f$ in $L^{3/2}(mathbb R)$. Prove that $f_nrightarrow f$ converges in $L^2mathbb (R)$.



I have and idea to first use the H$ddot {o}$lder inequality:



$int_{Omega} |f_n-f|^2 dx leq ||f_n-f||_3||f_n-f||_{3/2}$



Since $3$ and ${3/2}$ are conjugates. The latter term is convergent, but $||f_n-f||_3$ had to be worked out:



$||f_n-f||_3 leq ||f_n||_3+ ||f||_3 $



How do I obtain the bound for $f$ in $L^3 (mathbb R)$ ?



I have the following idea:



$int_Omega |f|^3dx=int_Omega lim_{n to infty} |f_n|^3 dx$



But how do I get the limit outside the integral or how can I use Fatou lemma here?










share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    Let $(f_n)_n$ be a bounded sequence in $L^3(mathbb R)$, such that $f_nrightarrow f$ in $L^{3/2}(mathbb R)$. Prove that $f_nrightarrow f$ converges in $L^2mathbb (R)$.



    I have and idea to first use the H$ddot {o}$lder inequality:



    $int_{Omega} |f_n-f|^2 dx leq ||f_n-f||_3||f_n-f||_{3/2}$



    Since $3$ and ${3/2}$ are conjugates. The latter term is convergent, but $||f_n-f||_3$ had to be worked out:



    $||f_n-f||_3 leq ||f_n||_3+ ||f||_3 $



    How do I obtain the bound for $f$ in $L^3 (mathbb R)$ ?



    I have the following idea:



    $int_Omega |f|^3dx=int_Omega lim_{n to infty} |f_n|^3 dx$



    But how do I get the limit outside the integral or how can I use Fatou lemma here?










    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      Let $(f_n)_n$ be a bounded sequence in $L^3(mathbb R)$, such that $f_nrightarrow f$ in $L^{3/2}(mathbb R)$. Prove that $f_nrightarrow f$ converges in $L^2mathbb (R)$.



      I have and idea to first use the H$ddot {o}$lder inequality:



      $int_{Omega} |f_n-f|^2 dx leq ||f_n-f||_3||f_n-f||_{3/2}$



      Since $3$ and ${3/2}$ are conjugates. The latter term is convergent, but $||f_n-f||_3$ had to be worked out:



      $||f_n-f||_3 leq ||f_n||_3+ ||f||_3 $



      How do I obtain the bound for $f$ in $L^3 (mathbb R)$ ?



      I have the following idea:



      $int_Omega |f|^3dx=int_Omega lim_{n to infty} |f_n|^3 dx$



      But how do I get the limit outside the integral or how can I use Fatou lemma here?










      share|cite|improve this question









      $endgroup$




      Let $(f_n)_n$ be a bounded sequence in $L^3(mathbb R)$, such that $f_nrightarrow f$ in $L^{3/2}(mathbb R)$. Prove that $f_nrightarrow f$ converges in $L^2mathbb (R)$.



      I have and idea to first use the H$ddot {o}$lder inequality:



      $int_{Omega} |f_n-f|^2 dx leq ||f_n-f||_3||f_n-f||_{3/2}$



      Since $3$ and ${3/2}$ are conjugates. The latter term is convergent, but $||f_n-f||_3$ had to be worked out:



      $||f_n-f||_3 leq ||f_n||_3+ ||f||_3 $



      How do I obtain the bound for $f$ in $L^3 (mathbb R)$ ?



      I have the following idea:



      $int_Omega |f|^3dx=int_Omega lim_{n to infty} |f_n|^3 dx$



      But how do I get the limit outside the integral or how can I use Fatou lemma here?







      convergence lp-spaces






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 27 at 9:06









      JohnnyJohnny

      406




      406






















          1 Answer
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          $begingroup$

          EDIT:I see now you had already pointed out the correct way to end. Sorry I had not read well your post(feel free to mark down my answer so to get a real one).



          You can say this to get a bound also for $f$:there is a subsequence $f_{n_k}$ which converges pointwise almost everywhere to $f$. Now, using Fatou lemma you could conclude.






          share|cite|improve this answer











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            1 Answer
            1






            active

            oldest

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            oldest

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            2












            $begingroup$

            EDIT:I see now you had already pointed out the correct way to end. Sorry I had not read well your post(feel free to mark down my answer so to get a real one).



            You can say this to get a bound also for $f$:there is a subsequence $f_{n_k}$ which converges pointwise almost everywhere to $f$. Now, using Fatou lemma you could conclude.






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              EDIT:I see now you had already pointed out the correct way to end. Sorry I had not read well your post(feel free to mark down my answer so to get a real one).



              You can say this to get a bound also for $f$:there is a subsequence $f_{n_k}$ which converges pointwise almost everywhere to $f$. Now, using Fatou lemma you could conclude.






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                EDIT:I see now you had already pointed out the correct way to end. Sorry I had not read well your post(feel free to mark down my answer so to get a real one).



                You can say this to get a bound also for $f$:there is a subsequence $f_{n_k}$ which converges pointwise almost everywhere to $f$. Now, using Fatou lemma you could conclude.






                share|cite|improve this answer











                $endgroup$



                EDIT:I see now you had already pointed out the correct way to end. Sorry I had not read well your post(feel free to mark down my answer so to get a real one).



                You can say this to get a bound also for $f$:there is a subsequence $f_{n_k}$ which converges pointwise almost everywhere to $f$. Now, using Fatou lemma you could conclude.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 27 at 9:27

























                answered Jan 27 at 9:19









                Tommaso ScognamiglioTommaso Scognamiglio

                527412




                527412






























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