Show that $phi(r) = ||f||_{L^r(0,1)}$, $rin [1,2]$ is continuous function of r.












0












$begingroup$


Show that the function $$phi(r) = ||f||_{L^r(0,1)}$$ for $rin [1,2]$ is a continuous function of $r$.



Do I need to use the definition of continuity to solve this question? I don't know how to start this question.






real-analysis analysis continuity lp-spaces







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Show that the function $$phi(r) = ||f||_{L^r(0,1)}$$ for $rin [1,2]$ is a continuous function of $r$.



    Do I need to use the definition of continuity to solve this question? I don't know how to start this question.






    real-analysis analysis continuity lp-spaces







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Show that the function $$phi(r) = ||f||_{L^r(0,1)}$$ for $rin [1,2]$ is a continuous function of $r$.



      Do I need to use the definition of continuity to solve this question? I don't know how to start this question.






      real-analysis analysis continuity lp-spaces







      share|cite|improve this question









      $endgroup$




      Show that the function $$phi(r) = ||f||_{L^r(0,1)}$$ for $rin [1,2]$ is a continuous function of $r$.



      Do I need to use the definition of continuity to solve this question? I don't know how to start this question.






      real-analysis analysis continuity lp-spaces




      real-analysis analysis continuity lp-spaces






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 3 at 8:04









      ijk2438ijk2438

      182




      182






















          1 Answer
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          $begingroup$

          We begin by assuming that the end point cases $phi(1)$ and $phi(2)$ are both finite. This guarantees $phi(r)<infty$ for all $1le rle 2$. Let $Phi(r) = phi(r)^r$. And note that $$F(x,y) = x^{frac{1}{y}}=e^{frac{ log x}{y}} $$ is jointly continuous on $xin [0,infty), yin (0,infty)$. Since $phi(r) = F(Phi(r), r)$, it suffices to show $Phi(r)$ is continuous. Note that it is sufficient to prove $Phi(r_n)to Phi(r)$ whenever $r_n to r$. This follows easily from dominated convergence theorem by noticing that
          $$
          0le |f(x)|^r = |f(x)|^r 1_{{|f(x)|<1}}+|f(x)|^r 1_{{|f(x)|ge 1}}le |f(x)| 1_{{|f(x)|<1}}+|f(x)|^2 1_{{|f(x)|ge 1}}in L^1(0,1).
          $$           and $r mapsto |f(x)|^r$ is continuous.



          Note: However, we can say more about $phi(r)$ actually. By Holder's inequality, we can show that
          $$
          |f|_{{p_t}}le |f|_{{p_0}}^{1-t}|f|_{{p_1}}^t
          $$           whenever $frac{1}{p_t}=frac{1-t}{p_0}+frac{t}{p_1}$, $p_0, p_1 in [1,infty]$. This shows that
          $$
          psi:frac{1}{p}mapsto log |f|_{p},
          $$           is convex, and thus is continuous on$ {psi(x) <infty}$. Therefore, it follows that $phi(r)=e^{psi(frac{1}{r})}$ is also continuous.






          share|cite|improve this answer











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            2












            $begingroup$

            We begin by assuming that the end point cases $phi(1)$ and $phi(2)$ are both finite. This guarantees $phi(r)<infty$ for all $1le rle 2$. Let $Phi(r) = phi(r)^r$. And note that $$F(x,y) = x^{frac{1}{y}}=e^{frac{ log x}{y}} $$ is jointly continuous on $xin [0,infty), yin (0,infty)$. Since $phi(r) = F(Phi(r), r)$, it suffices to show $Phi(r)$ is continuous. Note that it is sufficient to prove $Phi(r_n)to Phi(r)$ whenever $r_n to r$. This follows easily from dominated convergence theorem by noticing that
            $$
            0le |f(x)|^r = |f(x)|^r 1_{{|f(x)|<1}}+|f(x)|^r 1_{{|f(x)|ge 1}}le |f(x)| 1_{{|f(x)|<1}}+|f(x)|^2 1_{{|f(x)|ge 1}}in L^1(0,1).
            $$             and $r mapsto |f(x)|^r$ is continuous.



            Note: However, we can say more about $phi(r)$ actually. By Holder's inequality, we can show that
            $$
            |f|_{{p_t}}le |f|_{{p_0}}^{1-t}|f|_{{p_1}}^t
            $$             whenever $frac{1}{p_t}=frac{1-t}{p_0}+frac{t}{p_1}$, $p_0, p_1 in [1,infty]$. This shows that
            $$
            psi:frac{1}{p}mapsto log |f|_{p},
            $$             is convex, and thus is continuous on$ {psi(x) <infty}$. Therefore, it follows that $phi(r)=e^{psi(frac{1}{r})}$ is also continuous.






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              We begin by assuming that the end point cases $phi(1)$ and $phi(2)$ are both finite. This guarantees $phi(r)<infty$ for all $1le rle 2$. Let $Phi(r) = phi(r)^r$. And note that $$F(x,y) = x^{frac{1}{y}}=e^{frac{ log x}{y}} $$ is jointly continuous on $xin [0,infty), yin (0,infty)$. Since $phi(r) = F(Phi(r), r)$, it suffices to show $Phi(r)$ is continuous. Note that it is sufficient to prove $Phi(r_n)to Phi(r)$ whenever $r_n to r$. This follows easily from dominated convergence theorem by noticing that
              $$
              0le |f(x)|^r = |f(x)|^r 1_{{|f(x)|<1}}+|f(x)|^r 1_{{|f(x)|ge 1}}le |f(x)| 1_{{|f(x)|<1}}+|f(x)|^2 1_{{|f(x)|ge 1}}in L^1(0,1).
              $$               and $r mapsto |f(x)|^r$ is continuous.



              Note: However, we can say more about $phi(r)$ actually. By Holder's inequality, we can show that
              $$
              |f|_{{p_t}}le |f|_{{p_0}}^{1-t}|f|_{{p_1}}^t
              $$               whenever $frac{1}{p_t}=frac{1-t}{p_0}+frac{t}{p_1}$, $p_0, p_1 in [1,infty]$. This shows that
              $$
              psi:frac{1}{p}mapsto log |f|_{p},
              $$               is convex, and thus is continuous on$ {psi(x) <infty}$. Therefore, it follows that $phi(r)=e^{psi(frac{1}{r})}$ is also continuous.






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                We begin by assuming that the end point cases $phi(1)$ and $phi(2)$ are both finite. This guarantees $phi(r)<infty$ for all $1le rle 2$. Let $Phi(r) = phi(r)^r$. And note that $$F(x,y) = x^{frac{1}{y}}=e^{frac{ log x}{y}} $$ is jointly continuous on $xin [0,infty), yin (0,infty)$. Since $phi(r) = F(Phi(r), r)$, it suffices to show $Phi(r)$ is continuous. Note that it is sufficient to prove $Phi(r_n)to Phi(r)$ whenever $r_n to r$. This follows easily from dominated convergence theorem by noticing that
                $$
                0le |f(x)|^r = |f(x)|^r 1_{{|f(x)|<1}}+|f(x)|^r 1_{{|f(x)|ge 1}}le |f(x)| 1_{{|f(x)|<1}}+|f(x)|^2 1_{{|f(x)|ge 1}}in L^1(0,1).
                $$                 and $r mapsto |f(x)|^r$ is continuous.



                Note: However, we can say more about $phi(r)$ actually. By Holder's inequality, we can show that
                $$
                |f|_{{p_t}}le |f|_{{p_0}}^{1-t}|f|_{{p_1}}^t
                $$                 whenever $frac{1}{p_t}=frac{1-t}{p_0}+frac{t}{p_1}$, $p_0, p_1 in [1,infty]$. This shows that
                $$
                psi:frac{1}{p}mapsto log |f|_{p},
                $$                 is convex, and thus is continuous on$ {psi(x) <infty}$. Therefore, it follows that $phi(r)=e^{psi(frac{1}{r})}$ is also continuous.






                share|cite|improve this answer











                $endgroup$



                We begin by assuming that the end point cases $phi(1)$ and $phi(2)$ are both finite. This guarantees $phi(r)<infty$ for all $1le rle 2$. Let $Phi(r) = phi(r)^r$. And note that $$F(x,y) = x^{frac{1}{y}}=e^{frac{ log x}{y}} $$ is jointly continuous on $xin [0,infty), yin (0,infty)$. Since $phi(r) = F(Phi(r), r)$, it suffices to show $Phi(r)$ is continuous. Note that it is sufficient to prove $Phi(r_n)to Phi(r)$ whenever $r_n to r$. This follows easily from dominated convergence theorem by noticing that
                $$
                0le |f(x)|^r = |f(x)|^r 1_{{|f(x)|<1}}+|f(x)|^r 1_{{|f(x)|ge 1}}le |f(x)| 1_{{|f(x)|<1}}+|f(x)|^2 1_{{|f(x)|ge 1}}in L^1(0,1).
                $$                 and $r mapsto |f(x)|^r$ is continuous.



                Note: However, we can say more about $phi(r)$ actually. By Holder's inequality, we can show that
                $$
                |f|_{{p_t}}le |f|_{{p_0}}^{1-t}|f|_{{p_1}}^t
                $$                 whenever $frac{1}{p_t}=frac{1-t}{p_0}+frac{t}{p_1}$, $p_0, p_1 in [1,infty]$. This shows that
                $$
                psi:frac{1}{p}mapsto log |f|_{p},
                $$                 is convex, and thus is continuous on$ {psi(x) <infty}$. Therefore, it follows that $phi(r)=e^{psi(frac{1}{r})}$ is also continuous.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 3 at 9:14

























                answered Jan 3 at 8:39









                SongSong

                8,311625




                8,311625






























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