Continuous embeddings of $C(a,b)$ into itself with $L^p$ norms.












1












$begingroup$


Let $1leq p<qleqinfty$



Consider the spaces $X=(C(a,b), |cdot|_{L^q(a,b)})$ and $Y=(C(a,b), |cdot|_{L^p(a,b)})$. I want to show $X$ is continuously embedded in $Y$, but $Y$ is not continuously embedded in $X$.



Proof attempt:



Note an operator between normed spaces is continuous if and only if bounded, so it suffices to show that $exists C>0$ s.t.
$|x|_Yleq C|x|_X$



We want the inclusion map $iota(f) mapsto f$, to be bounded for $fin C(a,b)$ where $|f|_{L^q}$.



Note for a finite measure space $E$, (Lebesgue measure), there is a nesting of $L^p(E)$ spaces. We have the following relationship derived from Holder's inequality:
for $1leq p<qleq infty$
$|f|_{L^p}leq mu(E)^{1/p-1/q}|f|_{L^q}$



So we have $|f|_Y leq C|f|_X$ where $C=mu(E)^{1/p-1/q}$. So the inclusion map is continuous, and so $X$ is continuously embedded into $Y$.



However, $Y$ is not continuously embedded into $X$.



we can show the inverse inclusion map $iota^{-1}(f)mapsto f$ in general is not bounded by taking $fin L^p(a,b)setminus L^q(a,b)$, and noting the norm in $L^p$ is finite but the norm in $L^q$ is infinite, showing the inverse/backwards inclusion map is not continuous. I could use the explicit counterexample: $x^{-1/q}{chi(a,b)}$ to show this.






real-analysis functional-analysis operator-theory lp-spaces







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Let $1leq p<qleqinfty$



    Consider the spaces $X=(C(a,b), |cdot|_{L^q(a,b)})$ and $Y=(C(a,b), |cdot|_{L^p(a,b)})$. I want to show $X$ is continuously embedded in $Y$, but $Y$ is not continuously embedded in $X$.



    Proof attempt:



    Note an operator between normed spaces is continuous if and only if bounded, so it suffices to show that $exists C>0$ s.t.
    $|x|_Yleq C|x|_X$



    We want the inclusion map $iota(f) mapsto f$, to be bounded for $fin C(a,b)$ where $|f|_{L^q}$.



    Note for a finite measure space $E$, (Lebesgue measure), there is a nesting of $L^p(E)$ spaces. We have the following relationship derived from Holder's inequality:
    for $1leq p<qleq infty$
    $|f|_{L^p}leq mu(E)^{1/p-1/q}|f|_{L^q}$



    So we have $|f|_Y leq C|f|_X$ where $C=mu(E)^{1/p-1/q}$. So the inclusion map is continuous, and so $X$ is continuously embedded into $Y$.



    However, $Y$ is not continuously embedded into $X$.



    we can show the inverse inclusion map $iota^{-1}(f)mapsto f$ in general is not bounded by taking $fin L^p(a,b)setminus L^q(a,b)$, and noting the norm in $L^p$ is finite but the norm in $L^q$ is infinite, showing the inverse/backwards inclusion map is not continuous. I could use the explicit counterexample: $x^{-1/q}{chi(a,b)}$ to show this.






    real-analysis functional-analysis operator-theory lp-spaces







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Let $1leq p<qleqinfty$



      Consider the spaces $X=(C(a,b), |cdot|_{L^q(a,b)})$ and $Y=(C(a,b), |cdot|_{L^p(a,b)})$. I want to show $X$ is continuously embedded in $Y$, but $Y$ is not continuously embedded in $X$.



      Proof attempt:



      Note an operator between normed spaces is continuous if and only if bounded, so it suffices to show that $exists C>0$ s.t.
      $|x|_Yleq C|x|_X$



      We want the inclusion map $iota(f) mapsto f$, to be bounded for $fin C(a,b)$ where $|f|_{L^q}$.



      Note for a finite measure space $E$, (Lebesgue measure), there is a nesting of $L^p(E)$ spaces. We have the following relationship derived from Holder's inequality:
      for $1leq p<qleq infty$
      $|f|_{L^p}leq mu(E)^{1/p-1/q}|f|_{L^q}$



      So we have $|f|_Y leq C|f|_X$ where $C=mu(E)^{1/p-1/q}$. So the inclusion map is continuous, and so $X$ is continuously embedded into $Y$.



      However, $Y$ is not continuously embedded into $X$.



      we can show the inverse inclusion map $iota^{-1}(f)mapsto f$ in general is not bounded by taking $fin L^p(a,b)setminus L^q(a,b)$, and noting the norm in $L^p$ is finite but the norm in $L^q$ is infinite, showing the inverse/backwards inclusion map is not continuous. I could use the explicit counterexample: $x^{-1/q}{chi(a,b)}$ to show this.






      real-analysis functional-analysis operator-theory lp-spaces







      share|cite|improve this question











      $endgroup$




      Let $1leq p<qleqinfty$



      Consider the spaces $X=(C(a,b), |cdot|_{L^q(a,b)})$ and $Y=(C(a,b), |cdot|_{L^p(a,b)})$. I want to show $X$ is continuously embedded in $Y$, but $Y$ is not continuously embedded in $X$.



      Proof attempt:



      Note an operator between normed spaces is continuous if and only if bounded, so it suffices to show that $exists C>0$ s.t.
      $|x|_Yleq C|x|_X$



      We want the inclusion map $iota(f) mapsto f$, to be bounded for $fin C(a,b)$ where $|f|_{L^q}$.



      Note for a finite measure space $E$, (Lebesgue measure), there is a nesting of $L^p(E)$ spaces. We have the following relationship derived from Holder's inequality:
      for $1leq p<qleq infty$
      $|f|_{L^p}leq mu(E)^{1/p-1/q}|f|_{L^q}$



      So we have $|f|_Y leq C|f|_X$ where $C=mu(E)^{1/p-1/q}$. So the inclusion map is continuous, and so $X$ is continuously embedded into $Y$.



      However, $Y$ is not continuously embedded into $X$.



      we can show the inverse inclusion map $iota^{-1}(f)mapsto f$ in general is not bounded by taking $fin L^p(a,b)setminus L^q(a,b)$, and noting the norm in $L^p$ is finite but the norm in $L^q$ is infinite, showing the inverse/backwards inclusion map is not continuous. I could use the explicit counterexample: $x^{-1/q}{chi(a,b)}$ to show this.






      real-analysis functional-analysis operator-theory lp-spaces




      real-analysis functional-analysis operator-theory lp-spaces






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 15 '18 at 7:49









      Yiorgos S. Smyrlis

      63.7k1385165




      63.7k1385165










      asked Sep 21 '17 at 2:22









      Kernel_DirichletKernel_Dirichlet

      1,152416




      1,152416






















          1 Answer
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          $begingroup$

          Consider $(a,b)=(0,1)$, $rin (p,q)$ and
          $$
          f_varepsilon(x)=(x+varepsilon)^{-frac{1}{r}}.
          $$
          Then
          $$
          int_0^1 |,f_varepsilon|^p<int_0^1 x^{-p/r},dx=frac{1}{1-p/r}=frac{r}{r-p}
          $$
          and hence
          $$
          |,f_varepsilon|_p<left(frac{r}{r-p}right)^{1/p}.
          $$
          Meanwhile
          $$
          lim_{varepsilonto 0}int_0^1 |,f_varepsilon|^q=lim_{varepsilonto 0}int_0^1 (x+varepsilon)^{-frac{q}{r}},dx=lim_{varepsilonto 0}frac{1}{q/r-1}left(varepsilon^{-q/r}-(1+varepsilon)^{-q/r}right)=infty.
          $$






          share|cite|improve this answer









          $endgroup$














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            $begingroup$

            Consider $(a,b)=(0,1)$, $rin (p,q)$ and
            $$
            f_varepsilon(x)=(x+varepsilon)^{-frac{1}{r}}.
            $$
            Then
            $$
            int_0^1 |,f_varepsilon|^p<int_0^1 x^{-p/r},dx=frac{1}{1-p/r}=frac{r}{r-p}
            $$
            and hence
            $$
            |,f_varepsilon|_p<left(frac{r}{r-p}right)^{1/p}.
            $$
            Meanwhile
            $$
            lim_{varepsilonto 0}int_0^1 |,f_varepsilon|^q=lim_{varepsilonto 0}int_0^1 (x+varepsilon)^{-frac{q}{r}},dx=lim_{varepsilonto 0}frac{1}{q/r-1}left(varepsilon^{-q/r}-(1+varepsilon)^{-q/r}right)=infty.
            $$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Consider $(a,b)=(0,1)$, $rin (p,q)$ and
              $$
              f_varepsilon(x)=(x+varepsilon)^{-frac{1}{r}}.
              $$
              Then
              $$
              int_0^1 |,f_varepsilon|^p<int_0^1 x^{-p/r},dx=frac{1}{1-p/r}=frac{r}{r-p}
              $$
              and hence
              $$
              |,f_varepsilon|_p<left(frac{r}{r-p}right)^{1/p}.
              $$
              Meanwhile
              $$
              lim_{varepsilonto 0}int_0^1 |,f_varepsilon|^q=lim_{varepsilonto 0}int_0^1 (x+varepsilon)^{-frac{q}{r}},dx=lim_{varepsilonto 0}frac{1}{q/r-1}left(varepsilon^{-q/r}-(1+varepsilon)^{-q/r}right)=infty.
              $$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Consider $(a,b)=(0,1)$, $rin (p,q)$ and
                $$
                f_varepsilon(x)=(x+varepsilon)^{-frac{1}{r}}.
                $$
                Then
                $$
                int_0^1 |,f_varepsilon|^p<int_0^1 x^{-p/r},dx=frac{1}{1-p/r}=frac{r}{r-p}
                $$
                and hence
                $$
                |,f_varepsilon|_p<left(frac{r}{r-p}right)^{1/p}.
                $$
                Meanwhile
                $$
                lim_{varepsilonto 0}int_0^1 |,f_varepsilon|^q=lim_{varepsilonto 0}int_0^1 (x+varepsilon)^{-frac{q}{r}},dx=lim_{varepsilonto 0}frac{1}{q/r-1}left(varepsilon^{-q/r}-(1+varepsilon)^{-q/r}right)=infty.
                $$






                share|cite|improve this answer









                $endgroup$



                Consider $(a,b)=(0,1)$, $rin (p,q)$ and
                $$
                f_varepsilon(x)=(x+varepsilon)^{-frac{1}{r}}.
                $$
                Then
                $$
                int_0^1 |,f_varepsilon|^p<int_0^1 x^{-p/r},dx=frac{1}{1-p/r}=frac{r}{r-p}
                $$
                and hence
                $$
                |,f_varepsilon|_p<left(frac{r}{r-p}right)^{1/p}.
                $$
                Meanwhile
                $$
                lim_{varepsilonto 0}int_0^1 |,f_varepsilon|^q=lim_{varepsilonto 0}int_0^1 (x+varepsilon)^{-frac{q}{r}},dx=lim_{varepsilonto 0}frac{1}{q/r-1}left(varepsilon^{-q/r}-(1+varepsilon)^{-q/r}right)=infty.
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Sep 21 '17 at 9:27









                Yiorgos S. SmyrlisYiorgos S. Smyrlis

                63.7k1385165




                63.7k1385165






























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