Help showing a linear functional is bounded












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$begingroup$


Let $(X,M,mu)$ be a $sigma$-finite measure space, and $k:Xtimes X rightarrow mathbb{C}$ (or $mathbb{R}$) be $Xtimes X$ measurable. Suppose there are measurable functions $h,g:Xrightarrow(0,infty)$ and constants $c_1,c_2>0$ such that $$int_X |k(x,y)|g^q(y)dmu(y)leq c_1^qh^q(x)$$ a.e. and
$$int_X |k(x,y)|h^p(x)dmu(x)leq c_2^pg^p(y)$$ a.e.
Show that $(Tf)(x)=int_X k(x,y)f(y)dmu(y)$ defines a bounded linear operator $T:L^p(X)rightarrow L^p(X)$ with $||T||leq c_1c_2.$



I can show that $T$ is linear with not trouble, but I don't see how to get it is bounded. Any help would be most appriciated.






functional-analysis







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  • $begingroup$
    Your domain is a Banach space. So you can apply the closed graph theorem to prove a linear transformation is bounded.
    $endgroup$
    – GEdgar
    Dec 28 '18 at 21:44










  • $begingroup$
    @GEdgar I doubt that's helpful. I especially doubt it's helpful for showing $||T|| le c_1c_2$.
    $endgroup$
    – mathworker21
    Dec 28 '18 at 21:47










  • $begingroup$
    @mathworker21: you are probably right.
    $endgroup$
    – GEdgar
    Dec 28 '18 at 21:57
















2












$begingroup$


Let $(X,M,mu)$ be a $sigma$-finite measure space, and $k:Xtimes X rightarrow mathbb{C}$ (or $mathbb{R}$) be $Xtimes X$ measurable. Suppose there are measurable functions $h,g:Xrightarrow(0,infty)$ and constants $c_1,c_2>0$ such that $$int_X |k(x,y)|g^q(y)dmu(y)leq c_1^qh^q(x)$$ a.e. and
$$int_X |k(x,y)|h^p(x)dmu(x)leq c_2^pg^p(y)$$ a.e.
Show that $(Tf)(x)=int_X k(x,y)f(y)dmu(y)$ defines a bounded linear operator $T:L^p(X)rightarrow L^p(X)$ with $||T||leq c_1c_2.$



I can show that $T$ is linear with not trouble, but I don't see how to get it is bounded. Any help would be most appriciated.






functional-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    Your domain is a Banach space. So you can apply the closed graph theorem to prove a linear transformation is bounded.
    $endgroup$
    – GEdgar
    Dec 28 '18 at 21:44










  • $begingroup$
    @GEdgar I doubt that's helpful. I especially doubt it's helpful for showing $||T|| le c_1c_2$.
    $endgroup$
    – mathworker21
    Dec 28 '18 at 21:47










  • $begingroup$
    @mathworker21: you are probably right.
    $endgroup$
    – GEdgar
    Dec 28 '18 at 21:57














2












2








2





$begingroup$


Let $(X,M,mu)$ be a $sigma$-finite measure space, and $k:Xtimes X rightarrow mathbb{C}$ (or $mathbb{R}$) be $Xtimes X$ measurable. Suppose there are measurable functions $h,g:Xrightarrow(0,infty)$ and constants $c_1,c_2>0$ such that $$int_X |k(x,y)|g^q(y)dmu(y)leq c_1^qh^q(x)$$ a.e. and
$$int_X |k(x,y)|h^p(x)dmu(x)leq c_2^pg^p(y)$$ a.e.
Show that $(Tf)(x)=int_X k(x,y)f(y)dmu(y)$ defines a bounded linear operator $T:L^p(X)rightarrow L^p(X)$ with $||T||leq c_1c_2.$



I can show that $T$ is linear with not trouble, but I don't see how to get it is bounded. Any help would be most appriciated.






functional-analysis







share|cite|improve this question









$endgroup$




Let $(X,M,mu)$ be a $sigma$-finite measure space, and $k:Xtimes X rightarrow mathbb{C}$ (or $mathbb{R}$) be $Xtimes X$ measurable. Suppose there are measurable functions $h,g:Xrightarrow(0,infty)$ and constants $c_1,c_2>0$ such that $$int_X |k(x,y)|g^q(y)dmu(y)leq c_1^qh^q(x)$$ a.e. and
$$int_X |k(x,y)|h^p(x)dmu(x)leq c_2^pg^p(y)$$ a.e.
Show that $(Tf)(x)=int_X k(x,y)f(y)dmu(y)$ defines a bounded linear operator $T:L^p(X)rightarrow L^p(X)$ with $||T||leq c_1c_2.$



I can show that $T$ is linear with not trouble, but I don't see how to get it is bounded. Any help would be most appriciated.






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 28 '18 at 21:34









ScottScott

38918




38918












  • $begingroup$
    Your domain is a Banach space. So you can apply the closed graph theorem to prove a linear transformation is bounded.
    $endgroup$
    – GEdgar
    Dec 28 '18 at 21:44










  • $begingroup$
    @GEdgar I doubt that's helpful. I especially doubt it's helpful for showing $||T|| le c_1c_2$.
    $endgroup$
    – mathworker21
    Dec 28 '18 at 21:47










  • $begingroup$
    @mathworker21: you are probably right.
    $endgroup$
    – GEdgar
    Dec 28 '18 at 21:57


















  • $begingroup$
    Your domain is a Banach space. So you can apply the closed graph theorem to prove a linear transformation is bounded.
    $endgroup$
    – GEdgar
    Dec 28 '18 at 21:44










  • $begingroup$
    @GEdgar I doubt that's helpful. I especially doubt it's helpful for showing $||T|| le c_1c_2$.
    $endgroup$
    – mathworker21
    Dec 28 '18 at 21:47










  • $begingroup$
    @mathworker21: you are probably right.
    $endgroup$
    – GEdgar
    Dec 28 '18 at 21:57
















$begingroup$
Your domain is a Banach space. So you can apply the closed graph theorem to prove a linear transformation is bounded.
$endgroup$
– GEdgar
Dec 28 '18 at 21:44




$begingroup$
Your domain is a Banach space. So you can apply the closed graph theorem to prove a linear transformation is bounded.
$endgroup$
– GEdgar
Dec 28 '18 at 21:44












$begingroup$
@GEdgar I doubt that's helpful. I especially doubt it's helpful for showing $||T|| le c_1c_2$.
$endgroup$
– mathworker21
Dec 28 '18 at 21:47




$begingroup$
@GEdgar I doubt that's helpful. I especially doubt it's helpful for showing $||T|| le c_1c_2$.
$endgroup$
– mathworker21
Dec 28 '18 at 21:47












$begingroup$
@mathworker21: you are probably right.
$endgroup$
– GEdgar
Dec 28 '18 at 21:57




$begingroup$
@mathworker21: you are probably right.
$endgroup$
– GEdgar
Dec 28 '18 at 21:57










1 Answer
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$begingroup$

Here I'm assuming $1 < p < infty$, and $q$ is conjugate to $p$.



Fix $fin L^p(X)$. Note



$$lvert Tf(x)rvert le int_X lvert k(x,y)rvert lvert f(y)rvert, dmu(y) = int_X lvert k(x,y)rvert^{1/q} left(frac{g(y)}{h(x)}right)cdotlvert k(x,y)rvert^{1/p}left(frac{h(x)}{g(y)}right)lvert f(y)rvert, dmu(y)$$ and by Hölder's inequality, the latter integral is bounded by $$left[int_X lvert k(x,y)rvert left(frac{g(y)}{h(x)}right)^{q}, dmu(y)right]^{1/q} left[int_X lvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^p, lvert f(y)rvert^p dmu(y)right]^{1/p}$$ Thus
$$lvert Tf(x)rvert le c_1left[int_X lvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y)right]^{1/p}quad text{a.e.}$$ which implies $$|Tf|_p le c_1left[int_Xint_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y), dmu(x)right]^{1/p}$$ By Fubini and the hypothesis,
$$int_Xint_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y), dmu(x) =int_Xleft[int_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^p, dmu(x),right]lvert f(y)rvert^p dmu(y) le c_2^p|f|_p$$ Consequently, $$left[int_Xint_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y), dmu(x)right]^{1/p} le c_2|f|_p$$ yielding $|Tf|_p le c_1c_2|f|_p$. As $f$ was arbitrary, $T$ is bounded on $L^p(X)$ with $|T| le c_1 c_2$.






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    $begingroup$

    Here I'm assuming $1 < p < infty$, and $q$ is conjugate to $p$.



    Fix $fin L^p(X)$. Note



    $$lvert Tf(x)rvert le int_X lvert k(x,y)rvert lvert f(y)rvert, dmu(y) = int_X lvert k(x,y)rvert^{1/q} left(frac{g(y)}{h(x)}right)cdotlvert k(x,y)rvert^{1/p}left(frac{h(x)}{g(y)}right)lvert f(y)rvert, dmu(y)$$ and by Hölder's inequality, the latter integral is bounded by $$left[int_X lvert k(x,y)rvert left(frac{g(y)}{h(x)}right)^{q}, dmu(y)right]^{1/q} left[int_X lvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^p, lvert f(y)rvert^p dmu(y)right]^{1/p}$$ Thus
    $$lvert Tf(x)rvert le c_1left[int_X lvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y)right]^{1/p}quad text{a.e.}$$ which implies $$|Tf|_p le c_1left[int_Xint_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y), dmu(x)right]^{1/p}$$ By Fubini and the hypothesis,
    $$int_Xint_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y), dmu(x) =int_Xleft[int_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^p, dmu(x),right]lvert f(y)rvert^p dmu(y) le c_2^p|f|_p$$ Consequently, $$left[int_Xint_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y), dmu(x)right]^{1/p} le c_2|f|_p$$ yielding $|Tf|_p le c_1c_2|f|_p$. As $f$ was arbitrary, $T$ is bounded on $L^p(X)$ with $|T| le c_1 c_2$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Here I'm assuming $1 < p < infty$, and $q$ is conjugate to $p$.



      Fix $fin L^p(X)$. Note



      $$lvert Tf(x)rvert le int_X lvert k(x,y)rvert lvert f(y)rvert, dmu(y) = int_X lvert k(x,y)rvert^{1/q} left(frac{g(y)}{h(x)}right)cdotlvert k(x,y)rvert^{1/p}left(frac{h(x)}{g(y)}right)lvert f(y)rvert, dmu(y)$$ and by Hölder's inequality, the latter integral is bounded by $$left[int_X lvert k(x,y)rvert left(frac{g(y)}{h(x)}right)^{q}, dmu(y)right]^{1/q} left[int_X lvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^p, lvert f(y)rvert^p dmu(y)right]^{1/p}$$ Thus
      $$lvert Tf(x)rvert le c_1left[int_X lvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y)right]^{1/p}quad text{a.e.}$$ which implies $$|Tf|_p le c_1left[int_Xint_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y), dmu(x)right]^{1/p}$$ By Fubini and the hypothesis,
      $$int_Xint_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y), dmu(x) =int_Xleft[int_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^p, dmu(x),right]lvert f(y)rvert^p dmu(y) le c_2^p|f|_p$$ Consequently, $$left[int_Xint_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y), dmu(x)right]^{1/p} le c_2|f|_p$$ yielding $|Tf|_p le c_1c_2|f|_p$. As $f$ was arbitrary, $T$ is bounded on $L^p(X)$ with $|T| le c_1 c_2$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Here I'm assuming $1 < p < infty$, and $q$ is conjugate to $p$.



        Fix $fin L^p(X)$. Note



        $$lvert Tf(x)rvert le int_X lvert k(x,y)rvert lvert f(y)rvert, dmu(y) = int_X lvert k(x,y)rvert^{1/q} left(frac{g(y)}{h(x)}right)cdotlvert k(x,y)rvert^{1/p}left(frac{h(x)}{g(y)}right)lvert f(y)rvert, dmu(y)$$ and by Hölder's inequality, the latter integral is bounded by $$left[int_X lvert k(x,y)rvert left(frac{g(y)}{h(x)}right)^{q}, dmu(y)right]^{1/q} left[int_X lvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^p, lvert f(y)rvert^p dmu(y)right]^{1/p}$$ Thus
        $$lvert Tf(x)rvert le c_1left[int_X lvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y)right]^{1/p}quad text{a.e.}$$ which implies $$|Tf|_p le c_1left[int_Xint_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y), dmu(x)right]^{1/p}$$ By Fubini and the hypothesis,
        $$int_Xint_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y), dmu(x) =int_Xleft[int_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^p, dmu(x),right]lvert f(y)rvert^p dmu(y) le c_2^p|f|_p$$ Consequently, $$left[int_Xint_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y), dmu(x)right]^{1/p} le c_2|f|_p$$ yielding $|Tf|_p le c_1c_2|f|_p$. As $f$ was arbitrary, $T$ is bounded on $L^p(X)$ with $|T| le c_1 c_2$.






        share|cite|improve this answer









        $endgroup$



        Here I'm assuming $1 < p < infty$, and $q$ is conjugate to $p$.



        Fix $fin L^p(X)$. Note



        $$lvert Tf(x)rvert le int_X lvert k(x,y)rvert lvert f(y)rvert, dmu(y) = int_X lvert k(x,y)rvert^{1/q} left(frac{g(y)}{h(x)}right)cdotlvert k(x,y)rvert^{1/p}left(frac{h(x)}{g(y)}right)lvert f(y)rvert, dmu(y)$$ and by Hölder's inequality, the latter integral is bounded by $$left[int_X lvert k(x,y)rvert left(frac{g(y)}{h(x)}right)^{q}, dmu(y)right]^{1/q} left[int_X lvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^p, lvert f(y)rvert^p dmu(y)right]^{1/p}$$ Thus
        $$lvert Tf(x)rvert le c_1left[int_X lvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y)right]^{1/p}quad text{a.e.}$$ which implies $$|Tf|_p le c_1left[int_Xint_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y), dmu(x)right]^{1/p}$$ By Fubini and the hypothesis,
        $$int_Xint_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y), dmu(x) =int_Xleft[int_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^p, dmu(x),right]lvert f(y)rvert^p dmu(y) le c_2^p|f|_p$$ Consequently, $$left[int_Xint_Xlvert k(x,y)rvert left(frac{h(x)}{g(y)}right)^plvert f(y)rvert^p, dmu(y), dmu(x)right]^{1/p} le c_2|f|_p$$ yielding $|Tf|_p le c_1c_2|f|_p$. As $f$ was arbitrary, $T$ is bounded on $L^p(X)$ with $|T| le c_1 c_2$.







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        answered Dec 28 '18 at 23:41









        kobekobe

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