Let $T(f)(y)=int_0^infty K(x,y)cdot f(x)dx$, $,$ show $,$ $Vert T(f) Vert _p le Ccdot Vert f Vert _p$












1












$begingroup$


Now here is the full statement.




Let $K:(0, +infty) times (0, +infty) rightarrow Bbb R$ be a Lebesgue measurable function with $K(kx,ky)=k^{-1}K(x,y)$ for every $k>0$ and let
$$int_0^infty |K(x,1)|cdot x^{-frac 1p},dx=C<infty$$
For some $pin[1,infty]$. If $fin L^p$ show that the function
$$T(f)(y)=int_0^infty K(x,y)cdot f(x),dx$$
is well defined for almost every $yin (0,+infty)$ and satisfies $Vert T(f) Vert _p leq C Vert f Vert _p$.




I've been using inequalities for hours and got nowhere. I tried using Hölder's inequality and Tonelli's Theorem in many ways and got frustrated everytime. Any idea would help me a lot.






real-analysis measure-theory lebesgue-integral lebesgue-measure holder-inequality







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Now here is the full statement.




    Let $K:(0, +infty) times (0, +infty) rightarrow Bbb R$ be a Lebesgue measurable function with $K(kx,ky)=k^{-1}K(x,y)$ for every $k>0$ and let
    $$int_0^infty |K(x,1)|cdot x^{-frac 1p},dx=C<infty$$
    For some $pin[1,infty]$. If $fin L^p$ show that the function
    $$T(f)(y)=int_0^infty K(x,y)cdot f(x),dx$$
    is well defined for almost every $yin (0,+infty)$ and satisfies $Vert T(f) Vert _p leq C Vert f Vert _p$.




    I've been using inequalities for hours and got nowhere. I tried using Hölder's inequality and Tonelli's Theorem in many ways and got frustrated everytime. Any idea would help me a lot.






    real-analysis measure-theory lebesgue-integral lebesgue-measure holder-inequality







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Now here is the full statement.




      Let $K:(0, +infty) times (0, +infty) rightarrow Bbb R$ be a Lebesgue measurable function with $K(kx,ky)=k^{-1}K(x,y)$ for every $k>0$ and let
      $$int_0^infty |K(x,1)|cdot x^{-frac 1p},dx=C<infty$$
      For some $pin[1,infty]$. If $fin L^p$ show that the function
      $$T(f)(y)=int_0^infty K(x,y)cdot f(x),dx$$
      is well defined for almost every $yin (0,+infty)$ and satisfies $Vert T(f) Vert _p leq C Vert f Vert _p$.




      I've been using inequalities for hours and got nowhere. I tried using Hölder's inequality and Tonelli's Theorem in many ways and got frustrated everytime. Any idea would help me a lot.






      real-analysis measure-theory lebesgue-integral lebesgue-measure holder-inequality







      share|cite|improve this question











      $endgroup$




      Now here is the full statement.




      Let $K:(0, +infty) times (0, +infty) rightarrow Bbb R$ be a Lebesgue measurable function with $K(kx,ky)=k^{-1}K(x,y)$ for every $k>0$ and let
      $$int_0^infty |K(x,1)|cdot x^{-frac 1p},dx=C<infty$$
      For some $pin[1,infty]$. If $fin L^p$ show that the function
      $$T(f)(y)=int_0^infty K(x,y)cdot f(x),dx$$
      is well defined for almost every $yin (0,+infty)$ and satisfies $Vert T(f) Vert _p leq C Vert f Vert _p$.




      I've been using inequalities for hours and got nowhere. I tried using Hölder's inequality and Tonelli's Theorem in many ways and got frustrated everytime. Any idea would help me a lot.






      real-analysis measure-theory lebesgue-integral lebesgue-measure holder-inequality




      real-analysis measure-theory lebesgue-integral lebesgue-measure holder-inequality






      share|cite|improve this question















      share|cite|improve this question













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      edited Dec 14 '18 at 6:58









      Jonathan

      16312




      16312










      asked Dec 14 '18 at 6:22









      Marcos Martínez WagnerMarcos Martínez Wagner

      1148




      1148






















          1 Answer
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          4












          $begingroup$

          Observe we have
          begin{align}
          |T(f)|_p =& left(int^infty_0 left|int^infty_0 K(x, y)f(x) dx right|^p dyright)^{1/p}\
          =& left(int^infty_0 left|int^infty_0 K(u, 1)f(yu) du right|^p dyright)^{1/p}.
          end{align}
          Then by Minkowski's integral inequality (essentially triangle inequality), we have
          begin{align}
          |T(f)|_p leq& int^infty_0 |K(u, 1)|left( int^infty_0 |f(yu)|^p dyright)^{1/p} du\
          =& int^infty_0|K(u, 1)|u^{-1/p}left(int^infty_0 |f(x)|^p dx right)^{1/p} du.
          end{align}



          Edit: If $p in [1, infty)$, then we know that $C^infty_0(mathbb{R})$ is dense in $L^p(mathbb{R})$.
          Since the above estimate holds on $C^infty_0(mathbb{R})$ then it extends uniquely
          to all of $L^p(mathbb{R})$. Since $T(f) in L^p$ then we know that the set of $y in mathbb{R}$ such that $T(f)(y)=infty$ is measure zero.



          In the case $p=infty$, we see that
          begin{align}
          |T(f)(y)| =& left|int^infty_0 K(u, 1) f(yu) duright|\
          leq & int^infty_0 |K(u, 1)| ducdot |f|_{L^infty}.
          end{align}
          Hence $T$ is well-defined for all $f in L^infty$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Wow, that's pretty clever. I don't wont to push to hard but have you thought about the "well defined" part? Im struggling with that too, the "almost everywhere" part leaves me clueless. Even if you don't, I'm very grateful for your elegant+simple answer
            $endgroup$
            – Marcos Martínez Wagner
            Dec 14 '18 at 7:23











          Your Answer





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          $begingroup$

          Observe we have
          begin{align}
          |T(f)|_p =& left(int^infty_0 left|int^infty_0 K(x, y)f(x) dx right|^p dyright)^{1/p}\
          =& left(int^infty_0 left|int^infty_0 K(u, 1)f(yu) du right|^p dyright)^{1/p}.
          end{align}
          Then by Minkowski's integral inequality (essentially triangle inequality), we have
          begin{align}
          |T(f)|_p leq& int^infty_0 |K(u, 1)|left( int^infty_0 |f(yu)|^p dyright)^{1/p} du\
          =& int^infty_0|K(u, 1)|u^{-1/p}left(int^infty_0 |f(x)|^p dx right)^{1/p} du.
          end{align}



          Edit: If $p in [1, infty)$, then we know that $C^infty_0(mathbb{R})$ is dense in $L^p(mathbb{R})$.
          Since the above estimate holds on $C^infty_0(mathbb{R})$ then it extends uniquely
          to all of $L^p(mathbb{R})$. Since $T(f) in L^p$ then we know that the set of $y in mathbb{R}$ such that $T(f)(y)=infty$ is measure zero.



          In the case $p=infty$, we see that
          begin{align}
          |T(f)(y)| =& left|int^infty_0 K(u, 1) f(yu) duright|\
          leq & int^infty_0 |K(u, 1)| ducdot |f|_{L^infty}.
          end{align}
          Hence $T$ is well-defined for all $f in L^infty$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Wow, that's pretty clever. I don't wont to push to hard but have you thought about the "well defined" part? Im struggling with that too, the "almost everywhere" part leaves me clueless. Even if you don't, I'm very grateful for your elegant+simple answer
            $endgroup$
            – Marcos Martínez Wagner
            Dec 14 '18 at 7:23
















          4












          $begingroup$

          Observe we have
          begin{align}
          |T(f)|_p =& left(int^infty_0 left|int^infty_0 K(x, y)f(x) dx right|^p dyright)^{1/p}\
          =& left(int^infty_0 left|int^infty_0 K(u, 1)f(yu) du right|^p dyright)^{1/p}.
          end{align}
          Then by Minkowski's integral inequality (essentially triangle inequality), we have
          begin{align}
          |T(f)|_p leq& int^infty_0 |K(u, 1)|left( int^infty_0 |f(yu)|^p dyright)^{1/p} du\
          =& int^infty_0|K(u, 1)|u^{-1/p}left(int^infty_0 |f(x)|^p dx right)^{1/p} du.
          end{align}



          Edit: If $p in [1, infty)$, then we know that $C^infty_0(mathbb{R})$ is dense in $L^p(mathbb{R})$.
          Since the above estimate holds on $C^infty_0(mathbb{R})$ then it extends uniquely
          to all of $L^p(mathbb{R})$. Since $T(f) in L^p$ then we know that the set of $y in mathbb{R}$ such that $T(f)(y)=infty$ is measure zero.



          In the case $p=infty$, we see that
          begin{align}
          |T(f)(y)| =& left|int^infty_0 K(u, 1) f(yu) duright|\
          leq & int^infty_0 |K(u, 1)| ducdot |f|_{L^infty}.
          end{align}
          Hence $T$ is well-defined for all $f in L^infty$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Wow, that's pretty clever. I don't wont to push to hard but have you thought about the "well defined" part? Im struggling with that too, the "almost everywhere" part leaves me clueless. Even if you don't, I'm very grateful for your elegant+simple answer
            $endgroup$
            – Marcos Martínez Wagner
            Dec 14 '18 at 7:23














          4












          4








          4





          $begingroup$

          Observe we have
          begin{align}
          |T(f)|_p =& left(int^infty_0 left|int^infty_0 K(x, y)f(x) dx right|^p dyright)^{1/p}\
          =& left(int^infty_0 left|int^infty_0 K(u, 1)f(yu) du right|^p dyright)^{1/p}.
          end{align}
          Then by Minkowski's integral inequality (essentially triangle inequality), we have
          begin{align}
          |T(f)|_p leq& int^infty_0 |K(u, 1)|left( int^infty_0 |f(yu)|^p dyright)^{1/p} du\
          =& int^infty_0|K(u, 1)|u^{-1/p}left(int^infty_0 |f(x)|^p dx right)^{1/p} du.
          end{align}



          Edit: If $p in [1, infty)$, then we know that $C^infty_0(mathbb{R})$ is dense in $L^p(mathbb{R})$.
          Since the above estimate holds on $C^infty_0(mathbb{R})$ then it extends uniquely
          to all of $L^p(mathbb{R})$. Since $T(f) in L^p$ then we know that the set of $y in mathbb{R}$ such that $T(f)(y)=infty$ is measure zero.



          In the case $p=infty$, we see that
          begin{align}
          |T(f)(y)| =& left|int^infty_0 K(u, 1) f(yu) duright|\
          leq & int^infty_0 |K(u, 1)| ducdot |f|_{L^infty}.
          end{align}
          Hence $T$ is well-defined for all $f in L^infty$.






          share|cite|improve this answer











          $endgroup$



          Observe we have
          begin{align}
          |T(f)|_p =& left(int^infty_0 left|int^infty_0 K(x, y)f(x) dx right|^p dyright)^{1/p}\
          =& left(int^infty_0 left|int^infty_0 K(u, 1)f(yu) du right|^p dyright)^{1/p}.
          end{align}
          Then by Minkowski's integral inequality (essentially triangle inequality), we have
          begin{align}
          |T(f)|_p leq& int^infty_0 |K(u, 1)|left( int^infty_0 |f(yu)|^p dyright)^{1/p} du\
          =& int^infty_0|K(u, 1)|u^{-1/p}left(int^infty_0 |f(x)|^p dx right)^{1/p} du.
          end{align}



          Edit: If $p in [1, infty)$, then we know that $C^infty_0(mathbb{R})$ is dense in $L^p(mathbb{R})$.
          Since the above estimate holds on $C^infty_0(mathbb{R})$ then it extends uniquely
          to all of $L^p(mathbb{R})$. Since $T(f) in L^p$ then we know that the set of $y in mathbb{R}$ such that $T(f)(y)=infty$ is measure zero.



          In the case $p=infty$, we see that
          begin{align}
          |T(f)(y)| =& left|int^infty_0 K(u, 1) f(yu) duright|\
          leq & int^infty_0 |K(u, 1)| ducdot |f|_{L^infty}.
          end{align}
          Hence $T$ is well-defined for all $f in L^infty$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 15 '18 at 6:04

























          answered Dec 14 '18 at 6:46









          Jacky ChongJacky Chong

          18.7k21128




          18.7k21128












          • $begingroup$
            Wow, that's pretty clever. I don't wont to push to hard but have you thought about the "well defined" part? Im struggling with that too, the "almost everywhere" part leaves me clueless. Even if you don't, I'm very grateful for your elegant+simple answer
            $endgroup$
            – Marcos Martínez Wagner
            Dec 14 '18 at 7:23


















          • $begingroup$
            Wow, that's pretty clever. I don't wont to push to hard but have you thought about the "well defined" part? Im struggling with that too, the "almost everywhere" part leaves me clueless. Even if you don't, I'm very grateful for your elegant+simple answer
            $endgroup$
            – Marcos Martínez Wagner
            Dec 14 '18 at 7:23
















          $begingroup$
          Wow, that's pretty clever. I don't wont to push to hard but have you thought about the "well defined" part? Im struggling with that too, the "almost everywhere" part leaves me clueless. Even if you don't, I'm very grateful for your elegant+simple answer
          $endgroup$
          – Marcos Martínez Wagner
          Dec 14 '18 at 7:23




          $begingroup$
          Wow, that's pretty clever. I don't wont to push to hard but have you thought about the "well defined" part? Im struggling with that too, the "almost everywhere" part leaves me clueless. Even if you don't, I'm very grateful for your elegant+simple answer
          $endgroup$
          – Marcos Martínez Wagner
          Dec 14 '18 at 7:23


















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