Lebesgue integrable function involving distance function












1












$begingroup$


$f$ is a bounded measurable function with compact support in $mathbb{R}, int_mathbb{R}f(x)dx=0$



$M_f(x)=sup_{t>0}{|int_mathbb{R}f(y)frac{t}{t^2+(x-y)^2}dy|}$



Prove $M_f(x)in L(mathbb{R})$



My attempt to solve the problem: I saw Lebesgue integral involving distance function. I tried to use Tonelli Theorem but didn't succeed.






real-analysis lebesgue-integral







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    $f$ is a bounded measurable function with compact support in $mathbb{R}, int_mathbb{R}f(x)dx=0$



    $M_f(x)=sup_{t>0}{|int_mathbb{R}f(y)frac{t}{t^2+(x-y)^2}dy|}$



    Prove $M_f(x)in L(mathbb{R})$



    My attempt to solve the problem: I saw Lebesgue integral involving distance function. I tried to use Tonelli Theorem but didn't succeed.






    real-analysis lebesgue-integral







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      $f$ is a bounded measurable function with compact support in $mathbb{R}, int_mathbb{R}f(x)dx=0$



      $M_f(x)=sup_{t>0}{|int_mathbb{R}f(y)frac{t}{t^2+(x-y)^2}dy|}$



      Prove $M_f(x)in L(mathbb{R})$



      My attempt to solve the problem: I saw Lebesgue integral involving distance function. I tried to use Tonelli Theorem but didn't succeed.






      real-analysis lebesgue-integral







      share|cite|improve this question











      $endgroup$




      $f$ is a bounded measurable function with compact support in $mathbb{R}, int_mathbb{R}f(x)dx=0$



      $M_f(x)=sup_{t>0}{|int_mathbb{R}f(y)frac{t}{t^2+(x-y)^2}dy|}$



      Prove $M_f(x)in L(mathbb{R})$



      My attempt to solve the problem: I saw Lebesgue integral involving distance function. I tried to use Tonelli Theorem but didn't succeed.






      real-analysis lebesgue-integral




      real-analysis lebesgue-integral






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 14 at 14:03









      David C. Ullrich

      60.5k43994




      60.5k43994










      asked Jan 10 at 12:08









      NickNick

      63




      63






















          1 Answer
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          active

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          2












          $begingroup$

          In case anyone wonders where the question comes from: It's a standard thing, essentially the trivial half of the "atomic decomposition" for the "real" Hardy space $H^1(Bbb R)$:





          Theorem. Suppose $fin L^1(Bbb R)$. Then $M_f$ is inegrable if and only if $f=sum c_j f_j$ where $sum|c_j|<infty$, $f_j$ is supported in $[a_j,b_j]$,$||f_j||_inftyle1/(b_j-a_j)$ and $int f_j=0$.





          Functions $f_j$ as above are "atoms"; for one direction it's enough to show that if $f$ is an atom then $||M_f||_1le c$, which is essentially proved below. The other direction is not so simple



          Anyway: Say $|f|le c_1$ and $f$ is supported in $[-A,A]$. If you can show



          (i) $|M_f|lepi c_1$



          (ii) $|M_f(x)|le c_2/x^2$ for $|x|ge 2A$



          it follows that $M_f$ is integrable. (i) should be easy. For (ii), note that since $int f=0$ you have $$int f(y)frac t{t^2+(x-y)^2}dy=
          int_{-A}^A f(y)left(frac t{t^2+(x-y)^2}-a_{t,x}right)dy$$          for any choice of constants $a_{t,x}$. I think $a_{t,x}=t/(t^2+x^2)$ works; subtract the fractions, and note that if $|y|le A$ and $|x|ge 2A$ then $|x-y|ge|x|/2$.



          Edit: Since the OP seems to have lost interest I may as well fill in the details. Suppose $t>0$, $|y|le A$ and $x>2A$. s noted above, $x-yge x-x/2=x/2$. Now $$begin{align}left|frac t{t^2+(x-y)^2}-frac t{t^2+x^2}right|&=tfrac{|2xy-y^2|}{(t^2+(x-y)^2)(t^2+x^2)}
          \&le tfrac{2Ax+A^2}{(t^2+(x-y)^2)(t^2+x^2)}.end{align}$$
          Since $A^2le Ax/2$ this gives $$begin{align}left|frac t{t^2+(x-y)^2}-frac t{t^2+x^2}right|&le cfrac{tx}{(t^2+(x-y)^2)(t^2+x^2)}
          \&le cfrac{tx}{(t^2+x^2)^2}.end{align}$$          Now $x/(t^2+x^2)le x/x^2=1/x$. If $t>x$ then $t/(t^2+x^2)le t/t^2=1/tle 1/x$, whhile if $t<x$ then $t/(t^2+x^2)le x/x^2=1/x$; so we have $$left|frac t{t^2+(x-y)^2}-frac t{t^2+x^2}right|lefrac{c}{x^2}quad(x>2A, |y|le A).$$Similarly for $x<-2A$; hence $$M_f(x)lefrac{c||f||_1}{x^2}quad(|x|>2A),$$as required.






          share|cite|improve this answer











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            $begingroup$

            In case anyone wonders where the question comes from: It's a standard thing, essentially the trivial half of the "atomic decomposition" for the "real" Hardy space $H^1(Bbb R)$:





            Theorem. Suppose $fin L^1(Bbb R)$. Then $M_f$ is inegrable if and only if $f=sum c_j f_j$ where $sum|c_j|<infty$, $f_j$ is supported in $[a_j,b_j]$,$||f_j||_inftyle1/(b_j-a_j)$ and $int f_j=0$.





            Functions $f_j$ as above are "atoms"; for one direction it's enough to show that if $f$ is an atom then $||M_f||_1le c$, which is essentially proved below. The other direction is not so simple



            Anyway: Say $|f|le c_1$ and $f$ is supported in $[-A,A]$. If you can show



            (i) $|M_f|lepi c_1$



            (ii) $|M_f(x)|le c_2/x^2$ for $|x|ge 2A$



            it follows that $M_f$ is integrable. (i) should be easy. For (ii), note that since $int f=0$ you have $$int f(y)frac t{t^2+(x-y)^2}dy=
            int_{-A}^A f(y)left(frac t{t^2+(x-y)^2}-a_{t,x}right)dy$$            for any choice of constants $a_{t,x}$. I think $a_{t,x}=t/(t^2+x^2)$ works; subtract the fractions, and note that if $|y|le A$ and $|x|ge 2A$ then $|x-y|ge|x|/2$.



            Edit: Since the OP seems to have lost interest I may as well fill in the details. Suppose $t>0$, $|y|le A$ and $x>2A$. s noted above, $x-yge x-x/2=x/2$. Now $$begin{align}left|frac t{t^2+(x-y)^2}-frac t{t^2+x^2}right|&=tfrac{|2xy-y^2|}{(t^2+(x-y)^2)(t^2+x^2)}
            \&le tfrac{2Ax+A^2}{(t^2+(x-y)^2)(t^2+x^2)}.end{align}$$
            Since $A^2le Ax/2$ this gives $$begin{align}left|frac t{t^2+(x-y)^2}-frac t{t^2+x^2}right|&le cfrac{tx}{(t^2+(x-y)^2)(t^2+x^2)}
            \&le cfrac{tx}{(t^2+x^2)^2}.end{align}$$            Now $x/(t^2+x^2)le x/x^2=1/x$. If $t>x$ then $t/(t^2+x^2)le t/t^2=1/tle 1/x$, whhile if $t<x$ then $t/(t^2+x^2)le x/x^2=1/x$; so we have $$left|frac t{t^2+(x-y)^2}-frac t{t^2+x^2}right|lefrac{c}{x^2}quad(x>2A, |y|le A).$$Similarly for $x<-2A$; hence $$M_f(x)lefrac{c||f||_1}{x^2}quad(|x|>2A),$$as required.






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              In case anyone wonders where the question comes from: It's a standard thing, essentially the trivial half of the "atomic decomposition" for the "real" Hardy space $H^1(Bbb R)$:





              Theorem. Suppose $fin L^1(Bbb R)$. Then $M_f$ is inegrable if and only if $f=sum c_j f_j$ where $sum|c_j|<infty$, $f_j$ is supported in $[a_j,b_j]$,$||f_j||_inftyle1/(b_j-a_j)$ and $int f_j=0$.





              Functions $f_j$ as above are "atoms"; for one direction it's enough to show that if $f$ is an atom then $||M_f||_1le c$, which is essentially proved below. The other direction is not so simple



              Anyway: Say $|f|le c_1$ and $f$ is supported in $[-A,A]$. If you can show



              (i) $|M_f|lepi c_1$



              (ii) $|M_f(x)|le c_2/x^2$ for $|x|ge 2A$



              it follows that $M_f$ is integrable. (i) should be easy. For (ii), note that since $int f=0$ you have $$int f(y)frac t{t^2+(x-y)^2}dy=
              int_{-A}^A f(y)left(frac t{t^2+(x-y)^2}-a_{t,x}right)dy$$              for any choice of constants $a_{t,x}$. I think $a_{t,x}=t/(t^2+x^2)$ works; subtract the fractions, and note that if $|y|le A$ and $|x|ge 2A$ then $|x-y|ge|x|/2$.



              Edit: Since the OP seems to have lost interest I may as well fill in the details. Suppose $t>0$, $|y|le A$ and $x>2A$. s noted above, $x-yge x-x/2=x/2$. Now $$begin{align}left|frac t{t^2+(x-y)^2}-frac t{t^2+x^2}right|&=tfrac{|2xy-y^2|}{(t^2+(x-y)^2)(t^2+x^2)}
              \&le tfrac{2Ax+A^2}{(t^2+(x-y)^2)(t^2+x^2)}.end{align}$$
              Since $A^2le Ax/2$ this gives $$begin{align}left|frac t{t^2+(x-y)^2}-frac t{t^2+x^2}right|&le cfrac{tx}{(t^2+(x-y)^2)(t^2+x^2)}
              \&le cfrac{tx}{(t^2+x^2)^2}.end{align}$$              Now $x/(t^2+x^2)le x/x^2=1/x$. If $t>x$ then $t/(t^2+x^2)le t/t^2=1/tle 1/x$, whhile if $t<x$ then $t/(t^2+x^2)le x/x^2=1/x$; so we have $$left|frac t{t^2+(x-y)^2}-frac t{t^2+x^2}right|lefrac{c}{x^2}quad(x>2A, |y|le A).$$Similarly for $x<-2A$; hence $$M_f(x)lefrac{c||f||_1}{x^2}quad(|x|>2A),$$as required.






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                In case anyone wonders where the question comes from: It's a standard thing, essentially the trivial half of the "atomic decomposition" for the "real" Hardy space $H^1(Bbb R)$:





                Theorem. Suppose $fin L^1(Bbb R)$. Then $M_f$ is inegrable if and only if $f=sum c_j f_j$ where $sum|c_j|<infty$, $f_j$ is supported in $[a_j,b_j]$,$||f_j||_inftyle1/(b_j-a_j)$ and $int f_j=0$.





                Functions $f_j$ as above are "atoms"; for one direction it's enough to show that if $f$ is an atom then $||M_f||_1le c$, which is essentially proved below. The other direction is not so simple



                Anyway: Say $|f|le c_1$ and $f$ is supported in $[-A,A]$. If you can show



                (i) $|M_f|lepi c_1$



                (ii) $|M_f(x)|le c_2/x^2$ for $|x|ge 2A$



                it follows that $M_f$ is integrable. (i) should be easy. For (ii), note that since $int f=0$ you have $$int f(y)frac t{t^2+(x-y)^2}dy=
                int_{-A}^A f(y)left(frac t{t^2+(x-y)^2}-a_{t,x}right)dy$$                for any choice of constants $a_{t,x}$. I think $a_{t,x}=t/(t^2+x^2)$ works; subtract the fractions, and note that if $|y|le A$ and $|x|ge 2A$ then $|x-y|ge|x|/2$.



                Edit: Since the OP seems to have lost interest I may as well fill in the details. Suppose $t>0$, $|y|le A$ and $x>2A$. s noted above, $x-yge x-x/2=x/2$. Now $$begin{align}left|frac t{t^2+(x-y)^2}-frac t{t^2+x^2}right|&=tfrac{|2xy-y^2|}{(t^2+(x-y)^2)(t^2+x^2)}
                \&le tfrac{2Ax+A^2}{(t^2+(x-y)^2)(t^2+x^2)}.end{align}$$
                Since $A^2le Ax/2$ this gives $$begin{align}left|frac t{t^2+(x-y)^2}-frac t{t^2+x^2}right|&le cfrac{tx}{(t^2+(x-y)^2)(t^2+x^2)}
                \&le cfrac{tx}{(t^2+x^2)^2}.end{align}$$                Now $x/(t^2+x^2)le x/x^2=1/x$. If $t>x$ then $t/(t^2+x^2)le t/t^2=1/tle 1/x$, whhile if $t<x$ then $t/(t^2+x^2)le x/x^2=1/x$; so we have $$left|frac t{t^2+(x-y)^2}-frac t{t^2+x^2}right|lefrac{c}{x^2}quad(x>2A, |y|le A).$$Similarly for $x<-2A$; hence $$M_f(x)lefrac{c||f||_1}{x^2}quad(|x|>2A),$$as required.






                share|cite|improve this answer











                $endgroup$



                In case anyone wonders where the question comes from: It's a standard thing, essentially the trivial half of the "atomic decomposition" for the "real" Hardy space $H^1(Bbb R)$:





                Theorem. Suppose $fin L^1(Bbb R)$. Then $M_f$ is inegrable if and only if $f=sum c_j f_j$ where $sum|c_j|<infty$, $f_j$ is supported in $[a_j,b_j]$,$||f_j||_inftyle1/(b_j-a_j)$ and $int f_j=0$.





                Functions $f_j$ as above are "atoms"; for one direction it's enough to show that if $f$ is an atom then $||M_f||_1le c$, which is essentially proved below. The other direction is not so simple



                Anyway: Say $|f|le c_1$ and $f$ is supported in $[-A,A]$. If you can show



                (i) $|M_f|lepi c_1$



                (ii) $|M_f(x)|le c_2/x^2$ for $|x|ge 2A$



                it follows that $M_f$ is integrable. (i) should be easy. For (ii), note that since $int f=0$ you have $$int f(y)frac t{t^2+(x-y)^2}dy=
                int_{-A}^A f(y)left(frac t{t^2+(x-y)^2}-a_{t,x}right)dy$$                for any choice of constants $a_{t,x}$. I think $a_{t,x}=t/(t^2+x^2)$ works; subtract the fractions, and note that if $|y|le A$ and $|x|ge 2A$ then $|x-y|ge|x|/2$.



                Edit: Since the OP seems to have lost interest I may as well fill in the details. Suppose $t>0$, $|y|le A$ and $x>2A$. s noted above, $x-yge x-x/2=x/2$. Now $$begin{align}left|frac t{t^2+(x-y)^2}-frac t{t^2+x^2}right|&=tfrac{|2xy-y^2|}{(t^2+(x-y)^2)(t^2+x^2)}
                \&le tfrac{2Ax+A^2}{(t^2+(x-y)^2)(t^2+x^2)}.end{align}$$
                Since $A^2le Ax/2$ this gives $$begin{align}left|frac t{t^2+(x-y)^2}-frac t{t^2+x^2}right|&le cfrac{tx}{(t^2+(x-y)^2)(t^2+x^2)}
                \&le cfrac{tx}{(t^2+x^2)^2}.end{align}$$                Now $x/(t^2+x^2)le x/x^2=1/x$. If $t>x$ then $t/(t^2+x^2)le t/t^2=1/tle 1/x$, whhile if $t<x$ then $t/(t^2+x^2)le x/x^2=1/x$; so we have $$left|frac t{t^2+(x-y)^2}-frac t{t^2+x^2}right|lefrac{c}{x^2}quad(x>2A, |y|le A).$$Similarly for $x<-2A$; hence $$M_f(x)lefrac{c||f||_1}{x^2}quad(|x|>2A),$$as required.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 14 at 14:23

























                answered Jan 10 at 15:01









                David C. UllrichDavid C. Ullrich

                60.5k43994




                60.5k43994






























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