Hardy Littlewood maximal function $|h(x)|<|h^*(x)|$ for almost every x











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Suppose $h: Bbb R to : Bbb R$; Hardy Littlewood maximal function $h^*: Bbb R to [0, infty]$ defined by $h^*(x)=sup_{t>0}frac 1{2t}int_{x-t}^{x+t}|h|$.




Now to prove $|h(x)|leq|h^*(x)|$ for almost every $x in Bbb R$




I am thinking that I have to use Lebesgue differentiation theorem, first version that states "Suppose $fin L^{1}Bbb R$. Then



$lim_{ t to 0}frac 1{2t}int_{x-t}^{x+t}|f-f(b)|=0$ for all most all $b in Bbb R$.



My attempt: $||h(x)|-|h^*(x)||leq |h(x)-h^*(x)|=|h(x)-sup_{t>0}frac 1{2t}int_{x-t}^{x+t}|h(y)|dy|leq sup_{t>0}frac 1{2t}int_{x-t}^{x+t}|h(x)-h(y)|dy=0$ for almost every $x$. So, we get $|h(x)|leq h^*(x)$ for allmost every $x$.



Is my attempt coorect or is there any other way to approach? Please help










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    Suppose $h: Bbb R to : Bbb R$; Hardy Littlewood maximal function $h^*: Bbb R to [0, infty]$ defined by $h^*(x)=sup_{t>0}frac 1{2t}int_{x-t}^{x+t}|h|$.




    Now to prove $|h(x)|leq|h^*(x)|$ for almost every $x in Bbb R$




    I am thinking that I have to use Lebesgue differentiation theorem, first version that states "Suppose $fin L^{1}Bbb R$. Then



    $lim_{ t to 0}frac 1{2t}int_{x-t}^{x+t}|f-f(b)|=0$ for all most all $b in Bbb R$.



    My attempt: $||h(x)|-|h^*(x)||leq |h(x)-h^*(x)|=|h(x)-sup_{t>0}frac 1{2t}int_{x-t}^{x+t}|h(y)|dy|leq sup_{t>0}frac 1{2t}int_{x-t}^{x+t}|h(x)-h(y)|dy=0$ for almost every $x$. So, we get $|h(x)|leq h^*(x)$ for allmost every $x$.



    Is my attempt coorect or is there any other way to approach? Please help










    share|cite|improve this question
























      up vote
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      down vote

      favorite









      up vote
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      down vote

      favorite











      Suppose $h: Bbb R to : Bbb R$; Hardy Littlewood maximal function $h^*: Bbb R to [0, infty]$ defined by $h^*(x)=sup_{t>0}frac 1{2t}int_{x-t}^{x+t}|h|$.




      Now to prove $|h(x)|leq|h^*(x)|$ for almost every $x in Bbb R$




      I am thinking that I have to use Lebesgue differentiation theorem, first version that states "Suppose $fin L^{1}Bbb R$. Then



      $lim_{ t to 0}frac 1{2t}int_{x-t}^{x+t}|f-f(b)|=0$ for all most all $b in Bbb R$.



      My attempt: $||h(x)|-|h^*(x)||leq |h(x)-h^*(x)|=|h(x)-sup_{t>0}frac 1{2t}int_{x-t}^{x+t}|h(y)|dy|leq sup_{t>0}frac 1{2t}int_{x-t}^{x+t}|h(x)-h(y)|dy=0$ for almost every $x$. So, we get $|h(x)|leq h^*(x)$ for allmost every $x$.



      Is my attempt coorect or is there any other way to approach? Please help










      share|cite|improve this question













      Suppose $h: Bbb R to : Bbb R$; Hardy Littlewood maximal function $h^*: Bbb R to [0, infty]$ defined by $h^*(x)=sup_{t>0}frac 1{2t}int_{x-t}^{x+t}|h|$.




      Now to prove $|h(x)|leq|h^*(x)|$ for almost every $x in Bbb R$




      I am thinking that I have to use Lebesgue differentiation theorem, first version that states "Suppose $fin L^{1}Bbb R$. Then



      $lim_{ t to 0}frac 1{2t}int_{x-t}^{x+t}|f-f(b)|=0$ for all most all $b in Bbb R$.



      My attempt: $||h(x)|-|h^*(x)||leq |h(x)-h^*(x)|=|h(x)-sup_{t>0}frac 1{2t}int_{x-t}^{x+t}|h(y)|dy|leq sup_{t>0}frac 1{2t}int_{x-t}^{x+t}|h(x)-h(y)|dy=0$ for almost every $x$. So, we get $|h(x)|leq h^*(x)$ for allmost every $x$.



      Is my attempt coorect or is there any other way to approach? Please help







      real-analysis integration measure-theory lebesgue-integral lebesgue-measure






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      asked Nov 16 at 23:25









      Gimgim

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          This version of Lebesgue differentiation, which correctly stated is
          $$lim_{ t to 0}frac 1{2t}int_{x-t}^{x+t}|f-f(x)|=0 quad(text{a.e.} x)$$ immediately implies the weaker one ($frac1{2t}int_{x-t}^{x+t} f to f(x)$ a.e.) via triangle inequality
          $$ left| frac1{2t} int_{x-t}^{x+t} f(y) dy - f(x)right|le lim_{ t to 0}frac 1{2t}int_{x-t}^{x+t}|f-f(x)|to 0$$
          Therefore, for almost every $x$,
          $$ |f(x)| = lim_{tto 0}frac1{2t} int_{x-t}^{x+t } |f| le sup_{t> 0} frac1{2t} int_{x-t}^{x+t } |f| = f^*(x)$$






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            active

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            active

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            up vote
            0
            down vote



            accepted










            This version of Lebesgue differentiation, which correctly stated is
            $$lim_{ t to 0}frac 1{2t}int_{x-t}^{x+t}|f-f(x)|=0 quad(text{a.e.} x)$$ immediately implies the weaker one ($frac1{2t}int_{x-t}^{x+t} f to f(x)$ a.e.) via triangle inequality
            $$ left| frac1{2t} int_{x-t}^{x+t} f(y) dy - f(x)right|le lim_{ t to 0}frac 1{2t}int_{x-t}^{x+t}|f-f(x)|to 0$$
            Therefore, for almost every $x$,
            $$ |f(x)| = lim_{tto 0}frac1{2t} int_{x-t}^{x+t } |f| le sup_{t> 0} frac1{2t} int_{x-t}^{x+t } |f| = f^*(x)$$






            share|cite|improve this answer

























              up vote
              0
              down vote



              accepted










              This version of Lebesgue differentiation, which correctly stated is
              $$lim_{ t to 0}frac 1{2t}int_{x-t}^{x+t}|f-f(x)|=0 quad(text{a.e.} x)$$ immediately implies the weaker one ($frac1{2t}int_{x-t}^{x+t} f to f(x)$ a.e.) via triangle inequality
              $$ left| frac1{2t} int_{x-t}^{x+t} f(y) dy - f(x)right|le lim_{ t to 0}frac 1{2t}int_{x-t}^{x+t}|f-f(x)|to 0$$
              Therefore, for almost every $x$,
              $$ |f(x)| = lim_{tto 0}frac1{2t} int_{x-t}^{x+t } |f| le sup_{t> 0} frac1{2t} int_{x-t}^{x+t } |f| = f^*(x)$$






              share|cite|improve this answer























                up vote
                0
                down vote



                accepted







                up vote
                0
                down vote



                accepted






                This version of Lebesgue differentiation, which correctly stated is
                $$lim_{ t to 0}frac 1{2t}int_{x-t}^{x+t}|f-f(x)|=0 quad(text{a.e.} x)$$ immediately implies the weaker one ($frac1{2t}int_{x-t}^{x+t} f to f(x)$ a.e.) via triangle inequality
                $$ left| frac1{2t} int_{x-t}^{x+t} f(y) dy - f(x)right|le lim_{ t to 0}frac 1{2t}int_{x-t}^{x+t}|f-f(x)|to 0$$
                Therefore, for almost every $x$,
                $$ |f(x)| = lim_{tto 0}frac1{2t} int_{x-t}^{x+t } |f| le sup_{t> 0} frac1{2t} int_{x-t}^{x+t } |f| = f^*(x)$$






                share|cite|improve this answer












                This version of Lebesgue differentiation, which correctly stated is
                $$lim_{ t to 0}frac 1{2t}int_{x-t}^{x+t}|f-f(x)|=0 quad(text{a.e.} x)$$ immediately implies the weaker one ($frac1{2t}int_{x-t}^{x+t} f to f(x)$ a.e.) via triangle inequality
                $$ left| frac1{2t} int_{x-t}^{x+t} f(y) dy - f(x)right|le lim_{ t to 0}frac 1{2t}int_{x-t}^{x+t}|f-f(x)|to 0$$
                Therefore, for almost every $x$,
                $$ |f(x)| = lim_{tto 0}frac1{2t} int_{x-t}^{x+t } |f| le sup_{t> 0} frac1{2t} int_{x-t}^{x+t } |f| = f^*(x)$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 17 at 10:07









                Calvin Khor

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                10.7k21436






























                     

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