$fin L^1$ , $F_h(x) = frac{1}{2h} int_{-h}^h f(x-y)dlambda(y)$ is absolutely continuous












2












$begingroup$


$fin L^1(mathbb{R})$ I wish to show that $F_h(x) = frac{1}{2h} int_{-h}^h f(x-y)dlambda(y)$ is absolutely continuous over any close interval $[a,b]$



What I tried so far:



$|F_h(d)-F_h(c)| = |frac{1}{2h}int_{-h}^hf(d-y)-f(c-y)dlambda(y)|
=$



$=|frac{1}{2h}int_{h}^{-h}f(d+y)-f(c+y)dlambda(y)| $



I changed integration order just for convenience.
Due to translation invariance we can shift by $c$ and get:
$|frac{1}{2h}int_{h+c}^{-h+c}f((d-c)+y)-f(y)dlambda(y)|$
$le frac{1}{2h}int_{h+c}^{-h+c}|f((d-c)+y)-f(y)dlambda(y)|$



Using the translation continuity for $f$ : $|| f(y+delta)-f(y)||_{L^1}rightarrow0$ , we get that for $varepsilon >0$ exists $delta >0$ such that for $d-c<delta frac{1}{2h}int_{h+c}^{-h+c}|f((d-c)+y)-f(y)dlambda(y)|<varepsilon/2h$.



However when taking ${(c_i,d_i)}_{i=1} ^{i=n}$ interval is $[a,b]$ such that $sum_{i=1}^{i=n}(d_i-c_i) <delta$ we get $sum_{i=1}^{i=n}|F(d_i)-F(c_i)| < frac{1}{2h}varepsiloncdot n$. So it doesn't respond to absolutely continuous definition.



I am not so sure that this is the right approach for this problem, I even have a strong feeling that it should be more simple, but I don't find how to solve it.






real-analysis absolute-continuity







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    $fin L^1(mathbb{R})$ I wish to show that $F_h(x) = frac{1}{2h} int_{-h}^h f(x-y)dlambda(y)$ is absolutely continuous over any close interval $[a,b]$



    What I tried so far:



    $|F_h(d)-F_h(c)| = |frac{1}{2h}int_{-h}^hf(d-y)-f(c-y)dlambda(y)|
    =$



    $=|frac{1}{2h}int_{h}^{-h}f(d+y)-f(c+y)dlambda(y)| $



    I changed integration order just for convenience.
    Due to translation invariance we can shift by $c$ and get:
    $|frac{1}{2h}int_{h+c}^{-h+c}f((d-c)+y)-f(y)dlambda(y)|$
    $le frac{1}{2h}int_{h+c}^{-h+c}|f((d-c)+y)-f(y)dlambda(y)|$



    Using the translation continuity for $f$ : $|| f(y+delta)-f(y)||_{L^1}rightarrow0$ , we get that for $varepsilon >0$ exists $delta >0$ such that for $d-c<delta frac{1}{2h}int_{h+c}^{-h+c}|f((d-c)+y)-f(y)dlambda(y)|<varepsilon/2h$.



    However when taking ${(c_i,d_i)}_{i=1} ^{i=n}$ interval is $[a,b]$ such that $sum_{i=1}^{i=n}(d_i-c_i) <delta$ we get $sum_{i=1}^{i=n}|F(d_i)-F(c_i)| < frac{1}{2h}varepsiloncdot n$. So it doesn't respond to absolutely continuous definition.



    I am not so sure that this is the right approach for this problem, I even have a strong feeling that it should be more simple, but I don't find how to solve it.






    real-analysis absolute-continuity







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      $fin L^1(mathbb{R})$ I wish to show that $F_h(x) = frac{1}{2h} int_{-h}^h f(x-y)dlambda(y)$ is absolutely continuous over any close interval $[a,b]$



      What I tried so far:



      $|F_h(d)-F_h(c)| = |frac{1}{2h}int_{-h}^hf(d-y)-f(c-y)dlambda(y)|
      =$



      $=|frac{1}{2h}int_{h}^{-h}f(d+y)-f(c+y)dlambda(y)| $



      I changed integration order just for convenience.
      Due to translation invariance we can shift by $c$ and get:
      $|frac{1}{2h}int_{h+c}^{-h+c}f((d-c)+y)-f(y)dlambda(y)|$
      $le frac{1}{2h}int_{h+c}^{-h+c}|f((d-c)+y)-f(y)dlambda(y)|$



      Using the translation continuity for $f$ : $|| f(y+delta)-f(y)||_{L^1}rightarrow0$ , we get that for $varepsilon >0$ exists $delta >0$ such that for $d-c<delta frac{1}{2h}int_{h+c}^{-h+c}|f((d-c)+y)-f(y)dlambda(y)|<varepsilon/2h$.



      However when taking ${(c_i,d_i)}_{i=1} ^{i=n}$ interval is $[a,b]$ such that $sum_{i=1}^{i=n}(d_i-c_i) <delta$ we get $sum_{i=1}^{i=n}|F(d_i)-F(c_i)| < frac{1}{2h}varepsiloncdot n$. So it doesn't respond to absolutely continuous definition.



      I am not so sure that this is the right approach for this problem, I even have a strong feeling that it should be more simple, but I don't find how to solve it.






      real-analysis absolute-continuity







      share|cite|improve this question









      $endgroup$




      $fin L^1(mathbb{R})$ I wish to show that $F_h(x) = frac{1}{2h} int_{-h}^h f(x-y)dlambda(y)$ is absolutely continuous over any close interval $[a,b]$



      What I tried so far:



      $|F_h(d)-F_h(c)| = |frac{1}{2h}int_{-h}^hf(d-y)-f(c-y)dlambda(y)|
      =$



      $=|frac{1}{2h}int_{h}^{-h}f(d+y)-f(c+y)dlambda(y)| $



      I changed integration order just for convenience.
      Due to translation invariance we can shift by $c$ and get:
      $|frac{1}{2h}int_{h+c}^{-h+c}f((d-c)+y)-f(y)dlambda(y)|$
      $le frac{1}{2h}int_{h+c}^{-h+c}|f((d-c)+y)-f(y)dlambda(y)|$



      Using the translation continuity for $f$ : $|| f(y+delta)-f(y)||_{L^1}rightarrow0$ , we get that for $varepsilon >0$ exists $delta >0$ such that for $d-c<delta frac{1}{2h}int_{h+c}^{-h+c}|f((d-c)+y)-f(y)dlambda(y)|<varepsilon/2h$.



      However when taking ${(c_i,d_i)}_{i=1} ^{i=n}$ interval is $[a,b]$ such that $sum_{i=1}^{i=n}(d_i-c_i) <delta$ we get $sum_{i=1}^{i=n}|F(d_i)-F(c_i)| < frac{1}{2h}varepsiloncdot n$. So it doesn't respond to absolutely continuous definition.



      I am not so sure that this is the right approach for this problem, I even have a strong feeling that it should be more simple, but I don't find how to solve it.






      real-analysis absolute-continuity




      real-analysis absolute-continuity






      share|cite|improve this question













      share|cite|improve this question











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      asked Feb 2 at 8:31









      dandan

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      623613






















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          $begingroup$

          The following characterization might simplify the problem:




          A function $u:[a,b]to mathbb{R}$ is absolutely continuous if and only if there is a function $gin L^1(a,b)$ and a real constant $cin mathbb{R}$ such that
          $$int_a^x g(t),dt=u(x)+cqquad xin [a,b] $$




          Here you have
          begin{align*}u(x)=frac{1}{2h}int_{-h}^hf(x-y),dy=-frac{1}{2h}int_{x-h}^{x+h}f(y),dy
          end{align*}
          Heuristically, if $f$ were continuous then by the fundamental theorem of calculus we could say
          begin{align*}frac{mathrm{d}}{mathrm{d}x}left[-frac{1}{2h}int_{x-h}^{x+h}f(y),dyright]=-frac{1}{2h}left(f(x+h)-f(x-h)right)qquad xin [a,b]end{align*}
          This hints that a correct choice of $g$ is
          $$g(x):= -frac{1}{2h}left(f(x+h)-f(x-h)right)qquad xin [a,b]$$
          Clearly, $gin L^1(mathbb{R})$. Moreover,
          begin{align*}int_a^xg(t),dt&=-frac{1}{2h}int_a^x[f(t+h)-f(t-h)],dt=-frac{1}{2h}left[int_{a+h}^{x+h}f(t),dt-int_{a-h}^{x-h}f(t),dtright]=\
          &=-frac{1}{2h}left[c+int_{x-h}^{x+h}f(t),dtright]=-frac{c}{2h}+frac{1}{2h}int_{-h}^{h}f(x-t),dt
          end{align*}
          Where $c:=int_{a-h}^{a+h}f(t),dt$ is a real constant depending only on $h$ (and not on $x$). This proves that $u$ is absolutely continuous.






          share|cite|improve this answer











          $endgroup$














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            $begingroup$

            The following characterization might simplify the problem:




            A function $u:[a,b]to mathbb{R}$ is absolutely continuous if and only if there is a function $gin L^1(a,b)$ and a real constant $cin mathbb{R}$ such that
            $$int_a^x g(t),dt=u(x)+cqquad xin [a,b] $$




            Here you have
            begin{align*}u(x)=frac{1}{2h}int_{-h}^hf(x-y),dy=-frac{1}{2h}int_{x-h}^{x+h}f(y),dy
            end{align*}
            Heuristically, if $f$ were continuous then by the fundamental theorem of calculus we could say
            begin{align*}frac{mathrm{d}}{mathrm{d}x}left[-frac{1}{2h}int_{x-h}^{x+h}f(y),dyright]=-frac{1}{2h}left(f(x+h)-f(x-h)right)qquad xin [a,b]end{align*}
            This hints that a correct choice of $g$ is
            $$g(x):= -frac{1}{2h}left(f(x+h)-f(x-h)right)qquad xin [a,b]$$
            Clearly, $gin L^1(mathbb{R})$. Moreover,
            begin{align*}int_a^xg(t),dt&=-frac{1}{2h}int_a^x[f(t+h)-f(t-h)],dt=-frac{1}{2h}left[int_{a+h}^{x+h}f(t),dt-int_{a-h}^{x-h}f(t),dtright]=\
            &=-frac{1}{2h}left[c+int_{x-h}^{x+h}f(t),dtright]=-frac{c}{2h}+frac{1}{2h}int_{-h}^{h}f(x-t),dt
            end{align*}
            Where $c:=int_{a-h}^{a+h}f(t),dt$ is a real constant depending only on $h$ (and not on $x$). This proves that $u$ is absolutely continuous.






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              The following characterization might simplify the problem:




              A function $u:[a,b]to mathbb{R}$ is absolutely continuous if and only if there is a function $gin L^1(a,b)$ and a real constant $cin mathbb{R}$ such that
              $$int_a^x g(t),dt=u(x)+cqquad xin [a,b] $$




              Here you have
              begin{align*}u(x)=frac{1}{2h}int_{-h}^hf(x-y),dy=-frac{1}{2h}int_{x-h}^{x+h}f(y),dy
              end{align*}
              Heuristically, if $f$ were continuous then by the fundamental theorem of calculus we could say
              begin{align*}frac{mathrm{d}}{mathrm{d}x}left[-frac{1}{2h}int_{x-h}^{x+h}f(y),dyright]=-frac{1}{2h}left(f(x+h)-f(x-h)right)qquad xin [a,b]end{align*}
              This hints that a correct choice of $g$ is
              $$g(x):= -frac{1}{2h}left(f(x+h)-f(x-h)right)qquad xin [a,b]$$
              Clearly, $gin L^1(mathbb{R})$. Moreover,
              begin{align*}int_a^xg(t),dt&=-frac{1}{2h}int_a^x[f(t+h)-f(t-h)],dt=-frac{1}{2h}left[int_{a+h}^{x+h}f(t),dt-int_{a-h}^{x-h}f(t),dtright]=\
              &=-frac{1}{2h}left[c+int_{x-h}^{x+h}f(t),dtright]=-frac{c}{2h}+frac{1}{2h}int_{-h}^{h}f(x-t),dt
              end{align*}
              Where $c:=int_{a-h}^{a+h}f(t),dt$ is a real constant depending only on $h$ (and not on $x$). This proves that $u$ is absolutely continuous.






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                The following characterization might simplify the problem:




                A function $u:[a,b]to mathbb{R}$ is absolutely continuous if and only if there is a function $gin L^1(a,b)$ and a real constant $cin mathbb{R}$ such that
                $$int_a^x g(t),dt=u(x)+cqquad xin [a,b] $$




                Here you have
                begin{align*}u(x)=frac{1}{2h}int_{-h}^hf(x-y),dy=-frac{1}{2h}int_{x-h}^{x+h}f(y),dy
                end{align*}
                Heuristically, if $f$ were continuous then by the fundamental theorem of calculus we could say
                begin{align*}frac{mathrm{d}}{mathrm{d}x}left[-frac{1}{2h}int_{x-h}^{x+h}f(y),dyright]=-frac{1}{2h}left(f(x+h)-f(x-h)right)qquad xin [a,b]end{align*}
                This hints that a correct choice of $g$ is
                $$g(x):= -frac{1}{2h}left(f(x+h)-f(x-h)right)qquad xin [a,b]$$
                Clearly, $gin L^1(mathbb{R})$. Moreover,
                begin{align*}int_a^xg(t),dt&=-frac{1}{2h}int_a^x[f(t+h)-f(t-h)],dt=-frac{1}{2h}left[int_{a+h}^{x+h}f(t),dt-int_{a-h}^{x-h}f(t),dtright]=\
                &=-frac{1}{2h}left[c+int_{x-h}^{x+h}f(t),dtright]=-frac{c}{2h}+frac{1}{2h}int_{-h}^{h}f(x-t),dt
                end{align*}
                Where $c:=int_{a-h}^{a+h}f(t),dt$ is a real constant depending only on $h$ (and not on $x$). This proves that $u$ is absolutely continuous.






                share|cite|improve this answer











                $endgroup$



                The following characterization might simplify the problem:




                A function $u:[a,b]to mathbb{R}$ is absolutely continuous if and only if there is a function $gin L^1(a,b)$ and a real constant $cin mathbb{R}$ such that
                $$int_a^x g(t),dt=u(x)+cqquad xin [a,b] $$




                Here you have
                begin{align*}u(x)=frac{1}{2h}int_{-h}^hf(x-y),dy=-frac{1}{2h}int_{x-h}^{x+h}f(y),dy
                end{align*}
                Heuristically, if $f$ were continuous then by the fundamental theorem of calculus we could say
                begin{align*}frac{mathrm{d}}{mathrm{d}x}left[-frac{1}{2h}int_{x-h}^{x+h}f(y),dyright]=-frac{1}{2h}left(f(x+h)-f(x-h)right)qquad xin [a,b]end{align*}
                This hints that a correct choice of $g$ is
                $$g(x):= -frac{1}{2h}left(f(x+h)-f(x-h)right)qquad xin [a,b]$$
                Clearly, $gin L^1(mathbb{R})$. Moreover,
                begin{align*}int_a^xg(t),dt&=-frac{1}{2h}int_a^x[f(t+h)-f(t-h)],dt=-frac{1}{2h}left[int_{a+h}^{x+h}f(t),dt-int_{a-h}^{x-h}f(t),dtright]=\
                &=-frac{1}{2h}left[c+int_{x-h}^{x+h}f(t),dtright]=-frac{c}{2h}+frac{1}{2h}int_{-h}^{h}f(x-t),dt
                end{align*}
                Where $c:=int_{a-h}^{a+h}f(t),dt$ is a real constant depending only on $h$ (and not on $x$). This proves that $u$ is absolutely continuous.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Feb 2 at 9:45

























                answered Feb 2 at 9:36









                Lorenzo QuarisaLorenzo Quarisa

                3,765623




                3,765623






























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