Sufficient condition for $L^infty$-convergence in martingale convergence theorem.












1














Let $I$ denote the unit interval and $f:Ito I$ be a Borel measurable function.
For each $n$ let $P_n$ denote the partition of $I$ defined as
$$P_n= left{left[0, frac{1}{2^n}right), left[frac{1}{2^n}, frac{2}{2^n}right), left[frac{2}{2^n}, frac{3}{2^n}right), ldots, left[frac{2^n-2}{2^n}, frac{2^n-1}{2^n}right],left[frac{2^n-1}{2^n}, 1right]right}$$
and let $mathcal A_n$ be the $sigma$-algebra on $I$ generated by $P_n$.
Define $f_n=E[f|mathcal A_n]$, where the conditional expectation is with respect to the Lebesgue measure.



It is known by the martingale convergence theorem that $f_nto f$ pointwise a.e. and also in $L^1$.




Question. Are there some (nice) sufficient conditions on the nature of $f$ which ensure that the convergence above is in $L^infty$ also?







probability-theory measure-theory convergence lp-spaces martingales







share|cite|improve this question





























    1














    Let $I$ denote the unit interval and $f:Ito I$ be a Borel measurable function.
    For each $n$ let $P_n$ denote the partition of $I$ defined as
    $$P_n= left{left[0, frac{1}{2^n}right), left[frac{1}{2^n}, frac{2}{2^n}right), left[frac{2}{2^n}, frac{3}{2^n}right), ldots, left[frac{2^n-2}{2^n}, frac{2^n-1}{2^n}right],left[frac{2^n-1}{2^n}, 1right]right}$$
    and let $mathcal A_n$ be the $sigma$-algebra on $I$ generated by $P_n$.
    Define $f_n=E[f|mathcal A_n]$, where the conditional expectation is with respect to the Lebesgue measure.



    It is known by the martingale convergence theorem that $f_nto f$ pointwise a.e. and also in $L^1$.




    Question. Are there some (nice) sufficient conditions on the nature of $f$ which ensure that the convergence above is in $L^infty$ also?







    probability-theory measure-theory convergence lp-spaces martingales







    share|cite|improve this question



























      1












      1








      1


      1





      Let $I$ denote the unit interval and $f:Ito I$ be a Borel measurable function.
      For each $n$ let $P_n$ denote the partition of $I$ defined as
      $$P_n= left{left[0, frac{1}{2^n}right), left[frac{1}{2^n}, frac{2}{2^n}right), left[frac{2}{2^n}, frac{3}{2^n}right), ldots, left[frac{2^n-2}{2^n}, frac{2^n-1}{2^n}right],left[frac{2^n-1}{2^n}, 1right]right}$$
      and let $mathcal A_n$ be the $sigma$-algebra on $I$ generated by $P_n$.
      Define $f_n=E[f|mathcal A_n]$, where the conditional expectation is with respect to the Lebesgue measure.



      It is known by the martingale convergence theorem that $f_nto f$ pointwise a.e. and also in $L^1$.




      Question. Are there some (nice) sufficient conditions on the nature of $f$ which ensure that the convergence above is in $L^infty$ also?







      probability-theory measure-theory convergence lp-spaces martingales







      share|cite|improve this question















      Let $I$ denote the unit interval and $f:Ito I$ be a Borel measurable function.
      For each $n$ let $P_n$ denote the partition of $I$ defined as
      $$P_n= left{left[0, frac{1}{2^n}right), left[frac{1}{2^n}, frac{2}{2^n}right), left[frac{2}{2^n}, frac{3}{2^n}right), ldots, left[frac{2^n-2}{2^n}, frac{2^n-1}{2^n}right],left[frac{2^n-1}{2^n}, 1right]right}$$
      and let $mathcal A_n$ be the $sigma$-algebra on $I$ generated by $P_n$.
      Define $f_n=E[f|mathcal A_n]$, where the conditional expectation is with respect to the Lebesgue measure.



      It is known by the martingale convergence theorem that $f_nto f$ pointwise a.e. and also in $L^1$.




      Question. Are there some (nice) sufficient conditions on the nature of $f$ which ensure that the convergence above is in $L^infty$ also?







      probability-theory measure-theory convergence lp-spaces martingales




      probability-theory measure-theory convergence lp-spaces martingales






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 26 at 14:51









      Davide Giraudo

      125k16150259




      125k16150259










      asked Nov 23 at 19:38









      caffeinemachine

      6,49121250




      6,49121250






















          1 Answer
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          Denote by $I_{n,j}$ the interval $left[j2^{-n},(j+1)2^{-n}right)$. Then
          $$
          f_n(x)=2^nsum_{j=0}^{2^n-1}mathbb Eleft[fmathbf 1_{I_{n,j}}right]mathbf 1_{I_{n,j}}(x), xin I.
          $$
          If $f$ is continuous on $[0,1]$, then for all $xin I_{n,j}$, $$
          2^nmathbb Eleft[fmathbf 1_{I_{n,j}}right]-f(x)=frac1{2^{-n}}int_{j2^{-n}}^{(j+1)2^{-n}}left(f(t)-f(x)right)dt$$
          hence
          $$
          leftlvert f(x)-f_n(x)rightrvertleqslant sup_{substack{s,tin [0,1]\ leftlvert t-srightrvertleqslant 2^{-n}}}leftlvert f(t)-f(s)rightrvert
          $$
          and uniform continuity allows to conclude.



          However, in general, we cannot expect a convergence in $mathbb L^infty$ in the general case. Let $t$ be an element of $I$ such that for all $ngeqslant 1$ and $0leqslant jleqslant 2^n-1$, $xneq j2^{-n} $ (for example $t$ irrational).
          Let $n$ be fixed; then for some $j_0inleft{0,dots,2^n-1right}$, $tin I_{n.j_0}$. Therefore, letting $f=mathbf 1_{[0,t]}$, we have
          $$
          f_n(x)=mathbf 1_{left[0,j_02^{-n}right)}(x)+2^nleft(t-j_02^{-n}right)mathbf 1_{I_{n,j_0}}(x), xin I.$$
          Let $J$ and $J'$ be the intervals defined respectively by $J:=left[j_02^{-n},tright)$ and $J':=left[t, (j_0+1)2^{-n}right)$. By the assumption on $t$, these intervals have positive measure and letting $u:= 2^n left(t-j_02^{-n}right)$, it follows that
          $$
          leftlVert f-f_nrightrVert_inftygeqslant maxleft{sup_{xin J}leftlvert f(x)-f_n(x)rightrvert,sup_{xin J'}leftlvert f(x)-f_n(x)rightrvert
          right}geqslant max{u,1-u}geqslant 1/2.
          $$






          share|cite|improve this answer























          • In the display equation after "If $f$ is continuous on $[0, 1]$, then" I think the RHS is supposed to be $2^nsup_{s, tin [0, 1], |t-s|leq 1/2^n} |f(t)-f(s)|$. If this is indeed the case, then one cannot appeal to the uniform continuity of $f$ to argue that the $L^infty$ convergence occurs.
            – caffeinemachine
            Dec 10 at 13:03










          • @caffeinemachine There was a typo, but now I think it is correct.
            – Davide Giraudo
            Dec 10 at 17:04










          • I see. Thanks. It looks correct now.
            – caffeinemachine
            Dec 11 at 4:46










          • Actually I just realized that it was also correct before since $2^{-n}leq 1/n$, but not accurate.
            – Davide Giraudo
            Dec 11 at 9:22











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          Denote by $I_{n,j}$ the interval $left[j2^{-n},(j+1)2^{-n}right)$. Then
          $$
          f_n(x)=2^nsum_{j=0}^{2^n-1}mathbb Eleft[fmathbf 1_{I_{n,j}}right]mathbf 1_{I_{n,j}}(x), xin I.
          $$
          If $f$ is continuous on $[0,1]$, then for all $xin I_{n,j}$, $$
          2^nmathbb Eleft[fmathbf 1_{I_{n,j}}right]-f(x)=frac1{2^{-n}}int_{j2^{-n}}^{(j+1)2^{-n}}left(f(t)-f(x)right)dt$$
          hence
          $$
          leftlvert f(x)-f_n(x)rightrvertleqslant sup_{substack{s,tin [0,1]\ leftlvert t-srightrvertleqslant 2^{-n}}}leftlvert f(t)-f(s)rightrvert
          $$
          and uniform continuity allows to conclude.



          However, in general, we cannot expect a convergence in $mathbb L^infty$ in the general case. Let $t$ be an element of $I$ such that for all $ngeqslant 1$ and $0leqslant jleqslant 2^n-1$, $xneq j2^{-n} $ (for example $t$ irrational).
          Let $n$ be fixed; then for some $j_0inleft{0,dots,2^n-1right}$, $tin I_{n.j_0}$. Therefore, letting $f=mathbf 1_{[0,t]}$, we have
          $$
          f_n(x)=mathbf 1_{left[0,j_02^{-n}right)}(x)+2^nleft(t-j_02^{-n}right)mathbf 1_{I_{n,j_0}}(x), xin I.$$
          Let $J$ and $J'$ be the intervals defined respectively by $J:=left[j_02^{-n},tright)$ and $J':=left[t, (j_0+1)2^{-n}right)$. By the assumption on $t$, these intervals have positive measure and letting $u:= 2^n left(t-j_02^{-n}right)$, it follows that
          $$
          leftlVert f-f_nrightrVert_inftygeqslant maxleft{sup_{xin J}leftlvert f(x)-f_n(x)rightrvert,sup_{xin J'}leftlvert f(x)-f_n(x)rightrvert
          right}geqslant max{u,1-u}geqslant 1/2.
          $$






          share|cite|improve this answer























          • In the display equation after "If $f$ is continuous on $[0, 1]$, then" I think the RHS is supposed to be $2^nsup_{s, tin [0, 1], |t-s|leq 1/2^n} |f(t)-f(s)|$. If this is indeed the case, then one cannot appeal to the uniform continuity of $f$ to argue that the $L^infty$ convergence occurs.
            – caffeinemachine
            Dec 10 at 13:03










          • @caffeinemachine There was a typo, but now I think it is correct.
            – Davide Giraudo
            Dec 10 at 17:04










          • I see. Thanks. It looks correct now.
            – caffeinemachine
            Dec 11 at 4:46










          • Actually I just realized that it was also correct before since $2^{-n}leq 1/n$, but not accurate.
            – Davide Giraudo
            Dec 11 at 9:22
















          1














          Denote by $I_{n,j}$ the interval $left[j2^{-n},(j+1)2^{-n}right)$. Then
          $$
          f_n(x)=2^nsum_{j=0}^{2^n-1}mathbb Eleft[fmathbf 1_{I_{n,j}}right]mathbf 1_{I_{n,j}}(x), xin I.
          $$
          If $f$ is continuous on $[0,1]$, then for all $xin I_{n,j}$, $$
          2^nmathbb Eleft[fmathbf 1_{I_{n,j}}right]-f(x)=frac1{2^{-n}}int_{j2^{-n}}^{(j+1)2^{-n}}left(f(t)-f(x)right)dt$$
          hence
          $$
          leftlvert f(x)-f_n(x)rightrvertleqslant sup_{substack{s,tin [0,1]\ leftlvert t-srightrvertleqslant 2^{-n}}}leftlvert f(t)-f(s)rightrvert
          $$
          and uniform continuity allows to conclude.



          However, in general, we cannot expect a convergence in $mathbb L^infty$ in the general case. Let $t$ be an element of $I$ such that for all $ngeqslant 1$ and $0leqslant jleqslant 2^n-1$, $xneq j2^{-n} $ (for example $t$ irrational).
          Let $n$ be fixed; then for some $j_0inleft{0,dots,2^n-1right}$, $tin I_{n.j_0}$. Therefore, letting $f=mathbf 1_{[0,t]}$, we have
          $$
          f_n(x)=mathbf 1_{left[0,j_02^{-n}right)}(x)+2^nleft(t-j_02^{-n}right)mathbf 1_{I_{n,j_0}}(x), xin I.$$
          Let $J$ and $J'$ be the intervals defined respectively by $J:=left[j_02^{-n},tright)$ and $J':=left[t, (j_0+1)2^{-n}right)$. By the assumption on $t$, these intervals have positive measure and letting $u:= 2^n left(t-j_02^{-n}right)$, it follows that
          $$
          leftlVert f-f_nrightrVert_inftygeqslant maxleft{sup_{xin J}leftlvert f(x)-f_n(x)rightrvert,sup_{xin J'}leftlvert f(x)-f_n(x)rightrvert
          right}geqslant max{u,1-u}geqslant 1/2.
          $$






          share|cite|improve this answer























          • In the display equation after "If $f$ is continuous on $[0, 1]$, then" I think the RHS is supposed to be $2^nsup_{s, tin [0, 1], |t-s|leq 1/2^n} |f(t)-f(s)|$. If this is indeed the case, then one cannot appeal to the uniform continuity of $f$ to argue that the $L^infty$ convergence occurs.
            – caffeinemachine
            Dec 10 at 13:03










          • @caffeinemachine There was a typo, but now I think it is correct.
            – Davide Giraudo
            Dec 10 at 17:04










          • I see. Thanks. It looks correct now.
            – caffeinemachine
            Dec 11 at 4:46










          • Actually I just realized that it was also correct before since $2^{-n}leq 1/n$, but not accurate.
            – Davide Giraudo
            Dec 11 at 9:22














          1












          1








          1






          Denote by $I_{n,j}$ the interval $left[j2^{-n},(j+1)2^{-n}right)$. Then
          $$
          f_n(x)=2^nsum_{j=0}^{2^n-1}mathbb Eleft[fmathbf 1_{I_{n,j}}right]mathbf 1_{I_{n,j}}(x), xin I.
          $$
          If $f$ is continuous on $[0,1]$, then for all $xin I_{n,j}$, $$
          2^nmathbb Eleft[fmathbf 1_{I_{n,j}}right]-f(x)=frac1{2^{-n}}int_{j2^{-n}}^{(j+1)2^{-n}}left(f(t)-f(x)right)dt$$
          hence
          $$
          leftlvert f(x)-f_n(x)rightrvertleqslant sup_{substack{s,tin [0,1]\ leftlvert t-srightrvertleqslant 2^{-n}}}leftlvert f(t)-f(s)rightrvert
          $$
          and uniform continuity allows to conclude.



          However, in general, we cannot expect a convergence in $mathbb L^infty$ in the general case. Let $t$ be an element of $I$ such that for all $ngeqslant 1$ and $0leqslant jleqslant 2^n-1$, $xneq j2^{-n} $ (for example $t$ irrational).
          Let $n$ be fixed; then for some $j_0inleft{0,dots,2^n-1right}$, $tin I_{n.j_0}$. Therefore, letting $f=mathbf 1_{[0,t]}$, we have
          $$
          f_n(x)=mathbf 1_{left[0,j_02^{-n}right)}(x)+2^nleft(t-j_02^{-n}right)mathbf 1_{I_{n,j_0}}(x), xin I.$$
          Let $J$ and $J'$ be the intervals defined respectively by $J:=left[j_02^{-n},tright)$ and $J':=left[t, (j_0+1)2^{-n}right)$. By the assumption on $t$, these intervals have positive measure and letting $u:= 2^n left(t-j_02^{-n}right)$, it follows that
          $$
          leftlVert f-f_nrightrVert_inftygeqslant maxleft{sup_{xin J}leftlvert f(x)-f_n(x)rightrvert,sup_{xin J'}leftlvert f(x)-f_n(x)rightrvert
          right}geqslant max{u,1-u}geqslant 1/2.
          $$






          share|cite|improve this answer














          Denote by $I_{n,j}$ the interval $left[j2^{-n},(j+1)2^{-n}right)$. Then
          $$
          f_n(x)=2^nsum_{j=0}^{2^n-1}mathbb Eleft[fmathbf 1_{I_{n,j}}right]mathbf 1_{I_{n,j}}(x), xin I.
          $$
          If $f$ is continuous on $[0,1]$, then for all $xin I_{n,j}$, $$
          2^nmathbb Eleft[fmathbf 1_{I_{n,j}}right]-f(x)=frac1{2^{-n}}int_{j2^{-n}}^{(j+1)2^{-n}}left(f(t)-f(x)right)dt$$
          hence
          $$
          leftlvert f(x)-f_n(x)rightrvertleqslant sup_{substack{s,tin [0,1]\ leftlvert t-srightrvertleqslant 2^{-n}}}leftlvert f(t)-f(s)rightrvert
          $$
          and uniform continuity allows to conclude.



          However, in general, we cannot expect a convergence in $mathbb L^infty$ in the general case. Let $t$ be an element of $I$ such that for all $ngeqslant 1$ and $0leqslant jleqslant 2^n-1$, $xneq j2^{-n} $ (for example $t$ irrational).
          Let $n$ be fixed; then for some $j_0inleft{0,dots,2^n-1right}$, $tin I_{n.j_0}$. Therefore, letting $f=mathbf 1_{[0,t]}$, we have
          $$
          f_n(x)=mathbf 1_{left[0,j_02^{-n}right)}(x)+2^nleft(t-j_02^{-n}right)mathbf 1_{I_{n,j_0}}(x), xin I.$$
          Let $J$ and $J'$ be the intervals defined respectively by $J:=left[j_02^{-n},tright)$ and $J':=left[t, (j_0+1)2^{-n}right)$. By the assumption on $t$, these intervals have positive measure and letting $u:= 2^n left(t-j_02^{-n}right)$, it follows that
          $$
          leftlVert f-f_nrightrVert_inftygeqslant maxleft{sup_{xin J}leftlvert f(x)-f_n(x)rightrvert,sup_{xin J'}leftlvert f(x)-f_n(x)rightrvert
          right}geqslant max{u,1-u}geqslant 1/2.
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 10 at 17:03

























          answered Nov 26 at 10:04









          Davide Giraudo

          125k16150259




          125k16150259












          • In the display equation after "If $f$ is continuous on $[0, 1]$, then" I think the RHS is supposed to be $2^nsup_{s, tin [0, 1], |t-s|leq 1/2^n} |f(t)-f(s)|$. If this is indeed the case, then one cannot appeal to the uniform continuity of $f$ to argue that the $L^infty$ convergence occurs.
            – caffeinemachine
            Dec 10 at 13:03










          • @caffeinemachine There was a typo, but now I think it is correct.
            – Davide Giraudo
            Dec 10 at 17:04










          • I see. Thanks. It looks correct now.
            – caffeinemachine
            Dec 11 at 4:46










          • Actually I just realized that it was also correct before since $2^{-n}leq 1/n$, but not accurate.
            – Davide Giraudo
            Dec 11 at 9:22


















          • In the display equation after "If $f$ is continuous on $[0, 1]$, then" I think the RHS is supposed to be $2^nsup_{s, tin [0, 1], |t-s|leq 1/2^n} |f(t)-f(s)|$. If this is indeed the case, then one cannot appeal to the uniform continuity of $f$ to argue that the $L^infty$ convergence occurs.
            – caffeinemachine
            Dec 10 at 13:03










          • @caffeinemachine There was a typo, but now I think it is correct.
            – Davide Giraudo
            Dec 10 at 17:04










          • I see. Thanks. It looks correct now.
            – caffeinemachine
            Dec 11 at 4:46










          • Actually I just realized that it was also correct before since $2^{-n}leq 1/n$, but not accurate.
            – Davide Giraudo
            Dec 11 at 9:22
















          In the display equation after "If $f$ is continuous on $[0, 1]$, then" I think the RHS is supposed to be $2^nsup_{s, tin [0, 1], |t-s|leq 1/2^n} |f(t)-f(s)|$. If this is indeed the case, then one cannot appeal to the uniform continuity of $f$ to argue that the $L^infty$ convergence occurs.
          – caffeinemachine
          Dec 10 at 13:03




          In the display equation after "If $f$ is continuous on $[0, 1]$, then" I think the RHS is supposed to be $2^nsup_{s, tin [0, 1], |t-s|leq 1/2^n} |f(t)-f(s)|$. If this is indeed the case, then one cannot appeal to the uniform continuity of $f$ to argue that the $L^infty$ convergence occurs.
          – caffeinemachine
          Dec 10 at 13:03












          @caffeinemachine There was a typo, but now I think it is correct.
          – Davide Giraudo
          Dec 10 at 17:04




          @caffeinemachine There was a typo, but now I think it is correct.
          – Davide Giraudo
          Dec 10 at 17:04












          I see. Thanks. It looks correct now.
          – caffeinemachine
          Dec 11 at 4:46




          I see. Thanks. It looks correct now.
          – caffeinemachine
          Dec 11 at 4:46












          Actually I just realized that it was also correct before since $2^{-n}leq 1/n$, but not accurate.
          – Davide Giraudo
          Dec 11 at 9:22




          Actually I just realized that it was also correct before since $2^{-n}leq 1/n$, but not accurate.
          – Davide Giraudo
          Dec 11 at 9:22


















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          In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

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          .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
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          How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

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          1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
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