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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          Puebla de Zaragoza

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          Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
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          Musa

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          Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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