Sufficient condition for $L^infty$-convergence in martingale convergence theorem.
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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
edited Nov 26 at 14:51
Davide Giraudo
124k16150259
asked Nov 23 at 19:38
caffeinemachine
6,45521250
add a comment |
up vote
1
down vote
favorite
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
edited Nov 26 at 14:51
Davide Giraudo
124k16150259
asked Nov 23 at 19:38
caffeinemachine
6,45521250
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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
edited Nov 26 at 14:51
Davide Giraudo
124k16150259
asked Nov 23 at 19:38
caffeinemachine
6,45521250
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
edited Nov 26 at 14:51
Davide Giraudo
124k16150259
asked Nov 23 at 19:38
caffeinemachine
6,45521250
edited Nov 26 at 14:51
Davide Giraudo
124k16150259
asked Nov 23 at 19:38
caffeinemachine
6,45521250
edited Nov 26 at 14:51
Davide Giraudo
124k16150259
edited Nov 26 at 14:51
Davide Giraudo
124k16150259
edited Nov 26 at 14:51
Davide Giraudo
124k16150259
124k16150259
asked Nov 23 at 19:38
caffeinemachine
6,45521250
asked Nov 23 at 19:38
caffeinemachine
6,45521250
asked Nov 23 at 19:38
caffeinemachine
6,45521250
6,45521250
add a comment |
add a comment |
1 Answer
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oldest
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up vote
1
down vote
accepted
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
$$
leftlVert f-f_nrightrVert_inftygeqslant maxleft{sup_{xin J}leftlvert f(x)-f_n(x)rightrvert,sup_{xin J'}leftlvert f(x)-f_n(x)rightrvert
right}geqslant max{u,1-u}geqslant 1/2.
$$
answered Nov 26 at 10:04
Davide Giraudo
124k16150259
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
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
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
$$
leftlVert f-f_nrightrVert_inftygeqslant maxleft{sup_{xin J}leftlvert f(x)-f_n(x)rightrvert,sup_{xin J'}leftlvert f(x)-f_n(x)rightrvert
right}geqslant max{u,1-u}geqslant 1/2.
$$
answered Nov 26 at 10:04
Davide Giraudo
124k16150259
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
add a comment |
up vote
1
down vote
accepted
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
$$
leftlVert f-f_nrightrVert_inftygeqslant maxleft{sup_{xin J}leftlvert f(x)-f_n(x)rightrvert,sup_{xin J'}leftlvert f(x)-f_n(x)rightrvert
right}geqslant max{u,1-u}geqslant 1/2.
$$
answered Nov 26 at 10:04
Davide Giraudo
124k16150259
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
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
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
$$
leftlVert f-f_nrightrVert_inftygeqslant maxleft{sup_{xin J}leftlvert f(x)-f_n(x)rightrvert,sup_{xin J'}leftlvert f(x)-f_n(x)rightrvert
right}geqslant max{u,1-u}geqslant 1/2.
$$
answered Nov 26 at 10:04
Davide Giraudo
124k16150259
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
$$
leftlVert f-f_nrightrVert_inftygeqslant maxleft{sup_{xin J}leftlvert f(x)-f_n(x)rightrvert,sup_{xin J'}leftlvert f(x)-f_n(x)rightrvert
right}geqslant max{u,1-u}geqslant 1/2.
$$
answered Nov 26 at 10:04
Davide Giraudo
124k16150259
edited Dec 10 at 17:03
answered Nov 26 at 10:04
Davide Giraudo
124k16150259
answered Nov 26 at 10:04
Davide Giraudo
124k16150259
answered Nov 26 at 10:04
Davide Giraudo
124k16150259
124k16150259
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
add a comment |
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
add a comment |
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