Mixing
Almost everywhere bounded function
$begingroup$
Let $psi$ be measurable function such that for all $ f in L^{1}(mu)$, $ fpsi in L^{1}(mu)$. Then $psi$ is bounded almost everywhere(in sense of measure $mu$).
My attempt:
Use contraposition: if $psi $ isn't bounded everywhere then $forall M>0$, there exists a set $A$ such that $mu (A)>0 $ and $ psi(x) > M , forall x in A$. Now i need to prove that there is an $f in L^{1} $ such that $ fpsi notin L^{1} $. But I don't know how.
Any help or hint would be appreciated.Thanks in advance.
lebesgue-integral measurable-functions
asked Jan 29 at 21:14
MarkMark
278
$endgroup$
add a comment |
$begingroup$
Let $psi$ be measurable function such that for all $ f in L^{1}(mu)$, $ fpsi in L^{1}(mu)$. Then $psi$ is bounded almost everywhere(in sense of measure $mu$).
My attempt:
Use contraposition: if $psi $ isn't bounded everywhere then $forall M>0$, there exists a set $A$ such that $mu (A)>0 $ and $ psi(x) > M , forall x in A$. Now i need to prove that there is an $f in L^{1} $ such that $ fpsi notin L^{1} $. But I don't know how.
Any help or hint would be appreciated.Thanks in advance.
lebesgue-integral measurable-functions
asked Jan 29 at 21:14
MarkMark
278
$endgroup$
$begingroup$
I guess you rather mean $psi(x)>M$ $forall xin A$ ?
$endgroup$
– J.F
Jan 29 at 21:21
add a comment |
$begingroup$
Let $psi$ be measurable function such that for all $ f in L^{1}(mu)$, $ fpsi in L^{1}(mu)$. Then $psi$ is bounded almost everywhere(in sense of measure $mu$).
My attempt:
Use contraposition: if $psi $ isn't bounded everywhere then $forall M>0$, there exists a set $A$ such that $mu (A)>0 $ and $ psi(x) > M , forall x in A$. Now i need to prove that there is an $f in L^{1} $ such that $ fpsi notin L^{1} $. But I don't know how.
Any help or hint would be appreciated.Thanks in advance.
lebesgue-integral measurable-functions
asked Jan 29 at 21:14
MarkMark
278
$endgroup$
Let $psi$ be measurable function such that for all $ f in L^{1}(mu)$, $ fpsi in L^{1}(mu)$. Then $psi$ is bounded almost everywhere(in sense of measure $mu$).
My attempt:
Use contraposition: if $psi $ isn't bounded everywhere then $forall M>0$, there exists a set $A$ such that $mu (A)>0 $ and $ psi(x) > M , forall x in A$. Now i need to prove that there is an $f in L^{1} $ such that $ fpsi notin L^{1} $. But I don't know how.
Any help or hint would be appreciated.Thanks in advance.
lebesgue-integral measurable-functions
lebesgue-integral measurable-functions
asked Jan 29 at 21:14
MarkMark
278
asked Jan 29 at 21:14
MarkMark
278
edited Jan 29 at 21:23
Mark
asked Jan 29 at 21:14
MarkMark
278
asked Jan 29 at 21:14
MarkMark
278
asked Jan 29 at 21:14
MarkMark
278
278
$begingroup$
I guess you rather mean $psi(x)>M$ $forall xin A$ ?
$endgroup$
– J.F
Jan 29 at 21:21
add a comment |
$begingroup$
I guess you rather mean $psi(x)>M$ $forall xin A$ ?
$endgroup$
– J.F
Jan 29 at 21:21
$begingroup$
I guess you rather mean $psi(x)>M$ $forall xin A$ ?
$endgroup$
– J.F
Jan 29 at 21:21
$begingroup$
I guess you rather mean $psi(x)>M$ $forall xin A$ ?
$endgroup$
– J.F
Jan 29 at 21:21
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We can prove this statement using Baire Category argument. But here, I will argue by contrapositive showing that
$$
psi notin L^infty(mu) implies exists fin L^1(mu) : fpsinotin L^1(mu).
$$ Let $E_j ={ 2^jle |psi|<2^{j+1}}$ and $c_j = mu (E_j).$ Note that assumption implies $c_j>0$ for infinitely many $j$'s. Define $J={j:c_j>0}$ and $$
f(x) = sum_{jin J} frac{1}{c_j sqrt{2}^{j}} 1_{E_j}(x).
$$ Then $int |f|=int f =sum_{jin J} 2^{-j/2}<infty$. However,
$$
|psi(x)||f(x)| ge 2^jfrac{1}{c_j sqrt{2}^{j}}=frac{sqrt{2}^{j}}{c_j}
$$ on each $E_j, jin J$, which implies
$$
int |psi f|=sum_{jin J} int_{E_j} |psi f| ge sum_{jin J}int_{E_j}frac{sqrt{2}^{j}}{c_j}=sum_{jin J} sqrt{2}^j =infty.
$$ Thus $psi fnotin L^1(mu)$ and the claim is proved.
answered Jan 29 at 22:29
SongSong
18.5k21651
$endgroup$
$begingroup$
How do you know that measure of $E_{j}$ is positive ,they might be empty sets for some $j$
$endgroup$
– Mark
Jan 29 at 23:39
1
$begingroup$
@Mark Oops, you are absolutely right. I was silly. It should have been stated as there exist infinitely many $j$ such that $c_j>0$. I've corrected my answer.
$endgroup$
– Song
Jan 29 at 23:48
$begingroup$
Was this originally your idea or you did some similar problem ?
$endgroup$
– Mark
Jan 29 at 23:53
$begingroup$
@Mark I'm not sure about my memory, but maybe I learnt this or similar argument somewhere. The idea behind this argument is to find a sequence $alpha_j>0$ such that $sum_j alpha_j 2^j c_j =infty$ while $sum_j alpha_j c_j$ is bounded. Then define $f = sum_j alpha_j 1_{E_j}$. (Here $alpha_j = 2^{-j/2}(1/c_j)$.)
$endgroup$
– Song
Jan 29 at 23:59
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We can prove this statement using Baire Category argument. But here, I will argue by contrapositive showing that
$$
psi notin L^infty(mu) implies exists fin L^1(mu) : fpsinotin L^1(mu).
$$ Let $E_j ={ 2^jle |psi|<2^{j+1}}$ and $c_j = mu (E_j).$ Note that assumption implies $c_j>0$ for infinitely many $j$'s. Define $J={j:c_j>0}$ and $$
f(x) = sum_{jin J} frac{1}{c_j sqrt{2}^{j}} 1_{E_j}(x).
$$ Then $int |f|=int f =sum_{jin J} 2^{-j/2}<infty$. However,
$$
|psi(x)||f(x)| ge 2^jfrac{1}{c_j sqrt{2}^{j}}=frac{sqrt{2}^{j}}{c_j}
$$ on each $E_j, jin J$, which implies
$$
int |psi f|=sum_{jin J} int_{E_j} |psi f| ge sum_{jin J}int_{E_j}frac{sqrt{2}^{j}}{c_j}=sum_{jin J} sqrt{2}^j =infty.
$$ Thus $psi fnotin L^1(mu)$ and the claim is proved.
answered Jan 29 at 22:29
SongSong
18.5k21651
$endgroup$
$begingroup$
How do you know that measure of $E_{j}$ is positive ,they might be empty sets for some $j$
$endgroup$
– Mark
Jan 29 at 23:39
1
$begingroup$
@Mark Oops, you are absolutely right. I was silly. It should have been stated as there exist infinitely many $j$ such that $c_j>0$. I've corrected my answer.
$endgroup$
– Song
Jan 29 at 23:48
$begingroup$
Was this originally your idea or you did some similar problem ?
$endgroup$
– Mark
Jan 29 at 23:53
$begingroup$
@Mark I'm not sure about my memory, but maybe I learnt this or similar argument somewhere. The idea behind this argument is to find a sequence $alpha_j>0$ such that $sum_j alpha_j 2^j c_j =infty$ while $sum_j alpha_j c_j$ is bounded. Then define $f = sum_j alpha_j 1_{E_j}$. (Here $alpha_j = 2^{-j/2}(1/c_j)$.)
$endgroup$
– Song
Jan 29 at 23:59
add a comment |
$begingroup$
We can prove this statement using Baire Category argument. But here, I will argue by contrapositive showing that
$$
psi notin L^infty(mu) implies exists fin L^1(mu) : fpsinotin L^1(mu).
$$ Let $E_j ={ 2^jle |psi|<2^{j+1}}$ and $c_j = mu (E_j).$ Note that assumption implies $c_j>0$ for infinitely many $j$'s. Define $J={j:c_j>0}$ and $$
f(x) = sum_{jin J} frac{1}{c_j sqrt{2}^{j}} 1_{E_j}(x).
$$ Then $int |f|=int f =sum_{jin J} 2^{-j/2}<infty$. However,
$$
|psi(x)||f(x)| ge 2^jfrac{1}{c_j sqrt{2}^{j}}=frac{sqrt{2}^{j}}{c_j}
$$ on each $E_j, jin J$, which implies
$$
int |psi f|=sum_{jin J} int_{E_j} |psi f| ge sum_{jin J}int_{E_j}frac{sqrt{2}^{j}}{c_j}=sum_{jin J} sqrt{2}^j =infty.
$$ Thus $psi fnotin L^1(mu)$ and the claim is proved.
answered Jan 29 at 22:29
SongSong
18.5k21651
$endgroup$
$begingroup$
How do you know that measure of $E_{j}$ is positive ,they might be empty sets for some $j$
$endgroup$
– Mark
Jan 29 at 23:39
1
$begingroup$
@Mark Oops, you are absolutely right. I was silly. It should have been stated as there exist infinitely many $j$ such that $c_j>0$. I've corrected my answer.
$endgroup$
– Song
Jan 29 at 23:48
$begingroup$
Was this originally your idea or you did some similar problem ?
$endgroup$
– Mark
Jan 29 at 23:53
$begingroup$
@Mark I'm not sure about my memory, but maybe I learnt this or similar argument somewhere. The idea behind this argument is to find a sequence $alpha_j>0$ such that $sum_j alpha_j 2^j c_j =infty$ while $sum_j alpha_j c_j$ is bounded. Then define $f = sum_j alpha_j 1_{E_j}$. (Here $alpha_j = 2^{-j/2}(1/c_j)$.)
$endgroup$
– Song
Jan 29 at 23:59
add a comment |
$begingroup$
We can prove this statement using Baire Category argument. But here, I will argue by contrapositive showing that
$$
psi notin L^infty(mu) implies exists fin L^1(mu) : fpsinotin L^1(mu).
$$ Let $E_j ={ 2^jle |psi|<2^{j+1}}$ and $c_j = mu (E_j).$ Note that assumption implies $c_j>0$ for infinitely many $j$'s. Define $J={j:c_j>0}$ and $$
f(x) = sum_{jin J} frac{1}{c_j sqrt{2}^{j}} 1_{E_j}(x).
$$ Then $int |f|=int f =sum_{jin J} 2^{-j/2}<infty$. However,
$$
|psi(x)||f(x)| ge 2^jfrac{1}{c_j sqrt{2}^{j}}=frac{sqrt{2}^{j}}{c_j}
$$ on each $E_j, jin J$, which implies
$$
int |psi f|=sum_{jin J} int_{E_j} |psi f| ge sum_{jin J}int_{E_j}frac{sqrt{2}^{j}}{c_j}=sum_{jin J} sqrt{2}^j =infty.
$$ Thus $psi fnotin L^1(mu)$ and the claim is proved.
answered Jan 29 at 22:29
SongSong
18.5k21651
$endgroup$
We can prove this statement using Baire Category argument. But here, I will argue by contrapositive showing that
$$
psi notin L^infty(mu) implies exists fin L^1(mu) : fpsinotin L^1(mu).
$$ Let $E_j ={ 2^jle |psi|<2^{j+1}}$ and $c_j = mu (E_j).$ Note that assumption implies $c_j>0$ for infinitely many $j$'s. Define $J={j:c_j>0}$ and $$
f(x) = sum_{jin J} frac{1}{c_j sqrt{2}^{j}} 1_{E_j}(x).
$$ Then $int |f|=int f =sum_{jin J} 2^{-j/2}<infty$. However,
$$
|psi(x)||f(x)| ge 2^jfrac{1}{c_j sqrt{2}^{j}}=frac{sqrt{2}^{j}}{c_j}
$$ on each $E_j, jin J$, which implies
$$
int |psi f|=sum_{jin J} int_{E_j} |psi f| ge sum_{jin J}int_{E_j}frac{sqrt{2}^{j}}{c_j}=sum_{jin J} sqrt{2}^j =infty.
$$ Thus $psi fnotin L^1(mu)$ and the claim is proved.
answered Jan 29 at 22:29
SongSong
18.5k21651
edited Jan 29 at 23:52
answered Jan 29 at 22:29
SongSong
18.5k21651
answered Jan 29 at 22:29
SongSong
18.5k21651
answered Jan 29 at 22:29
SongSong
18.5k21651
18.5k21651
$begingroup$
How do you know that measure of $E_{j}$ is positive ,they might be empty sets for some $j$
$endgroup$
– Mark
Jan 29 at 23:39
1
$begingroup$
@Mark Oops, you are absolutely right. I was silly. It should have been stated as there exist infinitely many $j$ such that $c_j>0$. I've corrected my answer.
$endgroup$
– Song
Jan 29 at 23:48
$begingroup$
Was this originally your idea or you did some similar problem ?
$endgroup$
– Mark
Jan 29 at 23:53
$begingroup$
@Mark I'm not sure about my memory, but maybe I learnt this or similar argument somewhere. The idea behind this argument is to find a sequence $alpha_j>0$ such that $sum_j alpha_j 2^j c_j =infty$ while $sum_j alpha_j c_j$ is bounded. Then define $f = sum_j alpha_j 1_{E_j}$. (Here $alpha_j = 2^{-j/2}(1/c_j)$.)
$endgroup$
– Song
Jan 29 at 23:59
add a comment |
$begingroup$
How do you know that measure of $E_{j}$ is positive ,they might be empty sets for some $j$
$endgroup$
– Mark
Jan 29 at 23:39
1
$begingroup$
@Mark Oops, you are absolutely right. I was silly. It should have been stated as there exist infinitely many $j$ such that $c_j>0$. I've corrected my answer.
$endgroup$
– Song
Jan 29 at 23:48
$begingroup$
Was this originally your idea or you did some similar problem ?
$endgroup$
– Mark
Jan 29 at 23:53
$begingroup$
@Mark I'm not sure about my memory, but maybe I learnt this or similar argument somewhere. The idea behind this argument is to find a sequence $alpha_j>0$ such that $sum_j alpha_j 2^j c_j =infty$ while $sum_j alpha_j c_j$ is bounded. Then define $f = sum_j alpha_j 1_{E_j}$. (Here $alpha_j = 2^{-j/2}(1/c_j)$.)
$endgroup$
– Song
Jan 29 at 23:59
$begingroup$
How do you know that measure of $E_{j}$ is positive ,they might be empty sets for some $j$
$endgroup$
– Mark
Jan 29 at 23:39
$begingroup$
How do you know that measure of $E_{j}$ is positive ,they might be empty sets for some $j$
$endgroup$
– Mark
Jan 29 at 23:39
1
1
$begingroup$
@Mark Oops, you are absolutely right. I was silly. It should have been stated as there exist infinitely many $j$ such that $c_j>0$. I've corrected my answer.
$endgroup$
– Song
Jan 29 at 23:48
$begingroup$
@Mark Oops, you are absolutely right. I was silly. It should have been stated as there exist infinitely many $j$ such that $c_j>0$. I've corrected my answer.
$endgroup$
– Song
Jan 29 at 23:48
$begingroup$
Was this originally your idea or you did some similar problem ?
$endgroup$
– Mark
Jan 29 at 23:53
$begingroup$
Was this originally your idea or you did some similar problem ?
$endgroup$
– Mark
Jan 29 at 23:53
$begingroup$
@Mark I'm not sure about my memory, but maybe I learnt this or similar argument somewhere. The idea behind this argument is to find a sequence $alpha_j>0$ such that $sum_j alpha_j 2^j c_j =infty$ while $sum_j alpha_j c_j$ is bounded. Then define $f = sum_j alpha_j 1_{E_j}$. (Here $alpha_j = 2^{-j/2}(1/c_j)$.)
$endgroup$
– Song
Jan 29 at 23:59
$begingroup$
@Mark I'm not sure about my memory, but maybe I learnt this or similar argument somewhere. The idea behind this argument is to find a sequence $alpha_j>0$ such that $sum_j alpha_j 2^j c_j =infty$ while $sum_j alpha_j c_j$ is bounded. Then define $f = sum_j alpha_j 1_{E_j}$. (Here $alpha_j = 2^{-j/2}(1/c_j)$.)
$endgroup$
– Song
Jan 29 at 23:59
add a comment |
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$begingroup$
I guess you rather mean $psi(x)>M$ $forall xin A$ ?
$endgroup$
– J.F
Jan 29 at 21:21