$f, 1/f$ integrable implies that $f^2$, $1/f^2$ integrable?
up vote
1
down vote
favorite
on the measure space $(X,A,m)$ let $f: X to mathbb R$ be Borel-measurable, it is given that $f$, $1/f$ are well-defined and integrable.
I would like to prove that this implies that $f^2$ and $1/f^2$ are integrable but am stuck on this part. I thought about using Hölder's inequality, do you guys have any hints?
integration measure-theory borel-measures
asked Nov 27 at 14:49
mathsisrad
62
add a comment |
up vote
1
down vote
favorite
on the measure space $(X,A,m)$ let $f: X to mathbb R$ be Borel-measurable, it is given that $f$, $1/f$ are well-defined and integrable.
I would like to prove that this implies that $f^2$ and $1/f^2$ are integrable but am stuck on this part. I thought about using Hölder's inequality, do you guys have any hints?
integration measure-theory borel-measures
asked Nov 27 at 14:49
mathsisrad
62
MathJax is required for questions
– Akash Roy
Nov 27 at 14:50
Why do you suspect your statement is true?
– Umberto P.
Nov 27 at 14:58
I would actually like to prove that this implies that m(X) is finite. So I defined h:= ƒ*(1/ƒ) = 1. If I take the Lebesgue integral of h, I get m(X). If h is integrable, then m(X) is finite. I thought if I show that $f^2$, 1/$f^2$ is integrable, by Hölder's inequality this would imply that ƒ*1/ƒ = h = 1 is integrable
– mathsisrad
Nov 27 at 15:02
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
on the measure space $(X,A,m)$ let $f: X to mathbb R$ be Borel-measurable, it is given that $f$, $1/f$ are well-defined and integrable.
I would like to prove that this implies that $f^2$ and $1/f^2$ are integrable but am stuck on this part. I thought about using Hölder's inequality, do you guys have any hints?
integration measure-theory borel-measures
asked Nov 27 at 14:49
mathsisrad
62
on the measure space $(X,A,m)$ let $f: X to mathbb R$ be Borel-measurable, it is given that $f$, $1/f$ are well-defined and integrable.
I would like to prove that this implies that $f^2$ and $1/f^2$ are integrable but am stuck on this part. I thought about using Hölder's inequality, do you guys have any hints?
integration measure-theory borel-measures
integration measure-theory borel-measures
asked Nov 27 at 14:49
mathsisrad
62
asked Nov 27 at 14:49
mathsisrad
62
edited Nov 27 at 14:54
asked Nov 27 at 14:49
mathsisrad
62
asked Nov 27 at 14:49
mathsisrad
62
asked Nov 27 at 14:49
mathsisrad
62
62
MathJax is required for questions
– Akash Roy
Nov 27 at 14:50
Why do you suspect your statement is true?
– Umberto P.
Nov 27 at 14:58
I would actually like to prove that this implies that m(X) is finite. So I defined h:= ƒ*(1/ƒ) = 1. If I take the Lebesgue integral of h, I get m(X). If h is integrable, then m(X) is finite. I thought if I show that $f^2$, 1/$f^2$ is integrable, by Hölder's inequality this would imply that ƒ*1/ƒ = h = 1 is integrable
– mathsisrad
Nov 27 at 15:02
add a comment |
MathJax is required for questions
– Akash Roy
Nov 27 at 14:50
Why do you suspect your statement is true?
– Umberto P.
Nov 27 at 14:58
I would actually like to prove that this implies that m(X) is finite. So I defined h:= ƒ*(1/ƒ) = 1. If I take the Lebesgue integral of h, I get m(X). If h is integrable, then m(X) is finite. I thought if I show that $f^2$, 1/$f^2$ is integrable, by Hölder's inequality this would imply that ƒ*1/ƒ = h = 1 is integrable
– mathsisrad
Nov 27 at 15:02
MathJax is required for questions
– Akash Roy
Nov 27 at 14:50
MathJax is required for questions
– Akash Roy
Nov 27 at 14:50
Why do you suspect your statement is true?
– Umberto P.
Nov 27 at 14:58
Why do you suspect your statement is true?
– Umberto P.
Nov 27 at 14:58
I would actually like to prove that this implies that m(X) is finite. So I defined h:= ƒ*(1/ƒ) = 1. If I take the Lebesgue integral of h, I get m(X). If h is integrable, then m(X) is finite. I thought if I show that $f^2$, 1/$f^2$ is integrable, by Hölder's inequality this would imply that ƒ*1/ƒ = h = 1 is integrable
– mathsisrad
Nov 27 at 15:02
I would actually like to prove that this implies that m(X) is finite. So I defined h:= ƒ*(1/ƒ) = 1. If I take the Lebesgue integral of h, I get m(X). If h is integrable, then m(X) is finite. I thought if I show that $f^2$, 1/$f^2$ is integrable, by Hölder's inequality this would imply that ƒ*1/ƒ = h = 1 is integrable
– mathsisrad
Nov 27 at 15:02
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
$sqrt{x}$ and $1/sqrt{x}$ are integrable on $(0,1]$ with Lebesgue measure, but $1/x$ is not.
answered Nov 27 at 15:03
Robert Israel
317k23206457
yes but 1/x is not well-defined in 0, in my case f and 1/f are well-defined
– mathsisrad
Nov 27 at 15:04
1
$0$ is not part of the measure space; we don't need to define it there.
– Sambo
Nov 27 at 15:06
1
@mathsisrad Robert excluded zero from the domain, but even if you don't, you can arbitrarily define $f$ at $0$ and the problem persists. It's not clear to me what notion you're trying to iron out with "well-defined"; you might be assuming that they are both $L^infty$ but then the result is trivial.
– Ian
Nov 27 at 15:07
hmm but would the given criteria imply that f is in L^p, 1/f is in L^q with 1/p + 1/q = 1? All I want to show is that I can use Hölder's inequality to prove that f*(1/f) = 1 is integrable
– mathsisrad
Nov 27 at 15:09
Hint: If $f$ and $g$ are integrable, so is $sqrt{|f g|}$.
– Robert Israel
Nov 27 at 15:25
|
show 2 more comments
up vote
0
down vote
You might prefer trying to prove something true.
Consider $bigl((0,1), mathcal B((0,1)), mathcal L^1bigr)$ and $f(x)=sqrt x$.
Then $f$ and $1/f$ are integrable, but $1/f^2$ is not.
answered Nov 27 at 15:05
Federico
4,192512
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
$sqrt{x}$ and $1/sqrt{x}$ are integrable on $(0,1]$ with Lebesgue measure, but $1/x$ is not.
answered Nov 27 at 15:03
Robert Israel
317k23206457
yes but 1/x is not well-defined in 0, in my case f and 1/f are well-defined
– mathsisrad
Nov 27 at 15:04
1
$0$ is not part of the measure space; we don't need to define it there.
– Sambo
Nov 27 at 15:06
1
@mathsisrad Robert excluded zero from the domain, but even if you don't, you can arbitrarily define $f$ at $0$ and the problem persists. It's not clear to me what notion you're trying to iron out with "well-defined"; you might be assuming that they are both $L^infty$ but then the result is trivial.
– Ian
Nov 27 at 15:07
hmm but would the given criteria imply that f is in L^p, 1/f is in L^q with 1/p + 1/q = 1? All I want to show is that I can use Hölder's inequality to prove that f*(1/f) = 1 is integrable
– mathsisrad
Nov 27 at 15:09
Hint: If $f$ and $g$ are integrable, so is $sqrt{|f g|}$.
– Robert Israel
Nov 27 at 15:25
|
show 2 more comments
up vote
2
down vote
$sqrt{x}$ and $1/sqrt{x}$ are integrable on $(0,1]$ with Lebesgue measure, but $1/x$ is not.
answered Nov 27 at 15:03
Robert Israel
317k23206457
yes but 1/x is not well-defined in 0, in my case f and 1/f are well-defined
– mathsisrad
Nov 27 at 15:04
1
$0$ is not part of the measure space; we don't need to define it there.
– Sambo
Nov 27 at 15:06
1
@mathsisrad Robert excluded zero from the domain, but even if you don't, you can arbitrarily define $f$ at $0$ and the problem persists. It's not clear to me what notion you're trying to iron out with "well-defined"; you might be assuming that they are both $L^infty$ but then the result is trivial.
– Ian
Nov 27 at 15:07
hmm but would the given criteria imply that f is in L^p, 1/f is in L^q with 1/p + 1/q = 1? All I want to show is that I can use Hölder's inequality to prove that f*(1/f) = 1 is integrable
– mathsisrad
Nov 27 at 15:09
Hint: If $f$ and $g$ are integrable, so is $sqrt{|f g|}$.
– Robert Israel
Nov 27 at 15:25
|
show 2 more comments
up vote
2
down vote
up vote
2
down vote
$sqrt{x}$ and $1/sqrt{x}$ are integrable on $(0,1]$ with Lebesgue measure, but $1/x$ is not.
answered Nov 27 at 15:03
Robert Israel
317k23206457
$sqrt{x}$ and $1/sqrt{x}$ are integrable on $(0,1]$ with Lebesgue measure, but $1/x$ is not.
answered Nov 27 at 15:03
Robert Israel
317k23206457
answered Nov 27 at 15:03
Robert Israel
317k23206457
answered Nov 27 at 15:03
Robert Israel
317k23206457
answered Nov 27 at 15:03
Robert Israel
317k23206457
317k23206457
yes but 1/x is not well-defined in 0, in my case f and 1/f are well-defined
– mathsisrad
Nov 27 at 15:04
1
$0$ is not part of the measure space; we don't need to define it there.
– Sambo
Nov 27 at 15:06
1
@mathsisrad Robert excluded zero from the domain, but even if you don't, you can arbitrarily define $f$ at $0$ and the problem persists. It's not clear to me what notion you're trying to iron out with "well-defined"; you might be assuming that they are both $L^infty$ but then the result is trivial.
– Ian
Nov 27 at 15:07
hmm but would the given criteria imply that f is in L^p, 1/f is in L^q with 1/p + 1/q = 1? All I want to show is that I can use Hölder's inequality to prove that f*(1/f) = 1 is integrable
– mathsisrad
Nov 27 at 15:09
Hint: If $f$ and $g$ are integrable, so is $sqrt{|f g|}$.
– Robert Israel
Nov 27 at 15:25
|
show 2 more comments
yes but 1/x is not well-defined in 0, in my case f and 1/f are well-defined
– mathsisrad
Nov 27 at 15:04
1
$0$ is not part of the measure space; we don't need to define it there.
– Sambo
Nov 27 at 15:06
1
@mathsisrad Robert excluded zero from the domain, but even if you don't, you can arbitrarily define $f$ at $0$ and the problem persists. It's not clear to me what notion you're trying to iron out with "well-defined"; you might be assuming that they are both $L^infty$ but then the result is trivial.
– Ian
Nov 27 at 15:07
hmm but would the given criteria imply that f is in L^p, 1/f is in L^q with 1/p + 1/q = 1? All I want to show is that I can use Hölder's inequality to prove that f*(1/f) = 1 is integrable
– mathsisrad
Nov 27 at 15:09
Hint: If $f$ and $g$ are integrable, so is $sqrt{|f g|}$.
– Robert Israel
Nov 27 at 15:25
yes but 1/x is not well-defined in 0, in my case f and 1/f are well-defined
– mathsisrad
Nov 27 at 15:04
yes but 1/x is not well-defined in 0, in my case f and 1/f are well-defined
– mathsisrad
Nov 27 at 15:04
1
1
$0$ is not part of the measure space; we don't need to define it there.
– Sambo
Nov 27 at 15:06
$0$ is not part of the measure space; we don't need to define it there.
– Sambo
Nov 27 at 15:06
1
1
@mathsisrad Robert excluded zero from the domain, but even if you don't, you can arbitrarily define $f$ at $0$ and the problem persists. It's not clear to me what notion you're trying to iron out with "well-defined"; you might be assuming that they are both $L^infty$ but then the result is trivial.
– Ian
Nov 27 at 15:07
@mathsisrad Robert excluded zero from the domain, but even if you don't, you can arbitrarily define $f$ at $0$ and the problem persists. It's not clear to me what notion you're trying to iron out with "well-defined"; you might be assuming that they are both $L^infty$ but then the result is trivial.
– Ian
Nov 27 at 15:07
hmm but would the given criteria imply that f is in L^p, 1/f is in L^q with 1/p + 1/q = 1? All I want to show is that I can use Hölder's inequality to prove that f*(1/f) = 1 is integrable
– mathsisrad
Nov 27 at 15:09
hmm but would the given criteria imply that f is in L^p, 1/f is in L^q with 1/p + 1/q = 1? All I want to show is that I can use Hölder's inequality to prove that f*(1/f) = 1 is integrable
– mathsisrad
Nov 27 at 15:09
Hint: If $f$ and $g$ are integrable, so is $sqrt{|f g|}$.
– Robert Israel
Nov 27 at 15:25
Hint: If $f$ and $g$ are integrable, so is $sqrt{|f g|}$.
– Robert Israel
Nov 27 at 15:25
|
show 2 more comments
up vote
0
down vote
You might prefer trying to prove something true.
Consider $bigl((0,1), mathcal B((0,1)), mathcal L^1bigr)$ and $f(x)=sqrt x$.
Then $f$ and $1/f$ are integrable, but $1/f^2$ is not.
answered Nov 27 at 15:05
Federico
4,192512
add a comment |
up vote
0
down vote
You might prefer trying to prove something true.
Consider $bigl((0,1), mathcal B((0,1)), mathcal L^1bigr)$ and $f(x)=sqrt x$.
Then $f$ and $1/f$ are integrable, but $1/f^2$ is not.
answered Nov 27 at 15:05
Federico
4,192512
add a comment |
up vote
0
down vote
up vote
0
down vote
You might prefer trying to prove something true.
Consider $bigl((0,1), mathcal B((0,1)), mathcal L^1bigr)$ and $f(x)=sqrt x$.
Then $f$ and $1/f$ are integrable, but $1/f^2$ is not.
answered Nov 27 at 15:05
Federico
4,192512
You might prefer trying to prove something true.
Consider $bigl((0,1), mathcal B((0,1)), mathcal L^1bigr)$ and $f(x)=sqrt x$.
Then $f$ and $1/f$ are integrable, but $1/f^2$ is not.
answered Nov 27 at 15:05
Federico
4,192512
answered Nov 27 at 15:05
Federico
4,192512
answered Nov 27 at 15:05
Federico
4,192512
answered Nov 27 at 15:05
Federico
4,192512
4,192512
add a comment |
add a comment |
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– Akash Roy
Nov 27 at 14:50
Why do you suspect your statement is true?
– Umberto P.
Nov 27 at 14:58
I would actually like to prove that this implies that m(X) is finite. So I defined h:= ƒ*(1/ƒ) = 1. If I take the Lebesgue integral of h, I get m(X). If h is integrable, then m(X) is finite. I thought if I show that $f^2$, 1/$f^2$ is integrable, by Hölder's inequality this would imply that ƒ*1/ƒ = h = 1 is integrable
– mathsisrad
Nov 27 at 15:02