Does $fin L^1$ implies $logfin L^1$?
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Does $fin L^1$ implies $login L^1$?
I think it is not since $|log|f||$ can be arbitrarily big if $|f|$ is arbitrarily small.
But it seems $log|f|in L^1$ is true because of the following inequality,
$$int_Xlog|f|dmu leq logleft(int_X |f|dmu right)$$ if $mu(X)=1$ due to Jensen's inequality.
I am stuck with it... Thanks in advance for any hint and answers!
measure-theory inequality
asked Aug 26 at 19:46
Lev Ban
579116
add a comment |Â
up vote
1
down vote
favorite
Does $fin L^1$ implies $login L^1$?
I think it is not since $|log|f||$ can be arbitrarily big if $|f|$ is arbitrarily small.
But it seems $log|f|in L^1$ is true because of the following inequality,
$$int_Xlog|f|dmu leq logleft(int_X |f|dmu right)$$ if $mu(X)=1$ due to Jensen's inequality.
I am stuck with it... Thanks in advance for any hint and answers!
measure-theory inequality
asked Aug 26 at 19:46
Lev Ban
579116
3
What if $f$ is the zero function?
– Ashwin Trisal
Aug 26 at 19:51
5
Even if you don't allow $f=0$, the problem is you don't have a lower bound on $int_X log |f|; dmu$, which can be arbitrarily negative.
– Robert Israel
Aug 26 at 19:53
@AshwinTrisal Right! there was no restriction for $f$ not to be $0$ !Thanks!
– Lev Ban
Aug 26 at 19:53
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Does $fin L^1$ implies $login L^1$?
I think it is not since $|log|f||$ can be arbitrarily big if $|f|$ is arbitrarily small.
But it seems $log|f|in L^1$ is true because of the following inequality,
$$int_Xlog|f|dmu leq logleft(int_X |f|dmu right)$$ if $mu(X)=1$ due to Jensen's inequality.
I am stuck with it... Thanks in advance for any hint and answers!
measure-theory inequality
asked Aug 26 at 19:46
Lev Ban
579116
Does $fin L^1$ implies $login L^1$?
I think it is not since $|log|f||$ can be arbitrarily big if $|f|$ is arbitrarily small.
But it seems $log|f|in L^1$ is true because of the following inequality,
$$int_Xlog|f|dmu leq logleft(int_X |f|dmu right)$$ if $mu(X)=1$ due to Jensen's inequality.
I am stuck with it... Thanks in advance for any hint and answers!
measure-theory inequality
asked Aug 26 at 19:46
Lev Ban
579116
asked Aug 26 at 19:46
Lev Ban
579116
asked Aug 26 at 19:46
Lev Ban
579116
asked Aug 26 at 19:46
Lev Ban
579116
579116
3
What if $f$ is the zero function?
– Ashwin Trisal
Aug 26 at 19:51
5
Even if you don't allow $f=0$, the problem is you don't have a lower bound on $int_X log |f|; dmu$, which can be arbitrarily negative.
– Robert Israel
Aug 26 at 19:53
@AshwinTrisal Right! there was no restriction for $f$ not to be $0$ !Thanks!
– Lev Ban
Aug 26 at 19:53
add a comment |Â
3
What if $f$ is the zero function?
– Ashwin Trisal
Aug 26 at 19:51
5
Even if you don't allow $f=0$, the problem is you don't have a lower bound on $int_X log |f|; dmu$, which can be arbitrarily negative.
– Robert Israel
Aug 26 at 19:53
@AshwinTrisal Right! there was no restriction for $f$ not to be $0$ !Thanks!
– Lev Ban
Aug 26 at 19:53
3
3
What if $f$ is the zero function?
– Ashwin Trisal
Aug 26 at 19:51
What if $f$ is the zero function?
– Ashwin Trisal
Aug 26 at 19:51
5
5
Even if you don't allow $f=0$, the problem is you don't have a lower bound on $int_X log |f|; dmu$, which can be arbitrarily negative.
– Robert Israel
Aug 26 at 19:53
Even if you don't allow $f=0$, the problem is you don't have a lower bound on $int_X log |f|; dmu$, which can be arbitrarily negative.
– Robert Israel
Aug 26 at 19:53
@AshwinTrisal Right! there was no restriction for $f$ not to be $0$ !Thanks!
– Lev Ban
Aug 26 at 19:53
@AshwinTrisal Right! there was no restriction for $f$ not to be $0$ !Thanks!
– Lev Ban
Aug 26 at 19:53
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
What you said proves this holds for finite measure spaces, so instead consider
$f(x) = tfrac1x^2$ defined on $[1,infty)$. Then $$|f|_1 = int_1^infty x^-2,dx = left. x^-1right|_infty^1 = 1$$ so $fin L^1[0,infty)$ but
$$int_1^infty |log f|,dmu = int_1^infty 2log x,dmu = infty$$ so $log f notin L^1[0,infty)$. Zero function works too (see comments) but this is a nontrivial example.
answered Aug 26 at 19:53
cdipaolo
689311
Thank you for answer. But what if $mu(X)=1$? Is there still a counter example of nonzero $f$?
– Lev Ban
Aug 26 at 20:29
You are right, you'd need $|log f| = log f$, or $f geq 1$.
– cdipaolo
Aug 26 at 20:31
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
What you said proves this holds for finite measure spaces, so instead consider
$f(x) = tfrac1x^2$ defined on $[1,infty)$. Then $$|f|_1 = int_1^infty x^-2,dx = left. x^-1right|_infty^1 = 1$$ so $fin L^1[0,infty)$ but
$$int_1^infty |log f|,dmu = int_1^infty 2log x,dmu = infty$$ so $log f notin L^1[0,infty)$. Zero function works too (see comments) but this is a nontrivial example.
answered Aug 26 at 19:53
cdipaolo
689311
Thank you for answer. But what if $mu(X)=1$? Is there still a counter example of nonzero $f$?
– Lev Ban
Aug 26 at 20:29
You are right, you'd need $|log f| = log f$, or $f geq 1$.
– cdipaolo
Aug 26 at 20:31
add a comment |Â
up vote
1
down vote
accepted
What you said proves this holds for finite measure spaces, so instead consider
$f(x) = tfrac1x^2$ defined on $[1,infty)$. Then $$|f|_1 = int_1^infty x^-2,dx = left. x^-1right|_infty^1 = 1$$ so $fin L^1[0,infty)$ but
$$int_1^infty |log f|,dmu = int_1^infty 2log x,dmu = infty$$ so $log f notin L^1[0,infty)$. Zero function works too (see comments) but this is a nontrivial example.
answered Aug 26 at 19:53
cdipaolo
689311
Thank you for answer. But what if $mu(X)=1$? Is there still a counter example of nonzero $f$?
– Lev Ban
Aug 26 at 20:29
You are right, you'd need $|log f| = log f$, or $f geq 1$.
– cdipaolo
Aug 26 at 20:31
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
What you said proves this holds for finite measure spaces, so instead consider
$f(x) = tfrac1x^2$ defined on $[1,infty)$. Then $$|f|_1 = int_1^infty x^-2,dx = left. x^-1right|_infty^1 = 1$$ so $fin L^1[0,infty)$ but
$$int_1^infty |log f|,dmu = int_1^infty 2log x,dmu = infty$$ so $log f notin L^1[0,infty)$. Zero function works too (see comments) but this is a nontrivial example.
answered Aug 26 at 19:53
cdipaolo
689311
What you said proves this holds for finite measure spaces, so instead consider
$f(x) = tfrac1x^2$ defined on $[1,infty)$. Then $$|f|_1 = int_1^infty x^-2,dx = left. x^-1right|_infty^1 = 1$$ so $fin L^1[0,infty)$ but
$$int_1^infty |log f|,dmu = int_1^infty 2log x,dmu = infty$$ so $log f notin L^1[0,infty)$. Zero function works too (see comments) but this is a nontrivial example.
answered Aug 26 at 19:53
cdipaolo
689311
answered Aug 26 at 19:53
cdipaolo
689311
answered Aug 26 at 19:53
cdipaolo
689311
answered Aug 26 at 19:53
cdipaolo
689311
689311
Thank you for answer. But what if $mu(X)=1$? Is there still a counter example of nonzero $f$?
– Lev Ban
Aug 26 at 20:29
You are right, you'd need $|log f| = log f$, or $f geq 1$.
– cdipaolo
Aug 26 at 20:31
add a comment |Â
Thank you for answer. But what if $mu(X)=1$? Is there still a counter example of nonzero $f$?
– Lev Ban
Aug 26 at 20:29
You are right, you'd need $|log f| = log f$, or $f geq 1$.
– cdipaolo
Aug 26 at 20:31
Thank you for answer. But what if $mu(X)=1$? Is there still a counter example of nonzero $f$?
– Lev Ban
Aug 26 at 20:29
Thank you for answer. But what if $mu(X)=1$? Is there still a counter example of nonzero $f$?
– Lev Ban
Aug 26 at 20:29
You are right, you'd need $|log f| = log f$, or $f geq 1$.
– cdipaolo
Aug 26 at 20:31
You are right, you'd need $|log f| = log f$, or $f geq 1$.
– cdipaolo
Aug 26 at 20:31
add a comment |Â
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3
What if $f$ is the zero function?
– Ashwin Trisal
Aug 26 at 19:51
5
Even if you don't allow $f=0$, the problem is you don't have a lower bound on $int_X log |f|; dmu$, which can be arbitrarily negative.
– Robert Israel
Aug 26 at 19:53
@AshwinTrisal Right! there was no restriction for $f$ not to be $0$ !Thanks!
– Lev Ban
Aug 26 at 19:53