$limsup$ and integral inequality
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I have functions $f$, $g$ :$(0, infty) rightarrow (0, infty)$, and $g$ is invertible and decreasing (I don't know if it is relevant or not in this case). I know also that $limsup_{x searrow 0} frac {g(x)} {f(x)} < infty$. I should be able to deduce that $int_0^r g(x) dx leq int_0^r f(x) dx$ for some small enough $r in (0, 1)$. It should be easy to see, but the problem is, I don't see how.
real-analysis integration limsup-and-liminf
asked Jan 19 at 18:22
diipadaapadiipadaapa
62
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show 1 more comment
$begingroup$
I have functions $f$, $g$ :$(0, infty) rightarrow (0, infty)$, and $g$ is invertible and decreasing (I don't know if it is relevant or not in this case). I know also that $limsup_{x searrow 0} frac {g(x)} {f(x)} < infty$. I should be able to deduce that $int_0^r g(x) dx leq int_0^r f(x) dx$ for some small enough $r in (0, 1)$. It should be easy to see, but the problem is, I don't see how.
real-analysis integration limsup-and-liminf
asked Jan 19 at 18:22
diipadaapadiipadaapa
62
$endgroup$
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You have twice $g$ in your integral inequality.
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– mathcounterexamples.net
Jan 19 at 18:26
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True, fixed it.
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– diipadaapa
Jan 20 at 19:17
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Did you mean for the $limsup$ to be $<1$?
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– Umberto P.
Jan 20 at 19:26
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@UmbertoP. No, $<infty$ is what was given.
$endgroup$
– diipadaapa
Jan 21 at 22:14
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Take any decreasing $g$ (which is then invertible) and let $f(x) = dfrac 12 g(x)$. Then the $limsup$ equals $2$ but $int_0^r g(x) , dx = 2 int_0^r f(x) , dx > int_0^r f(x) , dx$.
$endgroup$
– Umberto P.
Jan 21 at 22:17
|
show 1 more comment
$begingroup$
I have functions $f$, $g$ :$(0, infty) rightarrow (0, infty)$, and $g$ is invertible and decreasing (I don't know if it is relevant or not in this case). I know also that $limsup_{x searrow 0} frac {g(x)} {f(x)} < infty$. I should be able to deduce that $int_0^r g(x) dx leq int_0^r f(x) dx$ for some small enough $r in (0, 1)$. It should be easy to see, but the problem is, I don't see how.
real-analysis integration limsup-and-liminf
asked Jan 19 at 18:22
diipadaapadiipadaapa
62
$endgroup$
I have functions $f$, $g$ :$(0, infty) rightarrow (0, infty)$, and $g$ is invertible and decreasing (I don't know if it is relevant or not in this case). I know also that $limsup_{x searrow 0} frac {g(x)} {f(x)} < infty$. I should be able to deduce that $int_0^r g(x) dx leq int_0^r f(x) dx$ for some small enough $r in (0, 1)$. It should be easy to see, but the problem is, I don't see how.
real-analysis integration limsup-and-liminf
real-analysis integration limsup-and-liminf
asked Jan 19 at 18:22
diipadaapadiipadaapa
62
asked Jan 19 at 18:22
diipadaapadiipadaapa
62
edited Jan 20 at 19:16
diipadaapa
asked Jan 19 at 18:22
diipadaapadiipadaapa
62
asked Jan 19 at 18:22
diipadaapadiipadaapa
62
asked Jan 19 at 18:22
diipadaapadiipadaapa
62
62
$begingroup$
You have twice $g$ in your integral inequality.
$endgroup$
– mathcounterexamples.net
Jan 19 at 18:26
$begingroup$
True, fixed it.
$endgroup$
– diipadaapa
Jan 20 at 19:17
$begingroup$
Did you mean for the $limsup$ to be $<1$?
$endgroup$
– Umberto P.
Jan 20 at 19:26
$begingroup$
@UmbertoP. No, $<infty$ is what was given.
$endgroup$
– diipadaapa
Jan 21 at 22:14
$begingroup$
Take any decreasing $g$ (which is then invertible) and let $f(x) = dfrac 12 g(x)$. Then the $limsup$ equals $2$ but $int_0^r g(x) , dx = 2 int_0^r f(x) , dx > int_0^r f(x) , dx$.
$endgroup$
– Umberto P.
Jan 21 at 22:17
|
show 1 more comment
$begingroup$
You have twice $g$ in your integral inequality.
$endgroup$
– mathcounterexamples.net
Jan 19 at 18:26
$begingroup$
True, fixed it.
$endgroup$
– diipadaapa
Jan 20 at 19:17
$begingroup$
Did you mean for the $limsup$ to be $<1$?
$endgroup$
– Umberto P.
Jan 20 at 19:26
$begingroup$
@UmbertoP. No, $<infty$ is what was given.
$endgroup$
– diipadaapa
Jan 21 at 22:14
$begingroup$
Take any decreasing $g$ (which is then invertible) and let $f(x) = dfrac 12 g(x)$. Then the $limsup$ equals $2$ but $int_0^r g(x) , dx = 2 int_0^r f(x) , dx > int_0^r f(x) , dx$.
$endgroup$
– Umberto P.
Jan 21 at 22:17
$begingroup$
You have twice $g$ in your integral inequality.
$endgroup$
– mathcounterexamples.net
Jan 19 at 18:26
$begingroup$
You have twice $g$ in your integral inequality.
$endgroup$
– mathcounterexamples.net
Jan 19 at 18:26
$begingroup$
True, fixed it.
$endgroup$
– diipadaapa
Jan 20 at 19:17
$begingroup$
True, fixed it.
$endgroup$
– diipadaapa
Jan 20 at 19:17
$begingroup$
Did you mean for the $limsup$ to be $<1$?
$endgroup$
– Umberto P.
Jan 20 at 19:26
$begingroup$
Did you mean for the $limsup$ to be $<1$?
$endgroup$
– Umberto P.
Jan 20 at 19:26
$begingroup$
@UmbertoP. No, $<infty$ is what was given.
$endgroup$
– diipadaapa
Jan 21 at 22:14
$begingroup$
@UmbertoP. No, $<infty$ is what was given.
$endgroup$
– diipadaapa
Jan 21 at 22:14
$begingroup$
Take any decreasing $g$ (which is then invertible) and let $f(x) = dfrac 12 g(x)$. Then the $limsup$ equals $2$ but $int_0^r g(x) , dx = 2 int_0^r f(x) , dx > int_0^r f(x) , dx$.
$endgroup$
– Umberto P.
Jan 21 at 22:17
$begingroup$
Take any decreasing $g$ (which is then invertible) and let $f(x) = dfrac 12 g(x)$. Then the $limsup$ equals $2$ but $int_0^r g(x) , dx = 2 int_0^r f(x) , dx > int_0^r f(x) , dx$.
$endgroup$
– Umberto P.
Jan 21 at 22:17
|
show 1 more comment
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$begingroup$
You have twice $g$ in your integral inequality.
$endgroup$
– mathcounterexamples.net
Jan 19 at 18:26
$begingroup$
True, fixed it.
$endgroup$
– diipadaapa
Jan 20 at 19:17
$begingroup$
Did you mean for the $limsup$ to be $<1$?
$endgroup$
– Umberto P.
Jan 20 at 19:26
$begingroup$
@UmbertoP. No, $<infty$ is what was given.
$endgroup$
– diipadaapa
Jan 21 at 22:14
$begingroup$
Take any decreasing $g$ (which is then invertible) and let $f(x) = dfrac 12 g(x)$. Then the $limsup$ equals $2$ but $int_0^r g(x) , dx = 2 int_0^r f(x) , dx > int_0^r f(x) , dx$.
$endgroup$
– Umberto P.
Jan 21 at 22:17