Comparing two limsup'sAnalogues of Luzin's theoremDo convex and decreasing functions preserve the semimartingale property?Inequality on probability distributionsThe Hardy Z-function and failure of the Riemann hypothesisA Normal Distribution InequalityQuantifying the “flatness” of functions which are the Fourier transforms of positive functionsLower Matuszewska index of positive increasing $O$-regular functionsSurprisingly simple minimum of a rational function on $mathbb R_+^n$Definite inequality comparing a product of sums?Proving two inequalities involving the gamma and digamma functions
Comparing two limsup's
Analogues of Luzin's theoremDo convex and decreasing functions preserve the semimartingale property?Inequality on probability distributionsThe Hardy Z-function and failure of the Riemann hypothesisA Normal Distribution InequalityQuantifying the “flatness” of functions which are the Fourier transforms of positive functionsLower Matuszewska index of positive increasing $O$-regular functionsSurprisingly simple minimum of a rational function on $mathbb R_+^n$Definite inequality comparing a product of sums?Proving two inequalities involving the gamma and digamma functions
$begingroup$
Let $fin L^2(0,infty)$ be a positive, decreasing function. Is it then true that
$$
limsup_xtoinfty xf(x) = limsup_xtoinfty frac1f(x)int_x^infty f^2(t), dt
$$
(and similarly for $liminf$)?
This looks strange at first; for example, the quotient of the two quantities can easily become both large or small. However, from the context in which this question arose, we have reason to believe that the statement is actually true.
real-analysis inequalities
asked 8 hours ago
Christian RemlingChristian Remling
12.7k2 gold badges24 silver badges45 bronze badges
$endgroup$
add a comment |
$begingroup$
Let $fin L^2(0,infty)$ be a positive, decreasing function. Is it then true that
$$
limsup_xtoinfty xf(x) = limsup_xtoinfty frac1f(x)int_x^infty f^2(t), dt
$$
(and similarly for $liminf$)?
This looks strange at first; for example, the quotient of the two quantities can easily become both large or small. However, from the context in which this question arose, we have reason to believe that the statement is actually true.
real-analysis inequalities
asked 8 hours ago
Christian RemlingChristian Remling
12.7k2 gold badges24 silver badges45 bronze badges
$endgroup$
add a comment |
$begingroup$
Let $fin L^2(0,infty)$ be a positive, decreasing function. Is it then true that
$$
limsup_xtoinfty xf(x) = limsup_xtoinfty frac1f(x)int_x^infty f^2(t), dt
$$
(and similarly for $liminf$)?
This looks strange at first; for example, the quotient of the two quantities can easily become both large or small. However, from the context in which this question arose, we have reason to believe that the statement is actually true.
real-analysis inequalities
asked 8 hours ago
Christian RemlingChristian Remling
12.7k2 gold badges24 silver badges45 bronze badges
$endgroup$
Let $fin L^2(0,infty)$ be a positive, decreasing function. Is it then true that
$$
limsup_xtoinfty xf(x) = limsup_xtoinfty frac1f(x)int_x^infty f^2(t), dt
$$
(and similarly for $liminf$)?
This looks strange at first; for example, the quotient of the two quantities can easily become both large or small. However, from the context in which this question arose, we have reason to believe that the statement is actually true.
real-analysis inequalities
real-analysis inequalities
asked 8 hours ago
Christian RemlingChristian Remling
12.7k2 gold badges24 silver badges45 bronze badges
asked 8 hours ago
Christian RemlingChristian Remling
12.7k2 gold badges24 silver badges45 bronze badges
asked 8 hours ago
Christian RemlingChristian Remling
12.7k2 gold badges24 silver badges45 bronze badges
asked 8 hours ago
Christian RemlingChristian Remling
12.7k2 gold badges24 silver badges45 bronze badges
asked 8 hours ago
Christian RemlingChristian Remling
12.7k2 gold badges24 silver badges45 bronze badges
12.7k2 gold badges24 silver badges45 bronze badges
add a comment |
add a comment |
2 Answers
2
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oldest
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$begingroup$
No, this equality does not hold in general. What is true is that rhs is at most twice as large as lhs and this is sharp.
For the sake of brevity I will only show an example where lhs is $1$ and rhs is greater than $1.9$. Choose a very fast growing sequence $a_n ge 1$. We will construct our function on each interval $I_n = [a_n, a_n+1]$ separately.
Let $f(x) = frac1100a_n$ on $[a_n, 100a_n)$ and $f(x) = frac1x$ on $[100a_n, a_n+1)$ (on $[0, a_1]$ say $f(x) = 1$).
Obviously $limsup xf(x) = 1$ ($f(x) le frac1x$ for $x > a_1$).
We have $$frac1f(a_n)int_a_n^infty f(x)^2dx ge 100a_n(99a_nfrac110000a_n^2 + frac1100a_n - frac1a_n+1) = 1.99 - frac100a_na_n+1 ge 1.9,$$
where in the frist inequality we truncated integral at $a_n+1$ and the last ineqaulity is true if $a_n+1 ge 10^9a_n$.
Therefore rhs is at least $1.9$. It is easy to make rhs to be equal to $2$ with the same method. Turns out this is sharp:
Again for brevity I will prove that if lhs is less than $1$ then rhs is at most $2$. For big enough $x$ we have $f(x) le frac1x$. Therefore
$$frac1f(x)int_x^infty f(t)^2dt le frac1f(x)(int_x^frac1f(x)f(x)^2dt + int_frac1f(x)^infty frac1t^2dt) le frac1f(x)(f(x) + f(x)) = 2.$$
edited 5 hours ago
Christian Remling
12.7k2 gold badges24 silver badges45 bronze badges
answered 6 hours ago
Aleksei KulikovAleksei Kulikov
1,4161 gold badge4 silver badges11 bronze badges
$endgroup$
2
$begingroup$
If I don't mistake, it is also true that $RHSge frac 12 LHS$ and that without extra assumptions nothing interesting can be said about $liminf$'s.
$endgroup$
– fedja
6 hours ago
add a comment |
$begingroup$
Assuming lots of stuff, just to get the ball rolling:
$$frac1f(x)int_x^inftyf^2(t)dt=frac1xf(x)fracint_0^frac1xf^2(frac1t)frac1t^2dtfrac1x.$$
So we claim that
$$(xf(x))^2sim fracint_0^frac1xf^2(frac1t)frac1t^2dtfrac1x.$$
If $f^2(1/t)/t^2$ was continuous, then
$$fracint_0^frac1xf^2(frac1t)frac1t^2dtfrac1xsim f^2(x)x^2frac1/x1/x=f^2(x)x^2.$$
So maybe getting a counterexample by making $f^2(1/t)/t^2$ not well-behaved eg. blowing up at $t=0$ i.e. $f(x)>frac1x$.
answered 7 hours ago
Thomas KojarThomas Kojar
6104 silver badges15 bronze badges
$endgroup$
add a comment |
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2 Answers
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2 Answers
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active
oldest
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$begingroup$
No, this equality does not hold in general. What is true is that rhs is at most twice as large as lhs and this is sharp.
For the sake of brevity I will only show an example where lhs is $1$ and rhs is greater than $1.9$. Choose a very fast growing sequence $a_n ge 1$. We will construct our function on each interval $I_n = [a_n, a_n+1]$ separately.
Let $f(x) = frac1100a_n$ on $[a_n, 100a_n)$ and $f(x) = frac1x$ on $[100a_n, a_n+1)$ (on $[0, a_1]$ say $f(x) = 1$).
Obviously $limsup xf(x) = 1$ ($f(x) le frac1x$ for $x > a_1$).
We have $$frac1f(a_n)int_a_n^infty f(x)^2dx ge 100a_n(99a_nfrac110000a_n^2 + frac1100a_n - frac1a_n+1) = 1.99 - frac100a_na_n+1 ge 1.9,$$
where in the frist inequality we truncated integral at $a_n+1$ and the last ineqaulity is true if $a_n+1 ge 10^9a_n$.
Therefore rhs is at least $1.9$. It is easy to make rhs to be equal to $2$ with the same method. Turns out this is sharp:
Again for brevity I will prove that if lhs is less than $1$ then rhs is at most $2$. For big enough $x$ we have $f(x) le frac1x$. Therefore
$$frac1f(x)int_x^infty f(t)^2dt le frac1f(x)(int_x^frac1f(x)f(x)^2dt + int_frac1f(x)^infty frac1t^2dt) le frac1f(x)(f(x) + f(x)) = 2.$$
edited 5 hours ago
Christian Remling
12.7k2 gold badges24 silver badges45 bronze badges
answered 6 hours ago
Aleksei KulikovAleksei Kulikov
1,4161 gold badge4 silver badges11 bronze badges
$endgroup$
2
$begingroup$
If I don't mistake, it is also true that $RHSge frac 12 LHS$ and that without extra assumptions nothing interesting can be said about $liminf$'s.
$endgroup$
– fedja
6 hours ago
add a comment |
$begingroup$
No, this equality does not hold in general. What is true is that rhs is at most twice as large as lhs and this is sharp.
For the sake of brevity I will only show an example where lhs is $1$ and rhs is greater than $1.9$. Choose a very fast growing sequence $a_n ge 1$. We will construct our function on each interval $I_n = [a_n, a_n+1]$ separately.
Let $f(x) = frac1100a_n$ on $[a_n, 100a_n)$ and $f(x) = frac1x$ on $[100a_n, a_n+1)$ (on $[0, a_1]$ say $f(x) = 1$).
Obviously $limsup xf(x) = 1$ ($f(x) le frac1x$ for $x > a_1$).
We have $$frac1f(a_n)int_a_n^infty f(x)^2dx ge 100a_n(99a_nfrac110000a_n^2 + frac1100a_n - frac1a_n+1) = 1.99 - frac100a_na_n+1 ge 1.9,$$
where in the frist inequality we truncated integral at $a_n+1$ and the last ineqaulity is true if $a_n+1 ge 10^9a_n$.
Therefore rhs is at least $1.9$. It is easy to make rhs to be equal to $2$ with the same method. Turns out this is sharp:
Again for brevity I will prove that if lhs is less than $1$ then rhs is at most $2$. For big enough $x$ we have $f(x) le frac1x$. Therefore
$$frac1f(x)int_x^infty f(t)^2dt le frac1f(x)(int_x^frac1f(x)f(x)^2dt + int_frac1f(x)^infty frac1t^2dt) le frac1f(x)(f(x) + f(x)) = 2.$$
edited 5 hours ago
Christian Remling
12.7k2 gold badges24 silver badges45 bronze badges
answered 6 hours ago
Aleksei KulikovAleksei Kulikov
1,4161 gold badge4 silver badges11 bronze badges
$endgroup$
2
$begingroup$
If I don't mistake, it is also true that $RHSge frac 12 LHS$ and that without extra assumptions nothing interesting can be said about $liminf$'s.
$endgroup$
– fedja
6 hours ago
add a comment |
$begingroup$
No, this equality does not hold in general. What is true is that rhs is at most twice as large as lhs and this is sharp.
For the sake of brevity I will only show an example where lhs is $1$ and rhs is greater than $1.9$. Choose a very fast growing sequence $a_n ge 1$. We will construct our function on each interval $I_n = [a_n, a_n+1]$ separately.
Let $f(x) = frac1100a_n$ on $[a_n, 100a_n)$ and $f(x) = frac1x$ on $[100a_n, a_n+1)$ (on $[0, a_1]$ say $f(x) = 1$).
Obviously $limsup xf(x) = 1$ ($f(x) le frac1x$ for $x > a_1$).
We have $$frac1f(a_n)int_a_n^infty f(x)^2dx ge 100a_n(99a_nfrac110000a_n^2 + frac1100a_n - frac1a_n+1) = 1.99 - frac100a_na_n+1 ge 1.9,$$
where in the frist inequality we truncated integral at $a_n+1$ and the last ineqaulity is true if $a_n+1 ge 10^9a_n$.
Therefore rhs is at least $1.9$. It is easy to make rhs to be equal to $2$ with the same method. Turns out this is sharp:
Again for brevity I will prove that if lhs is less than $1$ then rhs is at most $2$. For big enough $x$ we have $f(x) le frac1x$. Therefore
$$frac1f(x)int_x^infty f(t)^2dt le frac1f(x)(int_x^frac1f(x)f(x)^2dt + int_frac1f(x)^infty frac1t^2dt) le frac1f(x)(f(x) + f(x)) = 2.$$
edited 5 hours ago
Christian Remling
12.7k2 gold badges24 silver badges45 bronze badges
answered 6 hours ago
Aleksei KulikovAleksei Kulikov
1,4161 gold badge4 silver badges11 bronze badges
$endgroup$
No, this equality does not hold in general. What is true is that rhs is at most twice as large as lhs and this is sharp.
For the sake of brevity I will only show an example where lhs is $1$ and rhs is greater than $1.9$. Choose a very fast growing sequence $a_n ge 1$. We will construct our function on each interval $I_n = [a_n, a_n+1]$ separately.
Let $f(x) = frac1100a_n$ on $[a_n, 100a_n)$ and $f(x) = frac1x$ on $[100a_n, a_n+1)$ (on $[0, a_1]$ say $f(x) = 1$).
Obviously $limsup xf(x) = 1$ ($f(x) le frac1x$ for $x > a_1$).
We have $$frac1f(a_n)int_a_n^infty f(x)^2dx ge 100a_n(99a_nfrac110000a_n^2 + frac1100a_n - frac1a_n+1) = 1.99 - frac100a_na_n+1 ge 1.9,$$
where in the frist inequality we truncated integral at $a_n+1$ and the last ineqaulity is true if $a_n+1 ge 10^9a_n$.
Therefore rhs is at least $1.9$. It is easy to make rhs to be equal to $2$ with the same method. Turns out this is sharp:
Again for brevity I will prove that if lhs is less than $1$ then rhs is at most $2$. For big enough $x$ we have $f(x) le frac1x$. Therefore
$$frac1f(x)int_x^infty f(t)^2dt le frac1f(x)(int_x^frac1f(x)f(x)^2dt + int_frac1f(x)^infty frac1t^2dt) le frac1f(x)(f(x) + f(x)) = 2.$$
edited 5 hours ago
Christian Remling
12.7k2 gold badges24 silver badges45 bronze badges
answered 6 hours ago
Aleksei KulikovAleksei Kulikov
1,4161 gold badge4 silver badges11 bronze badges
edited 5 hours ago
Christian Remling
12.7k2 gold badges24 silver badges45 bronze badges
edited 5 hours ago
Christian Remling
12.7k2 gold badges24 silver badges45 bronze badges
edited 5 hours ago
Christian Remling
12.7k2 gold badges24 silver badges45 bronze badges
12.7k2 gold badges24 silver badges45 bronze badges
answered 6 hours ago
Aleksei KulikovAleksei Kulikov
1,4161 gold badge4 silver badges11 bronze badges
answered 6 hours ago
Aleksei KulikovAleksei Kulikov
1,4161 gold badge4 silver badges11 bronze badges
answered 6 hours ago
Aleksei KulikovAleksei Kulikov
1,4161 gold badge4 silver badges11 bronze badges
1,4161 gold badge4 silver badges11 bronze badges
2
$begingroup$
If I don't mistake, it is also true that $RHSge frac 12 LHS$ and that without extra assumptions nothing interesting can be said about $liminf$'s.
$endgroup$
– fedja
6 hours ago
add a comment |
2
$begingroup$
If I don't mistake, it is also true that $RHSge frac 12 LHS$ and that without extra assumptions nothing interesting can be said about $liminf$'s.
$endgroup$
– fedja
6 hours ago
2
2
$begingroup$
If I don't mistake, it is also true that $RHSge frac 12 LHS$ and that without extra assumptions nothing interesting can be said about $liminf$'s.
$endgroup$
– fedja
6 hours ago
$begingroup$
If I don't mistake, it is also true that $RHSge frac 12 LHS$ and that without extra assumptions nothing interesting can be said about $liminf$'s.
$endgroup$
– fedja
6 hours ago
add a comment |
$begingroup$
Assuming lots of stuff, just to get the ball rolling:
$$frac1f(x)int_x^inftyf^2(t)dt=frac1xf(x)fracint_0^frac1xf^2(frac1t)frac1t^2dtfrac1x.$$
So we claim that
$$(xf(x))^2sim fracint_0^frac1xf^2(frac1t)frac1t^2dtfrac1x.$$
If $f^2(1/t)/t^2$ was continuous, then
$$fracint_0^frac1xf^2(frac1t)frac1t^2dtfrac1xsim f^2(x)x^2frac1/x1/x=f^2(x)x^2.$$
So maybe getting a counterexample by making $f^2(1/t)/t^2$ not well-behaved eg. blowing up at $t=0$ i.e. $f(x)>frac1x$.
answered 7 hours ago
Thomas KojarThomas Kojar
6104 silver badges15 bronze badges
$endgroup$
add a comment |
$begingroup$
Assuming lots of stuff, just to get the ball rolling:
$$frac1f(x)int_x^inftyf^2(t)dt=frac1xf(x)fracint_0^frac1xf^2(frac1t)frac1t^2dtfrac1x.$$
So we claim that
$$(xf(x))^2sim fracint_0^frac1xf^2(frac1t)frac1t^2dtfrac1x.$$
If $f^2(1/t)/t^2$ was continuous, then
$$fracint_0^frac1xf^2(frac1t)frac1t^2dtfrac1xsim f^2(x)x^2frac1/x1/x=f^2(x)x^2.$$
So maybe getting a counterexample by making $f^2(1/t)/t^2$ not well-behaved eg. blowing up at $t=0$ i.e. $f(x)>frac1x$.
answered 7 hours ago
Thomas KojarThomas Kojar
6104 silver badges15 bronze badges
$endgroup$
add a comment |
$begingroup$
Assuming lots of stuff, just to get the ball rolling:
$$frac1f(x)int_x^inftyf^2(t)dt=frac1xf(x)fracint_0^frac1xf^2(frac1t)frac1t^2dtfrac1x.$$
So we claim that
$$(xf(x))^2sim fracint_0^frac1xf^2(frac1t)frac1t^2dtfrac1x.$$
If $f^2(1/t)/t^2$ was continuous, then
$$fracint_0^frac1xf^2(frac1t)frac1t^2dtfrac1xsim f^2(x)x^2frac1/x1/x=f^2(x)x^2.$$
So maybe getting a counterexample by making $f^2(1/t)/t^2$ not well-behaved eg. blowing up at $t=0$ i.e. $f(x)>frac1x$.
answered 7 hours ago
Thomas KojarThomas Kojar
6104 silver badges15 bronze badges
$endgroup$
Assuming lots of stuff, just to get the ball rolling:
$$frac1f(x)int_x^inftyf^2(t)dt=frac1xf(x)fracint_0^frac1xf^2(frac1t)frac1t^2dtfrac1x.$$
So we claim that
$$(xf(x))^2sim fracint_0^frac1xf^2(frac1t)frac1t^2dtfrac1x.$$
If $f^2(1/t)/t^2$ was continuous, then
$$fracint_0^frac1xf^2(frac1t)frac1t^2dtfrac1xsim f^2(x)x^2frac1/x1/x=f^2(x)x^2.$$
So maybe getting a counterexample by making $f^2(1/t)/t^2$ not well-behaved eg. blowing up at $t=0$ i.e. $f(x)>frac1x$.
answered 7 hours ago
Thomas KojarThomas Kojar
6104 silver badges15 bronze badges
edited 6 hours ago
answered 7 hours ago
Thomas KojarThomas Kojar
6104 silver badges15 bronze badges
answered 7 hours ago
Thomas KojarThomas Kojar
6104 silver badges15 bronze badges
answered 7 hours ago
Thomas KojarThomas Kojar
6104 silver badges15 bronze badges
6104 silver badges15 bronze badges
add a comment |
add a comment |
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