Is there a version of the rearrangement inequality for nonnegative functions on [0,1]?
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First, let us recall the sequence version of the rearrangement inequality.
Rearrangement inequality for nonnegative finite sequences: Let $Ninmathbb{N}$ with $x_1leq x_2leqcdotsleq x_N$ and $y_1leq y_2leqcdotsleq y_N$ nondecreasing sequences of nonnegative real numbers. If $sigma$ is a permutation of ${1,2,cdots,N}$ then we have
$$sum_{i=1}^Nx_{N+1-i}y_ileqsum_{i=1}^Nx_{sigma(i)}y_ileqsum_{i=1}^Nx_iy_i.$$
(Actually, this is valid for finite real-valued sequences, not just nonnegative ones. But I wish to restrict my attention to the nonnegative case for now.)
A rearrangement inequality for measurable nonnegative functions on $[0,1]$ may be conjectured as follows. If $f:[0,1]to[-infty,infty]$ is measurable then we define its distribution function $mu_f:[0,infty]to[0,infty]$ by the rule
$$mu_f(t)=muleft{xin[0,1]:|f(x)|>tright}$$
where $mu$ is the standard Lebesgue measure. Two functions are called equimeasurable of they have the same distribution function. It is known, in particular, that $f:[0,1]to[-infty,infty]$ is equimeasurable with its nonincreasing rearrangement $f^*:[0,1]to[0,infty]$ defined by
$$f^*(x)=infleft{t:mu_f(t)leq xright}.$$
Hence, it is also equimeasurable with what we might call its nondecreasing rearrangement $f_*:[0,1]to[0,infty]$ defined by the rule
$$f_*(x)=f^*(1-x).$$
Observe that the rearrangement inequality for nonnegative finite sequences is equivalent to the following: If $f:{1,2,cdots,N}to[0,infty]$ and $g:{1,2,cdots,N}to[0,infty]$ are measurable functions, then
$$sum_{i=1}^Nf_*(i)g^*(i)leq sum_{i=1}^Nf(i)g^*(i)leqsum_{i=1}^Nf^*(i)g^*(i),$$
where the definitions above are extended in the obvious way to measurable functions $f,g:{1,cdots,N}to[0,infty]$. I conjecture the following.
Conjecture. Let $f,g:[0,1]to[0,infty]$ be nonnegative measurable functions. Then
$$int_0^1f_*(x)g^*(x);dxleq int_0^1f(x)g^*(x);dxleqint_0^1f^*(x)g^*(x);dx.$$
The last inequality is just Hardy-Littlewood. But I am curious if it is known whether the first inequality holds, i.e. whether
$$int_0^1f_*(x)g^*(x);dxleq int_0^1f(x)g^*(x);dx.$$
Thanks!
real-analysis measure-theory
asked Nov 21 at 4:53
Ben W
1,263513
add a comment |
up vote
2
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First, let us recall the sequence version of the rearrangement inequality.
Rearrangement inequality for nonnegative finite sequences: Let $Ninmathbb{N}$ with $x_1leq x_2leqcdotsleq x_N$ and $y_1leq y_2leqcdotsleq y_N$ nondecreasing sequences of nonnegative real numbers. If $sigma$ is a permutation of ${1,2,cdots,N}$ then we have
$$sum_{i=1}^Nx_{N+1-i}y_ileqsum_{i=1}^Nx_{sigma(i)}y_ileqsum_{i=1}^Nx_iy_i.$$
(Actually, this is valid for finite real-valued sequences, not just nonnegative ones. But I wish to restrict my attention to the nonnegative case for now.)
A rearrangement inequality for measurable nonnegative functions on $[0,1]$ may be conjectured as follows. If $f:[0,1]to[-infty,infty]$ is measurable then we define its distribution function $mu_f:[0,infty]to[0,infty]$ by the rule
$$mu_f(t)=muleft{xin[0,1]:|f(x)|>tright}$$
where $mu$ is the standard Lebesgue measure. Two functions are called equimeasurable of they have the same distribution function. It is known, in particular, that $f:[0,1]to[-infty,infty]$ is equimeasurable with its nonincreasing rearrangement $f^*:[0,1]to[0,infty]$ defined by
$$f^*(x)=infleft{t:mu_f(t)leq xright}.$$
Hence, it is also equimeasurable with what we might call its nondecreasing rearrangement $f_*:[0,1]to[0,infty]$ defined by the rule
$$f_*(x)=f^*(1-x).$$
Observe that the rearrangement inequality for nonnegative finite sequences is equivalent to the following: If $f:{1,2,cdots,N}to[0,infty]$ and $g:{1,2,cdots,N}to[0,infty]$ are measurable functions, then
$$sum_{i=1}^Nf_*(i)g^*(i)leq sum_{i=1}^Nf(i)g^*(i)leqsum_{i=1}^Nf^*(i)g^*(i),$$
where the definitions above are extended in the obvious way to measurable functions $f,g:{1,cdots,N}to[0,infty]$. I conjecture the following.
Conjecture. Let $f,g:[0,1]to[0,infty]$ be nonnegative measurable functions. Then
$$int_0^1f_*(x)g^*(x);dxleq int_0^1f(x)g^*(x);dxleqint_0^1f^*(x)g^*(x);dx.$$
The last inequality is just Hardy-Littlewood. But I am curious if it is known whether the first inequality holds, i.e. whether
$$int_0^1f_*(x)g^*(x);dxleq int_0^1f(x)g^*(x);dx.$$
Thanks!
real-analysis measure-theory
asked Nov 21 at 4:53
Ben W
1,263513
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
First, let us recall the sequence version of the rearrangement inequality.
Rearrangement inequality for nonnegative finite sequences: Let $Ninmathbb{N}$ with $x_1leq x_2leqcdotsleq x_N$ and $y_1leq y_2leqcdotsleq y_N$ nondecreasing sequences of nonnegative real numbers. If $sigma$ is a permutation of ${1,2,cdots,N}$ then we have
$$sum_{i=1}^Nx_{N+1-i}y_ileqsum_{i=1}^Nx_{sigma(i)}y_ileqsum_{i=1}^Nx_iy_i.$$
(Actually, this is valid for finite real-valued sequences, not just nonnegative ones. But I wish to restrict my attention to the nonnegative case for now.)
A rearrangement inequality for measurable nonnegative functions on $[0,1]$ may be conjectured as follows. If $f:[0,1]to[-infty,infty]$ is measurable then we define its distribution function $mu_f:[0,infty]to[0,infty]$ by the rule
$$mu_f(t)=muleft{xin[0,1]:|f(x)|>tright}$$
where $mu$ is the standard Lebesgue measure. Two functions are called equimeasurable of they have the same distribution function. It is known, in particular, that $f:[0,1]to[-infty,infty]$ is equimeasurable with its nonincreasing rearrangement $f^*:[0,1]to[0,infty]$ defined by
$$f^*(x)=infleft{t:mu_f(t)leq xright}.$$
Hence, it is also equimeasurable with what we might call its nondecreasing rearrangement $f_*:[0,1]to[0,infty]$ defined by the rule
$$f_*(x)=f^*(1-x).$$
Observe that the rearrangement inequality for nonnegative finite sequences is equivalent to the following: If $f:{1,2,cdots,N}to[0,infty]$ and $g:{1,2,cdots,N}to[0,infty]$ are measurable functions, then
$$sum_{i=1}^Nf_*(i)g^*(i)leq sum_{i=1}^Nf(i)g^*(i)leqsum_{i=1}^Nf^*(i)g^*(i),$$
where the definitions above are extended in the obvious way to measurable functions $f,g:{1,cdots,N}to[0,infty]$. I conjecture the following.
Conjecture. Let $f,g:[0,1]to[0,infty]$ be nonnegative measurable functions. Then
$$int_0^1f_*(x)g^*(x);dxleq int_0^1f(x)g^*(x);dxleqint_0^1f^*(x)g^*(x);dx.$$
The last inequality is just Hardy-Littlewood. But I am curious if it is known whether the first inequality holds, i.e. whether
$$int_0^1f_*(x)g^*(x);dxleq int_0^1f(x)g^*(x);dx.$$
Thanks!
real-analysis measure-theory
asked Nov 21 at 4:53
Ben W
1,263513
First, let us recall the sequence version of the rearrangement inequality.
Rearrangement inequality for nonnegative finite sequences: Let $Ninmathbb{N}$ with $x_1leq x_2leqcdotsleq x_N$ and $y_1leq y_2leqcdotsleq y_N$ nondecreasing sequences of nonnegative real numbers. If $sigma$ is a permutation of ${1,2,cdots,N}$ then we have
$$sum_{i=1}^Nx_{N+1-i}y_ileqsum_{i=1}^Nx_{sigma(i)}y_ileqsum_{i=1}^Nx_iy_i.$$
(Actually, this is valid for finite real-valued sequences, not just nonnegative ones. But I wish to restrict my attention to the nonnegative case for now.)
A rearrangement inequality for measurable nonnegative functions on $[0,1]$ may be conjectured as follows. If $f:[0,1]to[-infty,infty]$ is measurable then we define its distribution function $mu_f:[0,infty]to[0,infty]$ by the rule
$$mu_f(t)=muleft{xin[0,1]:|f(x)|>tright}$$
where $mu$ is the standard Lebesgue measure. Two functions are called equimeasurable of they have the same distribution function. It is known, in particular, that $f:[0,1]to[-infty,infty]$ is equimeasurable with its nonincreasing rearrangement $f^*:[0,1]to[0,infty]$ defined by
$$f^*(x)=infleft{t:mu_f(t)leq xright}.$$
Hence, it is also equimeasurable with what we might call its nondecreasing rearrangement $f_*:[0,1]to[0,infty]$ defined by the rule
$$f_*(x)=f^*(1-x).$$
Observe that the rearrangement inequality for nonnegative finite sequences is equivalent to the following: If $f:{1,2,cdots,N}to[0,infty]$ and $g:{1,2,cdots,N}to[0,infty]$ are measurable functions, then
$$sum_{i=1}^Nf_*(i)g^*(i)leq sum_{i=1}^Nf(i)g^*(i)leqsum_{i=1}^Nf^*(i)g^*(i),$$
where the definitions above are extended in the obvious way to measurable functions $f,g:{1,cdots,N}to[0,infty]$. I conjecture the following.
Conjecture. Let $f,g:[0,1]to[0,infty]$ be nonnegative measurable functions. Then
$$int_0^1f_*(x)g^*(x);dxleq int_0^1f(x)g^*(x);dxleqint_0^1f^*(x)g^*(x);dx.$$
The last inequality is just Hardy-Littlewood. But I am curious if it is known whether the first inequality holds, i.e. whether
$$int_0^1f_*(x)g^*(x);dxleq int_0^1f(x)g^*(x);dx.$$
Thanks!
real-analysis measure-theory
real-analysis measure-theory
asked Nov 21 at 4:53
Ben W
1,263513
asked Nov 21 at 4:53
Ben W
1,263513
asked Nov 21 at 4:53
Ben W
1,263513
asked Nov 21 at 4:53
Ben W
1,263513
asked Nov 21 at 4:53
Ben W
1,263513
1,263513
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