Is there a version of the rearrangement inequality for nonnegative functions on [0,1]?
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
add a comment |
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
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
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
1,148510
1,148510
add a comment |
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3007277%2fis-there-a-version-of-the-rearrangement-inequality-for-nonnegative-functions-on%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown