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Schwarz symmetrization is equimeasurable
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Suppose $OmegasubsetmathbbR^2$ is open and bounded, and let $f:Omegarightarrow [0,infty)$ be measurable. Moreover, let $Omega^ast$ denote the closed disk with midpoint $0inmathbbR^2$ and the same area as $Omega$ (i.e. $vertOmegavert=vert Omega^ast vert$).
The Schwarz symmetrization is defined as the function
$$f^ast:Omega^astrightarrow[0,infty),quad f(x):=suplbrace cgeq 0:xin Omega_c^ast rbrace,$$
where $Omega_c:=f^-1([c,infty))subsetOmega$ and $Omega_c^ast $ denotes the closed disk with midpoint $0inmathbbR^2$ and $vertOmega_c^ast vert= vertOmega_c vert$.
I would like to understand the following two problems:
1) Why are $f$ and $f^ast$ equimeasurable, i.e. $vert lbrace xinOmega^ast: f^ast(x)geq crbrace vert=vert Omega_c vert$ for all $cgeq 0$ ?
2) Sometimes, the function $f^ast$ is defined alternatively as
$$f^ast(x):=int_0^infty chi_ lbrace f geq t rbrace^ast (x) dt.$$
I am wondering if that definition is equivalent to the first one?
Unfortunately, I do not know how to answer 1) or 2). I think that if we use $vertOmega_c^ast vert= vertOmega_c vert$, it might be helpful. Then it remains to show:
$$vert lbrace xinOmega^ast: f^ast(x)geq crbrace vert = vert lbrace xinOmega: f(x)geq crbrace ^ast vert$$
Are both sets within $vertcdots vert$ in the last equation the same? That is, do we have:
$$lbrace xinOmega^ast: f^ast(x)geq crbrace = lbrace xinOmega: f(x)geq crbrace ^ast$$
If the last statement is true, then 1) and 2) becomes trivial. But I do not know how to prove any of the three statements.
real-analysis measure-theory geometric-measure-theory decreasing-rearrangements
edited Aug 2 at 20:27
user357151
13.7k31140
asked Aug 1 at 14:36
Oliver Watt
1926
add a comment |Â
up vote
3
down vote
favorite
Suppose $OmegasubsetmathbbR^2$ is open and bounded, and let $f:Omegarightarrow [0,infty)$ be measurable. Moreover, let $Omega^ast$ denote the closed disk with midpoint $0inmathbbR^2$ and the same area as $Omega$ (i.e. $vertOmegavert=vert Omega^ast vert$).
The Schwarz symmetrization is defined as the function
$$f^ast:Omega^astrightarrow[0,infty),quad f(x):=suplbrace cgeq 0:xin Omega_c^ast rbrace,$$
where $Omega_c:=f^-1([c,infty))subsetOmega$ and $Omega_c^ast $ denotes the closed disk with midpoint $0inmathbbR^2$ and $vertOmega_c^ast vert= vertOmega_c vert$.
I would like to understand the following two problems:
1) Why are $f$ and $f^ast$ equimeasurable, i.e. $vert lbrace xinOmega^ast: f^ast(x)geq crbrace vert=vert Omega_c vert$ for all $cgeq 0$ ?
2) Sometimes, the function $f^ast$ is defined alternatively as
$$f^ast(x):=int_0^infty chi_ lbrace f geq t rbrace^ast (x) dt.$$
I am wondering if that definition is equivalent to the first one?
Unfortunately, I do not know how to answer 1) or 2). I think that if we use $vertOmega_c^ast vert= vertOmega_c vert$, it might be helpful. Then it remains to show:
$$vert lbrace xinOmega^ast: f^ast(x)geq crbrace vert = vert lbrace xinOmega: f(x)geq crbrace ^ast vert$$
Are both sets within $vertcdots vert$ in the last equation the same? That is, do we have:
$$lbrace xinOmega^ast: f^ast(x)geq crbrace = lbrace xinOmega: f(x)geq crbrace ^ast$$
If the last statement is true, then 1) and 2) becomes trivial. But I do not know how to prove any of the three statements.
real-analysis measure-theory geometric-measure-theory decreasing-rearrangements
edited Aug 2 at 20:27
user357151
13.7k31140
asked Aug 1 at 14:36
Oliver Watt
1926
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Suppose $OmegasubsetmathbbR^2$ is open and bounded, and let $f:Omegarightarrow [0,infty)$ be measurable. Moreover, let $Omega^ast$ denote the closed disk with midpoint $0inmathbbR^2$ and the same area as $Omega$ (i.e. $vertOmegavert=vert Omega^ast vert$).
The Schwarz symmetrization is defined as the function
$$f^ast:Omega^astrightarrow[0,infty),quad f(x):=suplbrace cgeq 0:xin Omega_c^ast rbrace,$$
where $Omega_c:=f^-1([c,infty))subsetOmega$ and $Omega_c^ast $ denotes the closed disk with midpoint $0inmathbbR^2$ and $vertOmega_c^ast vert= vertOmega_c vert$.
I would like to understand the following two problems:
1) Why are $f$ and $f^ast$ equimeasurable, i.e. $vert lbrace xinOmega^ast: f^ast(x)geq crbrace vert=vert Omega_c vert$ for all $cgeq 0$ ?
2) Sometimes, the function $f^ast$ is defined alternatively as
$$f^ast(x):=int_0^infty chi_ lbrace f geq t rbrace^ast (x) dt.$$
I am wondering if that definition is equivalent to the first one?
Unfortunately, I do not know how to answer 1) or 2). I think that if we use $vertOmega_c^ast vert= vertOmega_c vert$, it might be helpful. Then it remains to show:
$$vert lbrace xinOmega^ast: f^ast(x)geq crbrace vert = vert lbrace xinOmega: f(x)geq crbrace ^ast vert$$
Are both sets within $vertcdots vert$ in the last equation the same? That is, do we have:
$$lbrace xinOmega^ast: f^ast(x)geq crbrace = lbrace xinOmega: f(x)geq crbrace ^ast$$
If the last statement is true, then 1) and 2) becomes trivial. But I do not know how to prove any of the three statements.
real-analysis measure-theory geometric-measure-theory decreasing-rearrangements
edited Aug 2 at 20:27
user357151
13.7k31140
asked Aug 1 at 14:36
Oliver Watt
1926
Suppose $OmegasubsetmathbbR^2$ is open and bounded, and let $f:Omegarightarrow [0,infty)$ be measurable. Moreover, let $Omega^ast$ denote the closed disk with midpoint $0inmathbbR^2$ and the same area as $Omega$ (i.e. $vertOmegavert=vert Omega^ast vert$).
The Schwarz symmetrization is defined as the function
$$f^ast:Omega^astrightarrow[0,infty),quad f(x):=suplbrace cgeq 0:xin Omega_c^ast rbrace,$$
where $Omega_c:=f^-1([c,infty))subsetOmega$ and $Omega_c^ast $ denotes the closed disk with midpoint $0inmathbbR^2$ and $vertOmega_c^ast vert= vertOmega_c vert$.
I would like to understand the following two problems:
1) Why are $f$ and $f^ast$ equimeasurable, i.e. $vert lbrace xinOmega^ast: f^ast(x)geq crbrace vert=vert Omega_c vert$ for all $cgeq 0$ ?
2) Sometimes, the function $f^ast$ is defined alternatively as
$$f^ast(x):=int_0^infty chi_ lbrace f geq t rbrace^ast (x) dt.$$
I am wondering if that definition is equivalent to the first one?
Unfortunately, I do not know how to answer 1) or 2). I think that if we use $vertOmega_c^ast vert= vertOmega_c vert$, it might be helpful. Then it remains to show:
$$vert lbrace xinOmega^ast: f^ast(x)geq crbrace vert = vert lbrace xinOmega: f(x)geq crbrace ^ast vert$$
Are both sets within $vertcdots vert$ in the last equation the same? That is, do we have:
$$lbrace xinOmega^ast: f^ast(x)geq crbrace = lbrace xinOmega: f(x)geq crbrace ^ast$$
If the last statement is true, then 1) and 2) becomes trivial. But I do not know how to prove any of the three statements.
real-analysis measure-theory geometric-measure-theory decreasing-rearrangements
edited Aug 2 at 20:27
user357151
13.7k31140
asked Aug 1 at 14:36
Oliver Watt
1926
edited Aug 2 at 20:27
user357151
13.7k31140
edited Aug 2 at 20:27
user357151
13.7k31140
edited Aug 2 at 20:27
user357151
13.7k31140
13.7k31140
asked Aug 1 at 14:36
Oliver Watt
1926
asked Aug 1 at 14:36
Oliver Watt
1926
asked Aug 1 at 14:36
Oliver Watt
1926
1926
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Yes, the equality
$$ lbrace xinOmega^ast: f^ast(x)geq crbrace = lbrace xinOmega: f(x)geq crbrace ^ast $$
is true, which implies affirmative answers to your other questions. Note that the set on the right is $Omega_c^*$. So the claim is that
$$
f^*(x)ge c iff xin Omega_c^* tag1
$$
The implication $impliedby$ is immediate because $f^*(x):=suplbrace cgeq 0:xin Omega_c^ast rbrace$.
Conversely, if $f^*(x)ge c$ then for every $ninmathbbN$ it holds that $xin Omega_c-1/n^ast$. Note that
$$
Omega_c = bigcap_ninmathbbN Omega_c-1/n
$$
which by the continuity of measure implies (this is where we need the measure to be finite, which it is because of boundedness):
$$
left|Omega_cright| = inf_ninmathbbN|Omega_c-1/n|
$$
Since the measure determines the Schwarz symmetrization of these sets, we have
$$
Omega_c^* = bigcap_ninmathbbN Omega_c-1/n^*
$$
(the intersection of closed balls of radii $r_nsearrow r$ is the closed ball of radius $r$).
Thus $xin Omega_x^*$, proving (1).
answered Aug 2 at 17:36
user357151
13.7k31140
Nice answer! I do not quite understand your last argument: "Since the measure determines the Schwarz symmetrization of these sets, we have $ Omega_c^* = bigcap_ninmathbbN Omega_c-1/n^* $". Is the following explanation correct?: $$ vert Omega_c^* vert = vert Omega_c vert = inf_ninmathbbN vert Omega_c-1/nvert = inf_ninmathbbN vert Omega_c-1/n^* vert = vert bigcap_ninmathbbN Omega_c-1/n^* vert$$. Thus, both symmetrizations must be the same.
– Denilson Orr
Aug 2 at 18:55
2
I meant that the intersection of closed balls of radii $r_nsearrow r$ is the closed ball of radius $r$
– user357151
Aug 2 at 19:03
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Yes, the equality
$$ lbrace xinOmega^ast: f^ast(x)geq crbrace = lbrace xinOmega: f(x)geq crbrace ^ast $$
is true, which implies affirmative answers to your other questions. Note that the set on the right is $Omega_c^*$. So the claim is that
$$
f^*(x)ge c iff xin Omega_c^* tag1
$$
The implication $impliedby$ is immediate because $f^*(x):=suplbrace cgeq 0:xin Omega_c^ast rbrace$.
Conversely, if $f^*(x)ge c$ then for every $ninmathbbN$ it holds that $xin Omega_c-1/n^ast$. Note that
$$
Omega_c = bigcap_ninmathbbN Omega_c-1/n
$$
which by the continuity of measure implies (this is where we need the measure to be finite, which it is because of boundedness):
$$
left|Omega_cright| = inf_ninmathbbN|Omega_c-1/n|
$$
Since the measure determines the Schwarz symmetrization of these sets, we have
$$
Omega_c^* = bigcap_ninmathbbN Omega_c-1/n^*
$$
(the intersection of closed balls of radii $r_nsearrow r$ is the closed ball of radius $r$).
Thus $xin Omega_x^*$, proving (1).
answered Aug 2 at 17:36
user357151
13.7k31140
Nice answer! I do not quite understand your last argument: "Since the measure determines the Schwarz symmetrization of these sets, we have $ Omega_c^* = bigcap_ninmathbbN Omega_c-1/n^* $". Is the following explanation correct?: $$ vert Omega_c^* vert = vert Omega_c vert = inf_ninmathbbN vert Omega_c-1/nvert = inf_ninmathbbN vert Omega_c-1/n^* vert = vert bigcap_ninmathbbN Omega_c-1/n^* vert$$. Thus, both symmetrizations must be the same.
– Denilson Orr
Aug 2 at 18:55
2
I meant that the intersection of closed balls of radii $r_nsearrow r$ is the closed ball of radius $r$
– user357151
Aug 2 at 19:03
add a comment |Â
up vote
2
down vote
accepted
Yes, the equality
$$ lbrace xinOmega^ast: f^ast(x)geq crbrace = lbrace xinOmega: f(x)geq crbrace ^ast $$
is true, which implies affirmative answers to your other questions. Note that the set on the right is $Omega_c^*$. So the claim is that
$$
f^*(x)ge c iff xin Omega_c^* tag1
$$
The implication $impliedby$ is immediate because $f^*(x):=suplbrace cgeq 0:xin Omega_c^ast rbrace$.
Conversely, if $f^*(x)ge c$ then for every $ninmathbbN$ it holds that $xin Omega_c-1/n^ast$. Note that
$$
Omega_c = bigcap_ninmathbbN Omega_c-1/n
$$
which by the continuity of measure implies (this is where we need the measure to be finite, which it is because of boundedness):
$$
left|Omega_cright| = inf_ninmathbbN|Omega_c-1/n|
$$
Since the measure determines the Schwarz symmetrization of these sets, we have
$$
Omega_c^* = bigcap_ninmathbbN Omega_c-1/n^*
$$
(the intersection of closed balls of radii $r_nsearrow r$ is the closed ball of radius $r$).
Thus $xin Omega_x^*$, proving (1).
answered Aug 2 at 17:36
user357151
13.7k31140
Nice answer! I do not quite understand your last argument: "Since the measure determines the Schwarz symmetrization of these sets, we have $ Omega_c^* = bigcap_ninmathbbN Omega_c-1/n^* $". Is the following explanation correct?: $$ vert Omega_c^* vert = vert Omega_c vert = inf_ninmathbbN vert Omega_c-1/nvert = inf_ninmathbbN vert Omega_c-1/n^* vert = vert bigcap_ninmathbbN Omega_c-1/n^* vert$$. Thus, both symmetrizations must be the same.
– Denilson Orr
Aug 2 at 18:55
2
I meant that the intersection of closed balls of radii $r_nsearrow r$ is the closed ball of radius $r$
– user357151
Aug 2 at 19:03
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Yes, the equality
$$ lbrace xinOmega^ast: f^ast(x)geq crbrace = lbrace xinOmega: f(x)geq crbrace ^ast $$
is true, which implies affirmative answers to your other questions. Note that the set on the right is $Omega_c^*$. So the claim is that
$$
f^*(x)ge c iff xin Omega_c^* tag1
$$
The implication $impliedby$ is immediate because $f^*(x):=suplbrace cgeq 0:xin Omega_c^ast rbrace$.
Conversely, if $f^*(x)ge c$ then for every $ninmathbbN$ it holds that $xin Omega_c-1/n^ast$. Note that
$$
Omega_c = bigcap_ninmathbbN Omega_c-1/n
$$
which by the continuity of measure implies (this is where we need the measure to be finite, which it is because of boundedness):
$$
left|Omega_cright| = inf_ninmathbbN|Omega_c-1/n|
$$
Since the measure determines the Schwarz symmetrization of these sets, we have
$$
Omega_c^* = bigcap_ninmathbbN Omega_c-1/n^*
$$
(the intersection of closed balls of radii $r_nsearrow r$ is the closed ball of radius $r$).
Thus $xin Omega_x^*$, proving (1).
answered Aug 2 at 17:36
user357151
13.7k31140
Yes, the equality
$$ lbrace xinOmega^ast: f^ast(x)geq crbrace = lbrace xinOmega: f(x)geq crbrace ^ast $$
is true, which implies affirmative answers to your other questions. Note that the set on the right is $Omega_c^*$. So the claim is that
$$
f^*(x)ge c iff xin Omega_c^* tag1
$$
The implication $impliedby$ is immediate because $f^*(x):=suplbrace cgeq 0:xin Omega_c^ast rbrace$.
Conversely, if $f^*(x)ge c$ then for every $ninmathbbN$ it holds that $xin Omega_c-1/n^ast$. Note that
$$
Omega_c = bigcap_ninmathbbN Omega_c-1/n
$$
which by the continuity of measure implies (this is where we need the measure to be finite, which it is because of boundedness):
$$
left|Omega_cright| = inf_ninmathbbN|Omega_c-1/n|
$$
Since the measure determines the Schwarz symmetrization of these sets, we have
$$
Omega_c^* = bigcap_ninmathbbN Omega_c-1/n^*
$$
(the intersection of closed balls of radii $r_nsearrow r$ is the closed ball of radius $r$).
Thus $xin Omega_x^*$, proving (1).
answered Aug 2 at 17:36
user357151
13.7k31140
edited Aug 2 at 19:03
answered Aug 2 at 17:36
user357151
13.7k31140
answered Aug 2 at 17:36
user357151
13.7k31140
answered Aug 2 at 17:36
user357151
13.7k31140
13.7k31140
Nice answer! I do not quite understand your last argument: "Since the measure determines the Schwarz symmetrization of these sets, we have $ Omega_c^* = bigcap_ninmathbbN Omega_c-1/n^* $". Is the following explanation correct?: $$ vert Omega_c^* vert = vert Omega_c vert = inf_ninmathbbN vert Omega_c-1/nvert = inf_ninmathbbN vert Omega_c-1/n^* vert = vert bigcap_ninmathbbN Omega_c-1/n^* vert$$. Thus, both symmetrizations must be the same.
– Denilson Orr
Aug 2 at 18:55
2
I meant that the intersection of closed balls of radii $r_nsearrow r$ is the closed ball of radius $r$
– user357151
Aug 2 at 19:03
add a comment |Â
Nice answer! I do not quite understand your last argument: "Since the measure determines the Schwarz symmetrization of these sets, we have $ Omega_c^* = bigcap_ninmathbbN Omega_c-1/n^* $". Is the following explanation correct?: $$ vert Omega_c^* vert = vert Omega_c vert = inf_ninmathbbN vert Omega_c-1/nvert = inf_ninmathbbN vert Omega_c-1/n^* vert = vert bigcap_ninmathbbN Omega_c-1/n^* vert$$. Thus, both symmetrizations must be the same.
– Denilson Orr
Aug 2 at 18:55
2
I meant that the intersection of closed balls of radii $r_nsearrow r$ is the closed ball of radius $r$
– user357151
Aug 2 at 19:03
Nice answer! I do not quite understand your last argument: "Since the measure determines the Schwarz symmetrization of these sets, we have $ Omega_c^* = bigcap_ninmathbbN Omega_c-1/n^* $". Is the following explanation correct?: $$ vert Omega_c^* vert = vert Omega_c vert = inf_ninmathbbN vert Omega_c-1/nvert = inf_ninmathbbN vert Omega_c-1/n^* vert = vert bigcap_ninmathbbN Omega_c-1/n^* vert$$. Thus, both symmetrizations must be the same.
– Denilson Orr
Aug 2 at 18:55
Nice answer! I do not quite understand your last argument: "Since the measure determines the Schwarz symmetrization of these sets, we have $ Omega_c^* = bigcap_ninmathbbN Omega_c-1/n^* $". Is the following explanation correct?: $$ vert Omega_c^* vert = vert Omega_c vert = inf_ninmathbbN vert Omega_c-1/nvert = inf_ninmathbbN vert Omega_c-1/n^* vert = vert bigcap_ninmathbbN Omega_c-1/n^* vert$$. Thus, both symmetrizations must be the same.
– Denilson Orr
Aug 2 at 18:55
2
2
I meant that the intersection of closed balls of radii $r_nsearrow r$ is the closed ball of radius $r$
– user357151
Aug 2 at 19:03
I meant that the intersection of closed balls of radii $r_nsearrow r$ is the closed ball of radius $r$
– user357151
Aug 2 at 19:03
add a comment |Â
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