Disjoint Union of Dyadic Cubes in Calderon-Zygmund Decomposition

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In proving the existence of a countable collection of disjoint dyadic cubes which we denote $B_j$ with $f geq 0, f in L^1(mathbbR^n), lambda > 0$ and we define $E_tf:= sum_B in textdyadic cubes(frac1mu(B)int_Bf)chi_B$ where $chi$ denotes the indicator function. satisfying:

1. $ f leq lambda a.e.$ on $(cup B_j)^c$


2. $mu(cup B_j) leq frac1lambda||f||_L^1$
   3.$lambda < frac1mu(B_j int_b_jf leq 2^n lambda. forall j$

One defines the set $Omega_t:= x in mathbbR^n: E_tf(x) > lambda, E_jf(x) leq lambda, forall j < t$ i.e. the larget dyadic cube whose average is geater than $ lambda$.

Question: The $Omega_t$ is a disjoint union of elements of $textdyadic cubes$. What does this mean? I'm only familiar with disjoint union in the sense given in the following link: https://en.wikipedia.org/wiki/Disjoint_union

real-analysis harmonic-analysis

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edited Jul 26 at 20:04

Asaf Karagila

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asked Jul 26 at 20:00

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In proving the existence of a countable collection of disjoint dyadic cubes which we denote $B_j$ with $f geq 0, f in L^1(mathbbR^n), lambda > 0$ and we define $E_tf:= sum_B in textdyadic cubes(frac1mu(B)int_Bf)chi_B$ where $chi$ denotes the indicator function. satisfying:

1. $ f leq lambda a.e.$ on $(cup B_j)^c$


2. $mu(cup B_j) leq frac1lambda||f||_L^1$
   3.$lambda < frac1mu(B_j int_b_jf leq 2^n lambda. forall j$

One defines the set $Omega_t:= x in mathbbR^n: E_tf(x) > lambda, E_jf(x) leq lambda, forall j < t$ i.e. the larget dyadic cube whose average is geater than $ lambda$.

Question: The $Omega_t$ is a disjoint union of elements of $textdyadic cubes$. What does this mean? I'm only familiar with disjoint union in the sense given in the following link: https://en.wikipedia.org/wiki/Disjoint_union

real-analysis harmonic-analysis

share|cite|improve this question

edited Jul 26 at 20:04

Asaf Karagila

291k31402732

asked Jul 26 at 20:00

VBACODER

748

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

In proving the existence of a countable collection of disjoint dyadic cubes which we denote $B_j$ with $f geq 0, f in L^1(mathbbR^n), lambda > 0$ and we define $E_tf:= sum_B in textdyadic cubes(frac1mu(B)int_Bf)chi_B$ where $chi$ denotes the indicator function. satisfying:

1. $ f leq lambda a.e.$ on $(cup B_j)^c$


2. $mu(cup B_j) leq frac1lambda||f||_L^1$
   3.$lambda < frac1mu(B_j int_b_jf leq 2^n lambda. forall j$

One defines the set $Omega_t:= x in mathbbR^n: E_tf(x) > lambda, E_jf(x) leq lambda, forall j < t$ i.e. the larget dyadic cube whose average is geater than $ lambda$.

Question: The $Omega_t$ is a disjoint union of elements of $textdyadic cubes$. What does this mean? I'm only familiar with disjoint union in the sense given in the following link: https://en.wikipedia.org/wiki/Disjoint_union

real-analysis harmonic-analysis

share|cite|improve this question

edited Jul 26 at 20:04

Asaf Karagila

291k31402732

asked Jul 26 at 20:00

VBACODER

748

In proving the existence of a countable collection of disjoint dyadic cubes which we denote $B_j$ with $f geq 0, f in L^1(mathbbR^n), lambda > 0$ and we define $E_tf:= sum_B in textdyadic cubes(frac1mu(B)int_Bf)chi_B$ where $chi$ denotes the indicator function. satisfying:

1. $ f leq lambda a.e.$ on $(cup B_j)^c$


2. $mu(cup B_j) leq frac1lambda||f||_L^1$
   3.$lambda < frac1mu(B_j int_b_jf leq 2^n lambda. forall j$

One defines the set $Omega_t:= x in mathbbR^n: E_tf(x) > lambda, E_jf(x) leq lambda, forall j < t$ i.e. the larget dyadic cube whose average is geater than $ lambda$.

Question: The $Omega_t$ is a disjoint union of elements of $textdyadic cubes$. What does this mean? I'm only familiar with disjoint union in the sense given in the following link: https://en.wikipedia.org/wiki/Disjoint_union

real-analysis harmonic-analysis

share|cite|improve this question

edited Jul 26 at 20:04

Asaf Karagila

291k31402732

asked Jul 26 at 20:00

VBACODER

748

share|cite|improve this question

share|cite|improve this question

edited Jul 26 at 20:04

Asaf Karagila

291k31402732

edited Jul 26 at 20:04

Asaf Karagila

291k31402732

edited Jul 26 at 20:04

Asaf Karagila

291k31402732

291k31402732

asked Jul 26 at 20:00

VBACODER

748

asked Jul 26 at 20:00

VBACODER

748

asked Jul 26 at 20:00

VBACODER

748

748

add a comment | 

add a comment | 

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There is something wrong in your definition of $E_t f$. It does not depend on $t$... I am copying below a proof of Calderon-Zygmund theorem.

Theorem Calderon-Zygmund
Let $fin L^1left(mathbbR^Nright) $ be a nonnegative function, and let $lambda>0$. Then there
exists a countable family $left Q_nright $ of open mutually
disjoint cubes such that
beginequation
fleft( xright) leq lambda quad text for mathcalL^Ntext a.e. %
xinmathbbR^Nsetminusbigcup_n=1^inftyoverlineQ_n,
endequation
and for every $ninmathbbN$,
beginequation
lambda<frac1leftvert Q_nrightvert int_Q_nfleft( xright)
,dxleq2^Nlambdatext.
endequation
Proof: Choose $L>0$ so large enough
$$
int_mathbbR^Nfleft( xright) ,dxleq lambda L^Ntext.%
$$
Decompose $mathbbR^N$ into a rectangular grid such that each cube $Q$ of
the partition has side length $L$. Since $fge 0$ we have
beginequation
frac1 int_Qfleft( xright) ,dx=frac1L^N int_Qfleft( xright) ,dxle frac1L^N int_mathbbR^Nfleft( xright) ,dx
leq lambda .
endequation
Fix one such cube $Q$ and subdivide it into $2^N$ congruent subcubes. Let
$Q^prime$ be one of these subcubes. If
$$
frac1 int_Q^primefleft( xright)
,dx>lambda,
$$
and in view of the fact that
$$
frac1 int_Q^primefleft( xright)
,dxleqfrac2^N int_Qfleft( xright)
,dxleq2^Nlambda,
$$
then $Q^prime$ will be selected
as one of the good cubes $Q_n$. On the other hand, if
$$
frac1 int_Q^primefleft( xright)
,dxleq lambda
$$
(note that since $int_mathbbR^Nfleft( xright) ,dxleq lambda L^N$ there is at least one), then we subdivide
$Q^prime$ into $2^N$ congruent subcubes and we repeat the process.

In this way we construct a family of cubes $left Q_nright $ for
which beginequation
lambda<frac1leftvert Q_nrightvert int_Q_nfleft( xright)
,dxleq2^Nlambdatext.
endequation is satisfied, and it remains to prove $fleft( xright) leq lambda$ for $mathcalL^N$ a.e. $
xinmathbbR^Nsetminusbigcup_n=1^inftyoverlineQ_n$. It
can be seen from the construction that the cubes that were not selected to
belong to the family $left Q_nright $ form a fine covering
$mathcalF$ of $mathbbR^Nsetminusbigcup_n=1^inftyoverline
Q_n$. Therefore if $xinmathbbR^Nsetminusbigcup_n=1^infty
overlineQ_n$ is a Lebesgue point for $f$, then by Lebesgue differentiation theorem we have
$$
fleft( xright) =limsup_operatorname*diamFrightarrow0text,%
,FinmathcalFtext, xin Ffrac1leftint_Ffleft(
yright) ,dyleq lambda
$$
and the proof is completed.

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answered Jul 27 at 2:48

Gio67

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There is something wrong in your definition of $E_t f$. It does not depend on $t$... I am copying below a proof of Calderon-Zygmund theorem.

Theorem Calderon-Zygmund
Let $fin L^1left(mathbbR^Nright) $ be a nonnegative function, and let $lambda>0$. Then there
exists a countable family $left Q_nright $ of open mutually
disjoint cubes such that
beginequation
fleft( xright) leq lambda quad text for mathcalL^Ntext a.e. %
xinmathbbR^Nsetminusbigcup_n=1^inftyoverlineQ_n,
endequation
and for every $ninmathbbN$,
beginequation
lambda<frac1leftvert Q_nrightvert int_Q_nfleft( xright)
,dxleq2^Nlambdatext.
endequation
Proof: Choose $L>0$ so large enough
$$
int_mathbbR^Nfleft( xright) ,dxleq lambda L^Ntext.%
$$
Decompose $mathbbR^N$ into a rectangular grid such that each cube $Q$ of
the partition has side length $L$. Since $fge 0$ we have
beginequation
frac1 int_Qfleft( xright) ,dx=frac1L^N int_Qfleft( xright) ,dxle frac1L^N int_mathbbR^Nfleft( xright) ,dx
leq lambda .
endequation
Fix one such cube $Q$ and subdivide it into $2^N$ congruent subcubes. Let
$Q^prime$ be one of these subcubes. If
$$
frac1 int_Q^primefleft( xright)
,dx>lambda,
$$
and in view of the fact that
$$
frac1 int_Q^primefleft( xright)
,dxleqfrac2^N int_Qfleft( xright)
,dxleq2^Nlambda,
$$
then $Q^prime$ will be selected
as one of the good cubes $Q_n$. On the other hand, if
$$
frac1 int_Q^primefleft( xright)
,dxleq lambda
$$
(note that since $int_mathbbR^Nfleft( xright) ,dxleq lambda L^N$ there is at least one), then we subdivide
$Q^prime$ into $2^N$ congruent subcubes and we repeat the process.

In this way we construct a family of cubes $left Q_nright $ for
which beginequation
lambda<frac1leftvert Q_nrightvert int_Q_nfleft( xright)
,dxleq2^Nlambdatext.
endequation is satisfied, and it remains to prove $fleft( xright) leq lambda$ for $mathcalL^N$ a.e. $
xinmathbbR^Nsetminusbigcup_n=1^inftyoverlineQ_n$. It
can be seen from the construction that the cubes that were not selected to
belong to the family $left Q_nright $ form a fine covering
$mathcalF$ of $mathbbR^Nsetminusbigcup_n=1^inftyoverline
Q_n$. Therefore if $xinmathbbR^Nsetminusbigcup_n=1^infty
overlineQ_n$ is a Lebesgue point for $f$, then by Lebesgue differentiation theorem we have
$$
fleft( xright) =limsup_operatorname*diamFrightarrow0text,%
,FinmathcalFtext, xin Ffrac1leftint_Ffleft(
yright) ,dyleq lambda
$$
and the proof is completed.

share|cite|improve this answer

answered Jul 27 at 2:48

Gio67

11.2k1526

add a comment | 

up vote
1
down vote



accepted

There is something wrong in your definition of $E_t f$. It does not depend on $t$... I am copying below a proof of Calderon-Zygmund theorem.

Theorem Calderon-Zygmund
Let $fin L^1left(mathbbR^Nright) $ be a nonnegative function, and let $lambda>0$. Then there
exists a countable family $left Q_nright $ of open mutually
disjoint cubes such that
beginequation
fleft( xright) leq lambda quad text for mathcalL^Ntext a.e. %
xinmathbbR^Nsetminusbigcup_n=1^inftyoverlineQ_n,
endequation
and for every $ninmathbbN$,
beginequation
lambda<frac1leftvert Q_nrightvert int_Q_nfleft( xright)
,dxleq2^Nlambdatext.
endequation
Proof: Choose $L>0$ so large enough
$$
int_mathbbR^Nfleft( xright) ,dxleq lambda L^Ntext.%
$$
Decompose $mathbbR^N$ into a rectangular grid such that each cube $Q$ of
the partition has side length $L$. Since $fge 0$ we have
beginequation
frac1 int_Qfleft( xright) ,dx=frac1L^N int_Qfleft( xright) ,dxle frac1L^N int_mathbbR^Nfleft( xright) ,dx
leq lambda .
endequation
Fix one such cube $Q$ and subdivide it into $2^N$ congruent subcubes. Let
$Q^prime$ be one of these subcubes. If
$$
frac1 int_Q^primefleft( xright)
,dx>lambda,
$$
and in view of the fact that
$$
frac1 int_Q^primefleft( xright)
,dxleqfrac2^N int_Qfleft( xright)
,dxleq2^Nlambda,
$$
then $Q^prime$ will be selected
as one of the good cubes $Q_n$. On the other hand, if
$$
frac1 int_Q^primefleft( xright)
,dxleq lambda
$$
(note that since $int_mathbbR^Nfleft( xright) ,dxleq lambda L^N$ there is at least one), then we subdivide
$Q^prime$ into $2^N$ congruent subcubes and we repeat the process.

In this way we construct a family of cubes $left Q_nright $ for
which beginequation
lambda<frac1leftvert Q_nrightvert int_Q_nfleft( xright)
,dxleq2^Nlambdatext.
endequation is satisfied, and it remains to prove $fleft( xright) leq lambda$ for $mathcalL^N$ a.e. $
xinmathbbR^Nsetminusbigcup_n=1^inftyoverlineQ_n$. It
can be seen from the construction that the cubes that were not selected to
belong to the family $left Q_nright $ form a fine covering
$mathcalF$ of $mathbbR^Nsetminusbigcup_n=1^inftyoverline
Q_n$. Therefore if $xinmathbbR^Nsetminusbigcup_n=1^infty
overlineQ_n$ is a Lebesgue point for $f$, then by Lebesgue differentiation theorem we have
$$
fleft( xright) =limsup_operatorname*diamFrightarrow0text,%
,FinmathcalFtext, xin Ffrac1leftint_Ffleft(
yright) ,dyleq lambda
$$
and the proof is completed.

share|cite|improve this answer

answered Jul 27 at 2:48

Gio67

11.2k1526

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

There is something wrong in your definition of $E_t f$. It does not depend on $t$... I am copying below a proof of Calderon-Zygmund theorem.

Theorem Calderon-Zygmund
Let $fin L^1left(mathbbR^Nright) $ be a nonnegative function, and let $lambda>0$. Then there
exists a countable family $left Q_nright $ of open mutually
disjoint cubes such that
beginequation
fleft( xright) leq lambda quad text for mathcalL^Ntext a.e. %
xinmathbbR^Nsetminusbigcup_n=1^inftyoverlineQ_n,
endequation
and for every $ninmathbbN$,
beginequation
lambda<frac1leftvert Q_nrightvert int_Q_nfleft( xright)
,dxleq2^Nlambdatext.
endequation
Proof: Choose $L>0$ so large enough
$$
int_mathbbR^Nfleft( xright) ,dxleq lambda L^Ntext.%
$$
Decompose $mathbbR^N$ into a rectangular grid such that each cube $Q$ of
the partition has side length $L$. Since $fge 0$ we have
beginequation
frac1 int_Qfleft( xright) ,dx=frac1L^N int_Qfleft( xright) ,dxle frac1L^N int_mathbbR^Nfleft( xright) ,dx
leq lambda .
endequation
Fix one such cube $Q$ and subdivide it into $2^N$ congruent subcubes. Let
$Q^prime$ be one of these subcubes. If
$$
frac1 int_Q^primefleft( xright)
,dx>lambda,
$$
and in view of the fact that
$$
frac1 int_Q^primefleft( xright)
,dxleqfrac2^N int_Qfleft( xright)
,dxleq2^Nlambda,
$$
then $Q^prime$ will be selected
as one of the good cubes $Q_n$. On the other hand, if
$$
frac1 int_Q^primefleft( xright)
,dxleq lambda
$$
(note that since $int_mathbbR^Nfleft( xright) ,dxleq lambda L^N$ there is at least one), then we subdivide
$Q^prime$ into $2^N$ congruent subcubes and we repeat the process.

In this way we construct a family of cubes $left Q_nright $ for
which beginequation
lambda<frac1leftvert Q_nrightvert int_Q_nfleft( xright)
,dxleq2^Nlambdatext.
endequation is satisfied, and it remains to prove $fleft( xright) leq lambda$ for $mathcalL^N$ a.e. $
xinmathbbR^Nsetminusbigcup_n=1^inftyoverlineQ_n$. It
can be seen from the construction that the cubes that were not selected to
belong to the family $left Q_nright $ form a fine covering
$mathcalF$ of $mathbbR^Nsetminusbigcup_n=1^inftyoverline
Q_n$. Therefore if $xinmathbbR^Nsetminusbigcup_n=1^infty
overlineQ_n$ is a Lebesgue point for $f$, then by Lebesgue differentiation theorem we have
$$
fleft( xright) =limsup_operatorname*diamFrightarrow0text,%
,FinmathcalFtext, xin Ffrac1leftint_Ffleft(
yright) ,dyleq lambda
$$
and the proof is completed.

share|cite|improve this answer

answered Jul 27 at 2:48

Gio67

11.2k1526

There is something wrong in your definition of $E_t f$. It does not depend on $t$... I am copying below a proof of Calderon-Zygmund theorem.

Theorem Calderon-Zygmund
Let $fin L^1left(mathbbR^Nright) $ be a nonnegative function, and let $lambda>0$. Then there
exists a countable family $left Q_nright $ of open mutually
disjoint cubes such that
beginequation
fleft( xright) leq lambda quad text for mathcalL^Ntext a.e. %
xinmathbbR^Nsetminusbigcup_n=1^inftyoverlineQ_n,
endequation
and for every $ninmathbbN$,
beginequation
lambda<frac1leftvert Q_nrightvert int_Q_nfleft( xright)
,dxleq2^Nlambdatext.
endequation
Proof: Choose $L>0$ so large enough
$$
int_mathbbR^Nfleft( xright) ,dxleq lambda L^Ntext.%
$$
Decompose $mathbbR^N$ into a rectangular grid such that each cube $Q$ of
the partition has side length $L$. Since $fge 0$ we have
beginequation
frac1 int_Qfleft( xright) ,dx=frac1L^N int_Qfleft( xright) ,dxle frac1L^N int_mathbbR^Nfleft( xright) ,dx
leq lambda .
endequation
Fix one such cube $Q$ and subdivide it into $2^N$ congruent subcubes. Let
$Q^prime$ be one of these subcubes. If
$$
frac1 int_Q^primefleft( xright)
,dx>lambda,
$$
and in view of the fact that
$$
frac1 int_Q^primefleft( xright)
,dxleqfrac2^N int_Qfleft( xright)
,dxleq2^Nlambda,
$$
then $Q^prime$ will be selected
as one of the good cubes $Q_n$. On the other hand, if
$$
frac1 int_Q^primefleft( xright)
,dxleq lambda
$$
(note that since $int_mathbbR^Nfleft( xright) ,dxleq lambda L^N$ there is at least one), then we subdivide
$Q^prime$ into $2^N$ congruent subcubes and we repeat the process.

In this way we construct a family of cubes $left Q_nright $ for
which beginequation
lambda<frac1leftvert Q_nrightvert int_Q_nfleft( xright)
,dxleq2^Nlambdatext.
endequation is satisfied, and it remains to prove $fleft( xright) leq lambda$ for $mathcalL^N$ a.e. $
xinmathbbR^Nsetminusbigcup_n=1^inftyoverlineQ_n$. It
can be seen from the construction that the cubes that were not selected to
belong to the family $left Q_nright $ form a fine covering
$mathcalF$ of $mathbbR^Nsetminusbigcup_n=1^inftyoverline
Q_n$. Therefore if $xinmathbbR^Nsetminusbigcup_n=1^infty
overlineQ_n$ is a Lebesgue point for $f$, then by Lebesgue differentiation theorem we have
$$
fleft( xright) =limsup_operatorname*diamFrightarrow0text,%
,FinmathcalFtext, xin Ffrac1leftint_Ffleft(
yright) ,dyleq lambda
$$
and the proof is completed.

share|cite|improve this answer

answered Jul 27 at 2:48

Gio67

11.2k1526

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 2:48

Gio67

11.2k1526

answered Jul 27 at 2:48

Gio67

11.2k1526

answered Jul 27 at 2:48

Gio67

11.2k1526

11.2k1526

add a comment | 

add a comment | 

 

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