Mixing
Number of circles with equal radius around a point
$begingroup$
Let $x_0$ be a point on the plane and $2^K$ be a large number. Given a family $mathcal C$ of circles with equal radius $R$ satisfying the following property:
- The distance $d_c$ of the centre of each circle $c$ to $x_0$ has $2^K R<d_c<2^{K+1}R$.
- The circles have bounded overlap, that is, there is an absoulute constant $M$ such that for all $x$
$$
sum_{cin mathcal C}1_{c}(x)leq M.
$$
Then what can we say about the cardinality of $mathcal C$, in terms of $K$?
Maybe it is easier to think of the case $M=1$ first. I came across this question when I am reading A study guide for the l 2 decoupling theorem by Bourgain and Demeter. It is one step for an inequality which they say it is easy to verify, but I found it looks like much deeper.
geometry euclidean-geometry circles harmonic-analysis
asked Jan 27 at 21:35
Dong LiDong Li
795413
$endgroup$
add a comment |
$begingroup$
Let $x_0$ be a point on the plane and $2^K$ be a large number. Given a family $mathcal C$ of circles with equal radius $R$ satisfying the following property:
- The distance $d_c$ of the centre of each circle $c$ to $x_0$ has $2^K R<d_c<2^{K+1}R$.
- The circles have bounded overlap, that is, there is an absoulute constant $M$ such that for all $x$
$$
sum_{cin mathcal C}1_{c}(x)leq M.
$$
Then what can we say about the cardinality of $mathcal C$, in terms of $K$?
Maybe it is easier to think of the case $M=1$ first. I came across this question when I am reading A study guide for the l 2 decoupling theorem by Bourgain and Demeter. It is one step for an inequality which they say it is easy to verify, but I found it looks like much deeper.
geometry euclidean-geometry circles harmonic-analysis
asked Jan 27 at 21:35
Dong LiDong Li
795413
$endgroup$
1
$begingroup$
By the bounded overlap property, have I understood correctly that you mean that each point is contained in at most $M$ circles?
$endgroup$
– Servaes
Jan 27 at 21:51
$begingroup$
And what exactly do they say is easy to verify? Do you have an explicit upper bound that you are looking to verify?
$endgroup$
– Servaes
Jan 27 at 22:02
add a comment |
$begingroup$
Let $x_0$ be a point on the plane and $2^K$ be a large number. Given a family $mathcal C$ of circles with equal radius $R$ satisfying the following property:
- The distance $d_c$ of the centre of each circle $c$ to $x_0$ has $2^K R<d_c<2^{K+1}R$.
- The circles have bounded overlap, that is, there is an absoulute constant $M$ such that for all $x$
$$
sum_{cin mathcal C}1_{c}(x)leq M.
$$
Then what can we say about the cardinality of $mathcal C$, in terms of $K$?
Maybe it is easier to think of the case $M=1$ first. I came across this question when I am reading A study guide for the l 2 decoupling theorem by Bourgain and Demeter. It is one step for an inequality which they say it is easy to verify, but I found it looks like much deeper.
geometry euclidean-geometry circles harmonic-analysis
asked Jan 27 at 21:35
Dong LiDong Li
795413
$endgroup$
Let $x_0$ be a point on the plane and $2^K$ be a large number. Given a family $mathcal C$ of circles with equal radius $R$ satisfying the following property:
- The distance $d_c$ of the centre of each circle $c$ to $x_0$ has $2^K R<d_c<2^{K+1}R$.
- The circles have bounded overlap, that is, there is an absoulute constant $M$ such that for all $x$
$$
sum_{cin mathcal C}1_{c}(x)leq M.
$$
Then what can we say about the cardinality of $mathcal C$, in terms of $K$?
Maybe it is easier to think of the case $M=1$ first. I came across this question when I am reading A study guide for the l 2 decoupling theorem by Bourgain and Demeter. It is one step for an inequality which they say it is easy to verify, but I found it looks like much deeper.
geometry euclidean-geometry circles harmonic-analysis
geometry euclidean-geometry circles harmonic-analysis
asked Jan 27 at 21:35
Dong LiDong Li
795413
asked Jan 27 at 21:35
Dong LiDong Li
795413
asked Jan 27 at 21:35
Dong LiDong Li
795413
asked Jan 27 at 21:35
Dong LiDong Li
795413
asked Jan 27 at 21:35
Dong LiDong Li
795413
795413
1
$begingroup$
By the bounded overlap property, have I understood correctly that you mean that each point is contained in at most $M$ circles?
$endgroup$
– Servaes
Jan 27 at 21:51
$begingroup$
And what exactly do they say is easy to verify? Do you have an explicit upper bound that you are looking to verify?
$endgroup$
– Servaes
Jan 27 at 22:02
add a comment |
1
$begingroup$
By the bounded overlap property, have I understood correctly that you mean that each point is contained in at most $M$ circles?
$endgroup$
– Servaes
Jan 27 at 21:51
$begingroup$
And what exactly do they say is easy to verify? Do you have an explicit upper bound that you are looking to verify?
$endgroup$
– Servaes
Jan 27 at 22:02
1
1
$begingroup$
By the bounded overlap property, have I understood correctly that you mean that each point is contained in at most $M$ circles?
$endgroup$
– Servaes
Jan 27 at 21:51
$begingroup$
By the bounded overlap property, have I understood correctly that you mean that each point is contained in at most $M$ circles?
$endgroup$
– Servaes
Jan 27 at 21:51
$begingroup$
And what exactly do they say is easy to verify? Do you have an explicit upper bound that you are looking to verify?
$endgroup$
– Servaes
Jan 27 at 22:02
$begingroup$
And what exactly do they say is easy to verify? Do you have an explicit upper bound that you are looking to verify?
$endgroup$
– Servaes
Jan 27 at 22:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A very rough upper bound on the cardinality of $mathcal{C}$ comes from the fact that each circle $cinmathcal{C}$ covers an area of at least $frac{pi}{2}R^2-varepsilon $ of the annulus
$$A:={xinBbb{R}^2: 2^K R<x<2^{K+1}R},$$
that its center must be inside, where $varepsilon>0$ quickly tends to $0$ as $K$ tends to infinity. The area of the annulus equals $2^{2K}R^2pi$ and the circles cover the annulus at most $M$ times, so there are at most
$$frac{M2^{2K}R^2pi}{frac{pi}{2}R^2-varepsilon}approx M2^{2K+1},$$
circles in $mathcal{C}$, where I just waved the $varepsilon$ away.
Covering the annulus by a radial grid of circles of radius $R$ yields a family $mathcal{C}$ of cardinality at least $M2^{2K-2}$, which can even easily be improved a bit, but already shows that the upper bound above has the same order of magnitude as the actual mininum.
answered Jan 27 at 22:01
ServaesServaes
28.5k34099
$endgroup$
$begingroup$
Thanks! I was not aware of using a measure-theoretic way of proving this as you did.
$endgroup$
– Dong Li
Jan 28 at 1:42
add a comment |
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1 Answer
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$begingroup$
A very rough upper bound on the cardinality of $mathcal{C}$ comes from the fact that each circle $cinmathcal{C}$ covers an area of at least $frac{pi}{2}R^2-varepsilon $ of the annulus
$$A:={xinBbb{R}^2: 2^K R<x<2^{K+1}R},$$
that its center must be inside, where $varepsilon>0$ quickly tends to $0$ as $K$ tends to infinity. The area of the annulus equals $2^{2K}R^2pi$ and the circles cover the annulus at most $M$ times, so there are at most
$$frac{M2^{2K}R^2pi}{frac{pi}{2}R^2-varepsilon}approx M2^{2K+1},$$
circles in $mathcal{C}$, where I just waved the $varepsilon$ away.
Covering the annulus by a radial grid of circles of radius $R$ yields a family $mathcal{C}$ of cardinality at least $M2^{2K-2}$, which can even easily be improved a bit, but already shows that the upper bound above has the same order of magnitude as the actual mininum.
answered Jan 27 at 22:01
ServaesServaes
28.5k34099
$endgroup$
$begingroup$
Thanks! I was not aware of using a measure-theoretic way of proving this as you did.
$endgroup$
– Dong Li
Jan 28 at 1:42
add a comment |
$begingroup$
A very rough upper bound on the cardinality of $mathcal{C}$ comes from the fact that each circle $cinmathcal{C}$ covers an area of at least $frac{pi}{2}R^2-varepsilon $ of the annulus
$$A:={xinBbb{R}^2: 2^K R<x<2^{K+1}R},$$
that its center must be inside, where $varepsilon>0$ quickly tends to $0$ as $K$ tends to infinity. The area of the annulus equals $2^{2K}R^2pi$ and the circles cover the annulus at most $M$ times, so there are at most
$$frac{M2^{2K}R^2pi}{frac{pi}{2}R^2-varepsilon}approx M2^{2K+1},$$
circles in $mathcal{C}$, where I just waved the $varepsilon$ away.
Covering the annulus by a radial grid of circles of radius $R$ yields a family $mathcal{C}$ of cardinality at least $M2^{2K-2}$, which can even easily be improved a bit, but already shows that the upper bound above has the same order of magnitude as the actual mininum.
answered Jan 27 at 22:01
ServaesServaes
28.5k34099
$endgroup$
$begingroup$
Thanks! I was not aware of using a measure-theoretic way of proving this as you did.
$endgroup$
– Dong Li
Jan 28 at 1:42
add a comment |
$begingroup$
A very rough upper bound on the cardinality of $mathcal{C}$ comes from the fact that each circle $cinmathcal{C}$ covers an area of at least $frac{pi}{2}R^2-varepsilon $ of the annulus
$$A:={xinBbb{R}^2: 2^K R<x<2^{K+1}R},$$
that its center must be inside, where $varepsilon>0$ quickly tends to $0$ as $K$ tends to infinity. The area of the annulus equals $2^{2K}R^2pi$ and the circles cover the annulus at most $M$ times, so there are at most
$$frac{M2^{2K}R^2pi}{frac{pi}{2}R^2-varepsilon}approx M2^{2K+1},$$
circles in $mathcal{C}$, where I just waved the $varepsilon$ away.
Covering the annulus by a radial grid of circles of radius $R$ yields a family $mathcal{C}$ of cardinality at least $M2^{2K-2}$, which can even easily be improved a bit, but already shows that the upper bound above has the same order of magnitude as the actual mininum.
answered Jan 27 at 22:01
ServaesServaes
28.5k34099
$endgroup$
A very rough upper bound on the cardinality of $mathcal{C}$ comes from the fact that each circle $cinmathcal{C}$ covers an area of at least $frac{pi}{2}R^2-varepsilon $ of the annulus
$$A:={xinBbb{R}^2: 2^K R<x<2^{K+1}R},$$
that its center must be inside, where $varepsilon>0$ quickly tends to $0$ as $K$ tends to infinity. The area of the annulus equals $2^{2K}R^2pi$ and the circles cover the annulus at most $M$ times, so there are at most
$$frac{M2^{2K}R^2pi}{frac{pi}{2}R^2-varepsilon}approx M2^{2K+1},$$
circles in $mathcal{C}$, where I just waved the $varepsilon$ away.
Covering the annulus by a radial grid of circles of radius $R$ yields a family $mathcal{C}$ of cardinality at least $M2^{2K-2}$, which can even easily be improved a bit, but already shows that the upper bound above has the same order of magnitude as the actual mininum.
answered Jan 27 at 22:01
ServaesServaes
28.5k34099
edited Jan 27 at 22:18
answered Jan 27 at 22:01
ServaesServaes
28.5k34099
answered Jan 27 at 22:01
ServaesServaes
28.5k34099
answered Jan 27 at 22:01
ServaesServaes
28.5k34099
28.5k34099
$begingroup$
Thanks! I was not aware of using a measure-theoretic way of proving this as you did.
$endgroup$
– Dong Li
Jan 28 at 1:42
add a comment |
$begingroup$
Thanks! I was not aware of using a measure-theoretic way of proving this as you did.
$endgroup$
– Dong Li
Jan 28 at 1:42
$begingroup$
Thanks! I was not aware of using a measure-theoretic way of proving this as you did.
$endgroup$
– Dong Li
Jan 28 at 1:42
$begingroup$
Thanks! I was not aware of using a measure-theoretic way of proving this as you did.
$endgroup$
– Dong Li
Jan 28 at 1:42
add a comment |
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1
$begingroup$
By the bounded overlap property, have I understood correctly that you mean that each point is contained in at most $M$ circles?
$endgroup$
– Servaes
Jan 27 at 21:51
$begingroup$
And what exactly do they say is easy to verify? Do you have an explicit upper bound that you are looking to verify?
$endgroup$
– Servaes
Jan 27 at 22:02