Number of circles with equal radius around a point












2












$begingroup$


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:




  1. The distance $d_c$ of the centre of each circle $c$ to $x_0$ has $2^K R<d_c<2^{K+1}R$.

  2. 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







share|cite|improve this question









$endgroup$








  • 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
















2












$begingroup$


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:




  1. The distance $d_c$ of the centre of each circle $c$ to $x_0$ has $2^K R<d_c<2^{K+1}R$.

  2. 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







share|cite|improve this question









$endgroup$








  • 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














2












2








2





$begingroup$


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:




  1. The distance $d_c$ of the centre of each circle $c$ to $x_0$ has $2^K R<d_c<2^{K+1}R$.

  2. 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







share|cite|improve this question









$endgroup$




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:




  1. The distance $d_c$ of the centre of each circle $c$ to $x_0$ has $2^K R<d_c<2^{K+1}R$.

  2. 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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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










1 Answer
1






active

oldest

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1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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













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1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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


















1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















1












1








1





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 27 at 22:18

























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




















  • $begingroup$
    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




















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