Specialized covering lemma for a Hardy-Littlewood maximal inequality












0












$begingroup$


In this question, a suggested approach is given for improving the constant in a Hardy-Littlewood maximal inequality from 3 to 2, and the following lemma is stated without proof:




Suppose $K$ is a compact set, and for every $x in K$, we are given an
open ball $B(x,r_x)$ that is centered at $x$ and of radius $r_x$.
Assume that $$R:= sup_{x in K} r_x < infty.$$ Let $mathcal{B}$ be
this collection of balls, i.e. $$mathcal{B} = {B(x,r_x) colon x in K}.$$ Then given any $varepsilon > 0$, there exists a finite
subcollection $mathcal{C}$ of balls from $mathcal{B}$, so that the
balls in $mathcal{C}$ are pairwise disjoint, and so that the
(concentric) dilates of balls in $mathcal{C}$ by $(2+varepsilon)$
times would cover $K$.




The hint in a comment is to include "sufficiently many" balls in the cover so that each epsilon neighborhood includes a center, but I am not sure how to get a finite cover with such a property. A further hint (or complete proof) would be appreciated.



Note: This is not homework.






real-analysis metric-spaces harmonic-analysis







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    In this question, a suggested approach is given for improving the constant in a Hardy-Littlewood maximal inequality from 3 to 2, and the following lemma is stated without proof:




    Suppose $K$ is a compact set, and for every $x in K$, we are given an
    open ball $B(x,r_x)$ that is centered at $x$ and of radius $r_x$.
    Assume that $$R:= sup_{x in K} r_x < infty.$$ Let $mathcal{B}$ be
    this collection of balls, i.e. $$mathcal{B} = {B(x,r_x) colon x in K}.$$ Then given any $varepsilon > 0$, there exists a finite
    subcollection $mathcal{C}$ of balls from $mathcal{B}$, so that the
    balls in $mathcal{C}$ are pairwise disjoint, and so that the
    (concentric) dilates of balls in $mathcal{C}$ by $(2+varepsilon)$
    times would cover $K$.




    The hint in a comment is to include "sufficiently many" balls in the cover so that each epsilon neighborhood includes a center, but I am not sure how to get a finite cover with such a property. A further hint (or complete proof) would be appreciated.



    Note: This is not homework.






    real-analysis metric-spaces harmonic-analysis







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      In this question, a suggested approach is given for improving the constant in a Hardy-Littlewood maximal inequality from 3 to 2, and the following lemma is stated without proof:




      Suppose $K$ is a compact set, and for every $x in K$, we are given an
      open ball $B(x,r_x)$ that is centered at $x$ and of radius $r_x$.
      Assume that $$R:= sup_{x in K} r_x < infty.$$ Let $mathcal{B}$ be
      this collection of balls, i.e. $$mathcal{B} = {B(x,r_x) colon x in K}.$$ Then given any $varepsilon > 0$, there exists a finite
      subcollection $mathcal{C}$ of balls from $mathcal{B}$, so that the
      balls in $mathcal{C}$ are pairwise disjoint, and so that the
      (concentric) dilates of balls in $mathcal{C}$ by $(2+varepsilon)$
      times would cover $K$.




      The hint in a comment is to include "sufficiently many" balls in the cover so that each epsilon neighborhood includes a center, but I am not sure how to get a finite cover with such a property. A further hint (or complete proof) would be appreciated.



      Note: This is not homework.






      real-analysis metric-spaces harmonic-analysis







      share|cite|improve this question









      $endgroup$




      In this question, a suggested approach is given for improving the constant in a Hardy-Littlewood maximal inequality from 3 to 2, and the following lemma is stated without proof:




      Suppose $K$ is a compact set, and for every $x in K$, we are given an
      open ball $B(x,r_x)$ that is centered at $x$ and of radius $r_x$.
      Assume that $$R:= sup_{x in K} r_x < infty.$$ Let $mathcal{B}$ be
      this collection of balls, i.e. $$mathcal{B} = {B(x,r_x) colon x in K}.$$ Then given any $varepsilon > 0$, there exists a finite
      subcollection $mathcal{C}$ of balls from $mathcal{B}$, so that the
      balls in $mathcal{C}$ are pairwise disjoint, and so that the
      (concentric) dilates of balls in $mathcal{C}$ by $(2+varepsilon)$
      times would cover $K$.




      The hint in a comment is to include "sufficiently many" balls in the cover so that each epsilon neighborhood includes a center, but I am not sure how to get a finite cover with such a property. A further hint (or complete proof) would be appreciated.



      Note: This is not homework.






      real-analysis metric-spaces harmonic-analysis




      real-analysis metric-spaces harmonic-analysis






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 14 '18 at 15:48









      Zach BoydZach Boyd

      7862517




      7862517






















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          $begingroup$

          Consider the open cover ${B(x,varepsilon r_x): xin K}$ and get a finite subcover, say ${B(x_j,varepsilon r_j): j=1,dots,N}$. WLOG, assume $r_1geq r_2geqdotsgeq r_N$. Then, for each $j=1,dots,N$, select the open ball $B(x_j,r_j)$ if it does not intersect with any selected ball, otherwise discard it. Say the selected balls form a set ${B(x_{j_k},r_{j_k}): k=1,dots,M}$ where $Mleq N$ and $j_{1}leq j_2leqdotsleq j_M$.



          For each $j$, if $B(x_j,varepsilon r_j)notsubsetbigcup_{k=1}^MB(x_{j_k},(2+varepsilon)r_{j_k})$, then $|x_j-x_{j_k}|+varepsilon r_j>(2+varepsilon)r_{j_k}$ for each $k$. Since $B(x_j,r_j)$ is not selected, we can also find some $k$ such that $B(x_j,r_j)cap B(x_{j_k},r_{j_k})neq varnothing$, so $|x_j-x_{j_k}|< r_j+r_{j_k}$. Note that we may choose the smallest $k$ with such a property, so we may assume $B(x_j,r_j)cap B(x_{j_l},r_{j_l})= varnothing$ if $l<k$. But this implies that $(2+varepsilon)r_{j_k}-varepsilon r_j<r_j+r_{j_k}$, i.e. $r_{j_k}<r_j$. Here is a contradiction, since we should have chosen $B(x_j,r_j)$ instead of $B(r_{j_k},r_{j_k})$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you! This seems to be a really sensible approach. I couldn't figure out how to force enough overlap. Interestingly, this seems to not satisfy the hint, in that not every epsilon neighborhood contains a center.
            $endgroup$
            – Zach Boyd
            Dec 16 '18 at 2:20










          • $begingroup$
            Yes I think there should be another way to prove this lemma using the hint. Here I didn't even use the assumption that sup r_x<infty, so I guess there should be a proof relying on this assumption.
            $endgroup$
            – ydx
            Dec 16 '18 at 2:49











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          $begingroup$

          Consider the open cover ${B(x,varepsilon r_x): xin K}$ and get a finite subcover, say ${B(x_j,varepsilon r_j): j=1,dots,N}$. WLOG, assume $r_1geq r_2geqdotsgeq r_N$. Then, for each $j=1,dots,N$, select the open ball $B(x_j,r_j)$ if it does not intersect with any selected ball, otherwise discard it. Say the selected balls form a set ${B(x_{j_k},r_{j_k}): k=1,dots,M}$ where $Mleq N$ and $j_{1}leq j_2leqdotsleq j_M$.



          For each $j$, if $B(x_j,varepsilon r_j)notsubsetbigcup_{k=1}^MB(x_{j_k},(2+varepsilon)r_{j_k})$, then $|x_j-x_{j_k}|+varepsilon r_j>(2+varepsilon)r_{j_k}$ for each $k$. Since $B(x_j,r_j)$ is not selected, we can also find some $k$ such that $B(x_j,r_j)cap B(x_{j_k},r_{j_k})neq varnothing$, so $|x_j-x_{j_k}|< r_j+r_{j_k}$. Note that we may choose the smallest $k$ with such a property, so we may assume $B(x_j,r_j)cap B(x_{j_l},r_{j_l})= varnothing$ if $l<k$. But this implies that $(2+varepsilon)r_{j_k}-varepsilon r_j<r_j+r_{j_k}$, i.e. $r_{j_k}<r_j$. Here is a contradiction, since we should have chosen $B(x_j,r_j)$ instead of $B(r_{j_k},r_{j_k})$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you! This seems to be a really sensible approach. I couldn't figure out how to force enough overlap. Interestingly, this seems to not satisfy the hint, in that not every epsilon neighborhood contains a center.
            $endgroup$
            – Zach Boyd
            Dec 16 '18 at 2:20










          • $begingroup$
            Yes I think there should be another way to prove this lemma using the hint. Here I didn't even use the assumption that sup r_x<infty, so I guess there should be a proof relying on this assumption.
            $endgroup$
            – ydx
            Dec 16 '18 at 2:49
















          1












          $begingroup$

          Consider the open cover ${B(x,varepsilon r_x): xin K}$ and get a finite subcover, say ${B(x_j,varepsilon r_j): j=1,dots,N}$. WLOG, assume $r_1geq r_2geqdotsgeq r_N$. Then, for each $j=1,dots,N$, select the open ball $B(x_j,r_j)$ if it does not intersect with any selected ball, otherwise discard it. Say the selected balls form a set ${B(x_{j_k},r_{j_k}): k=1,dots,M}$ where $Mleq N$ and $j_{1}leq j_2leqdotsleq j_M$.



          For each $j$, if $B(x_j,varepsilon r_j)notsubsetbigcup_{k=1}^MB(x_{j_k},(2+varepsilon)r_{j_k})$, then $|x_j-x_{j_k}|+varepsilon r_j>(2+varepsilon)r_{j_k}$ for each $k$. Since $B(x_j,r_j)$ is not selected, we can also find some $k$ such that $B(x_j,r_j)cap B(x_{j_k},r_{j_k})neq varnothing$, so $|x_j-x_{j_k}|< r_j+r_{j_k}$. Note that we may choose the smallest $k$ with such a property, so we may assume $B(x_j,r_j)cap B(x_{j_l},r_{j_l})= varnothing$ if $l<k$. But this implies that $(2+varepsilon)r_{j_k}-varepsilon r_j<r_j+r_{j_k}$, i.e. $r_{j_k}<r_j$. Here is a contradiction, since we should have chosen $B(x_j,r_j)$ instead of $B(r_{j_k},r_{j_k})$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you! This seems to be a really sensible approach. I couldn't figure out how to force enough overlap. Interestingly, this seems to not satisfy the hint, in that not every epsilon neighborhood contains a center.
            $endgroup$
            – Zach Boyd
            Dec 16 '18 at 2:20










          • $begingroup$
            Yes I think there should be another way to prove this lemma using the hint. Here I didn't even use the assumption that sup r_x<infty, so I guess there should be a proof relying on this assumption.
            $endgroup$
            – ydx
            Dec 16 '18 at 2:49














          1












          1








          1





          $begingroup$

          Consider the open cover ${B(x,varepsilon r_x): xin K}$ and get a finite subcover, say ${B(x_j,varepsilon r_j): j=1,dots,N}$. WLOG, assume $r_1geq r_2geqdotsgeq r_N$. Then, for each $j=1,dots,N$, select the open ball $B(x_j,r_j)$ if it does not intersect with any selected ball, otherwise discard it. Say the selected balls form a set ${B(x_{j_k},r_{j_k}): k=1,dots,M}$ where $Mleq N$ and $j_{1}leq j_2leqdotsleq j_M$.



          For each $j$, if $B(x_j,varepsilon r_j)notsubsetbigcup_{k=1}^MB(x_{j_k},(2+varepsilon)r_{j_k})$, then $|x_j-x_{j_k}|+varepsilon r_j>(2+varepsilon)r_{j_k}$ for each $k$. Since $B(x_j,r_j)$ is not selected, we can also find some $k$ such that $B(x_j,r_j)cap B(x_{j_k},r_{j_k})neq varnothing$, so $|x_j-x_{j_k}|< r_j+r_{j_k}$. Note that we may choose the smallest $k$ with such a property, so we may assume $B(x_j,r_j)cap B(x_{j_l},r_{j_l})= varnothing$ if $l<k$. But this implies that $(2+varepsilon)r_{j_k}-varepsilon r_j<r_j+r_{j_k}$, i.e. $r_{j_k}<r_j$. Here is a contradiction, since we should have chosen $B(x_j,r_j)$ instead of $B(r_{j_k},r_{j_k})$.






          share|cite|improve this answer











          $endgroup$



          Consider the open cover ${B(x,varepsilon r_x): xin K}$ and get a finite subcover, say ${B(x_j,varepsilon r_j): j=1,dots,N}$. WLOG, assume $r_1geq r_2geqdotsgeq r_N$. Then, for each $j=1,dots,N$, select the open ball $B(x_j,r_j)$ if it does not intersect with any selected ball, otherwise discard it. Say the selected balls form a set ${B(x_{j_k},r_{j_k}): k=1,dots,M}$ where $Mleq N$ and $j_{1}leq j_2leqdotsleq j_M$.



          For each $j$, if $B(x_j,varepsilon r_j)notsubsetbigcup_{k=1}^MB(x_{j_k},(2+varepsilon)r_{j_k})$, then $|x_j-x_{j_k}|+varepsilon r_j>(2+varepsilon)r_{j_k}$ for each $k$. Since $B(x_j,r_j)$ is not selected, we can also find some $k$ such that $B(x_j,r_j)cap B(x_{j_k},r_{j_k})neq varnothing$, so $|x_j-x_{j_k}|< r_j+r_{j_k}$. Note that we may choose the smallest $k$ with such a property, so we may assume $B(x_j,r_j)cap B(x_{j_l},r_{j_l})= varnothing$ if $l<k$. But this implies that $(2+varepsilon)r_{j_k}-varepsilon r_j<r_j+r_{j_k}$, i.e. $r_{j_k}<r_j$. Here is a contradiction, since we should have chosen $B(x_j,r_j)$ instead of $B(r_{j_k},r_{j_k})$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 16 '18 at 2:28

























          answered Dec 15 '18 at 0:38









          ydxydx

          73118




          73118












          • $begingroup$
            Thank you! This seems to be a really sensible approach. I couldn't figure out how to force enough overlap. Interestingly, this seems to not satisfy the hint, in that not every epsilon neighborhood contains a center.
            $endgroup$
            – Zach Boyd
            Dec 16 '18 at 2:20










          • $begingroup$
            Yes I think there should be another way to prove this lemma using the hint. Here I didn't even use the assumption that sup r_x<infty, so I guess there should be a proof relying on this assumption.
            $endgroup$
            – ydx
            Dec 16 '18 at 2:49


















          • $begingroup$
            Thank you! This seems to be a really sensible approach. I couldn't figure out how to force enough overlap. Interestingly, this seems to not satisfy the hint, in that not every epsilon neighborhood contains a center.
            $endgroup$
            – Zach Boyd
            Dec 16 '18 at 2:20










          • $begingroup$
            Yes I think there should be another way to prove this lemma using the hint. Here I didn't even use the assumption that sup r_x<infty, so I guess there should be a proof relying on this assumption.
            $endgroup$
            – ydx
            Dec 16 '18 at 2:49
















          $begingroup$
          Thank you! This seems to be a really sensible approach. I couldn't figure out how to force enough overlap. Interestingly, this seems to not satisfy the hint, in that not every epsilon neighborhood contains a center.
          $endgroup$
          – Zach Boyd
          Dec 16 '18 at 2:20




          $begingroup$
          Thank you! This seems to be a really sensible approach. I couldn't figure out how to force enough overlap. Interestingly, this seems to not satisfy the hint, in that not every epsilon neighborhood contains a center.
          $endgroup$
          – Zach Boyd
          Dec 16 '18 at 2:20












          $begingroup$
          Yes I think there should be another way to prove this lemma using the hint. Here I didn't even use the assumption that sup r_x<infty, so I guess there should be a proof relying on this assumption.
          $endgroup$
          – ydx
          Dec 16 '18 at 2:49




          $begingroup$
          Yes I think there should be another way to prove this lemma using the hint. Here I didn't even use the assumption that sup r_x<infty, so I guess there should be a proof relying on this assumption.
          $endgroup$
          – ydx
          Dec 16 '18 at 2:49


















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