In the book The Geometry of Domains in Spaces by Krantz and Parks, the authors proved the weak $(1,1)$-type estimate of the maximal function $M_\mu f$, where $\mu$ is a Radon measure, using their version of the Besicovitch covering theorem.

Let $d$ be a positive integer. Then there exists a constant $C=C(d)$ such that for any finite collection of balls $\mathcal B = \{B_i\}_{i=1}^m$ in $\Bbb R^d$ with the property that no ball contains the center of any other ball, we can partition the family $\mathcal B$ into $$ \mathcal B = \mathcal B_1 \cup\mathcal B_2 \cup \dots \cup \mathcal B_C, $$ where each subfamily $\mathcal B_j$ consists of disjoint balls.

This version of the covering theorem seems pretty restrictive, especially the part about no ball contains the center of any other balls. Indeed, in theor proof of the weak $(1,1)$-type estimate, they relied on a certain claim that they did not prove.

Edit As Skeeve mentioned in the comment, this claim is not explicitly stated in the book but more of a paraphrasing of the part the authors left out in a proof.

Claim: Let $K\subset \Bbb R^d$ be a compact set such that each $x\in K$ is associated with a real number $r_x>0$. Then $K$ can be covered by a family of balls $$ \mathcal B = \{ B(x_i,r_i) : i=1,\dots,k\ \}, $$ where $r_i := r_{x_i}$, such that for any distinct $i,j \le k$, we have $$ x_i\notin B(x_j,r_j) \quad\text{and}\quad x_j\notin B(x_i,r_i). $$

I don't find this claim to be trivial at all. In fact, I tried many different methods but failed to prove it. Note that the mapping $x\mapsto r_x$ doesn't enjoy any nice property like continuity of any kind.

While the usual version of Besicovitch covering theorem circumvents this problem, I still would like to know how to prove the above claim (or a counter example if it is actually false).

  • $\begingroup$ The claim you are proposing is definitely interesting, however I cannot find it in the cited book (which has 2 authors, by the way). Could you add, for completeness, a more precise reference? It seems to be related to Lemma 3.5.13, but your version of the claim is not stated there explicitly. $\endgroup$ – Skeeve May 17 '19 at 16:15
  • $\begingroup$ @Skeeve You are absolutely right, I was paraphrasing the implicit claim in the proof of Lemma 3.5.13. I should have mentioned explicitly that the claim is not stated verbatim in the book. $\endgroup$ – BigbearZzz May 17 '19 at 17:19
  • $\begingroup$ Possibly I misunderstood the question since it seems straightforward. Consider a maximal family of balls such that no center belongs to another ball so that any additional ball will have its center covered by the union. The maximal family exists by Kuratowski-Zorn (or transfinite induction). Then compactness gives you a finite subcover. (What's my misunderstanding?) $\endgroup$ – Wlod AA May 17 '19 at 18:53
  • $\begingroup$ @WlodAA but how would you prove that a maximal family is a cover of $K$? $\endgroup$ – BigbearZzz May 17 '19 at 18:58
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    $\begingroup$ Crosspost on MSE $\endgroup$ – YuiTo Cheng May 19 '19 at 4:36

This trivial counterexample in $\mathbb R^2$ should have taken me five minutes. Instead, I spent almost two days. The moral is the usual one: after 50 you'd better give up on mathematics.

Let $y,z$ be 2 points at distance $1$ from each other. We shall construct by induction a sequence of points $x_j$ and radii $r_j>\max(d(x_j,y),d(x_j,z))$ such that $x_j\to y$ when $j$ is odd, $x_j\to z$ when $j$ is even, $x_j$ do not lie on the line $yz$, $\max(d(x_j,y),d(x_j,z))<1$, the disk $D(x_j,\max(d(x_j,y),d(x_j,z)))$ contains $x_1,\dots, x_j$ but $x_{j+1}\notin D(x_j,r_j)$ . If you choose the radius $\rho$ for $y$ and $z$ small enough so that the corresponding disks do not contain $x_1$, you'll get a bad configuration.

Indeed, an attempt to choose $y$ or $z$ as one of the centers results in the exclusion of all the centers $x_j$, after which covering $x_1$ gets impossible.

Out of $x_i$, we can choose only one (if $i<j$, then $x_i\in D(x_j,r_j)$). But then, if we choose $x_i$, the point $x_{i+1}$ is not covered.

Now the sequences. Start with any $x_1$ very close to $y$ and not on the line $yz$ so that $d(x_1,z)<1$. Assume that $x_1,x_2,\dots x_j$ and $r_1,\dots,r_{j-1}$ are already constructed and, say $j$ is odd, so $x_j$ is close to $y$. Then the circles centered at $x_j$ and $y$ containing $z$ cross at an angle, so $D(y,1)\setminus \bar D(x_j,d(x_j,z))$ is an open set containing points arbitrarily close to $z$. Choose $x_{j+1}$ to be any point in that difference that doesn't lie on the line $yz$ and satisfies $d(y,x_{j+1})\ge 1-d(z,x_{j+1})>\max_{i\le j}d(z,x_i)+d(z,x_{j+1})\ge \max_{i\le j}d(x_i,x_{j+1})$ and choose $r_j$ anywhere between $d(x_j,z)$ and $d(x_j,x_{j+1})$.

Clearly, we can keep $x_j$ with odd indices at the distance $<1/3$ to $y$ and converging to $y$ and similarly for even indices and $z$.

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    $\begingroup$ Thank you (for your simplification of the proof for $1$-dim on MSE as well)! By the way, do you happen to have Krantz and Parks' book? Do you know if his statement in Lemma 3.5.13 can be interpreted in any other way that we don't need to prove the (false) claim, or relying on the standard Besicovitch covering theorem? $\endgroup$ – BigbearZzz May 19 '19 at 11:23
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    $\begingroup$ @BigbearZzz I don't have it, but I have a possibility to look at it ;-) I'm afraid that the mistake they make is beyond repair if you want to cover the entire $K$ and rely on nothing beyond 3.5.10. Of course, you can supplement it with some part from the proof of the standard version of the Besikovich theorem, but the elegance they were striving for (at the expense of correctness) will totally disappear in that case. $\endgroup$ – fedja May 19 '19 at 12:00
  • $\begingroup$ @BigbearZzz it seems to me that the only hope to repair that proof might be through continuity of $x \mapsto r_x$ (or maybe upper semicontinuity), which I asked in a separate question. $\endgroup$ – Skeeve May 19 '19 at 12:30
  • $\begingroup$ I am over 50 too so I could relate to what you are saying, and I propose a simplification (works pretty well for me :) You could give up on math well before you become 50 so you do not have to worry when you get 50 ... you missed the deadline apparently, poor lad, too late to do anything about it now! $\endgroup$ – Mirko May 19 '19 at 13:39
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    $\begingroup$ @Mirko Yes, the half-planes alternate and you can use linear approximations with bisectors though I do not see why being on a line sufficiently close to a point is any better or easier to visualize then being in a circular crescent. In my construction one may, indeed, pick all the points first and all radii afterwards. Note that, when constructing $x_{j+1}$, I never referred to any $r_i$. I just said that once $x_{j+1}$ is chosen, we are ready to define $r_j$, but we can also postpone the choice of $r_j$ for as long as we want if it makes things clearer. Did you draw pictures? It helps. $\endgroup$ – fedja May 19 '19 at 22:19

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