Let me restate Vitali covering lemma.

Let $\{B_i\}_{i\in F}$ be a finite collection of balls in the $\mathbb{R}^n$. Then there is $S\subset F$ such that the balls $\{B_i\}_{i\in S}$ are disjoint and

$$\mathop{\rm vol}\left(\bigcup_{i\in S}B_i\right)\ge \tfrac1{3^n} \mathop{\rm vol} \left(\bigcup_{i\in F}B_i\right),$$

Question. Can one make the constant better? In particular, is it OK to change $\tfrac1{3^n}$ to $\tfrac1{2^n}$ in the formulation?

It is likely that the answer is well known, but googling did not help.

Clearly the greedy algorithm which is used in the standard proof of Vitali covering lemma cannot give anything better than $\tfrac1{3^n}$.

One cannot make it better than $\tfrac{1}{2^n}$. (Consider a large collection of balls of the same radius such that each contains a fixed point.)

I see that for $n=1$ the optimal constant is $\tfrac1{2}$, but for higher dimensions I do not see a proof.

  • 1
    Is there a particular reason you are interested in improving the constant? This is a tempting thing to get into, but most of the time the constant is not important (save the case the constant can be 1, a special exception). – Daniel Spector Dec 5 '12 at 9:33
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    @Daniel, pure curiosity. – Stas Kuznetsov Dec 5 '12 at 19:57

It is not hard to see that we can improve the constant a bit. Denote the ball with the biggest radius by B and its radius by r. If there are two disjoint balls intersecting B with radius at least 0.99r, than put them in S, this will be "locally better" than $\frac 1{3^n}$ as the blow-up of the balls intersect. If there are no two such balls, then put B into S, this is again "locally better" than $\frac 1{3^n}$. Optimizing the constants will probably give something reasonable, but not 2, which I find to be a very nice conjecture.

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