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Let $(a_n,k_n) \in \mathbb{Z}_p \times \mathbb{N}$ for $n \in \mathbb{N}$ and consider the sequence of closed $p$-adic balls $B(a_n,k_n) = a_n + p^{k_n}\mathbb{Z}_p$. I assume that the $(a_n,k_n)$ are chosen so that each ball is disjoint. Is the sum $$\sum_{n = 1}^\infty \mu_p(B(a_n,k_n))^\varepsilon$$ convergent for any $\varepsilon > 0$? (Here $\mu_p$ denotes the usual $p$-adic Haar measure).

This is easy to show for $\varepsilon = 1$ since then it is simply the fact that $\mathbb{Z}_p$ has finite measure. But I cannot prove it for any $\varepsilon < 1$.

It is possible that $\mathbb{Z}_p$ is a red herring and there is some more general statement which holds in a compact probability space.

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No. Choose $p=2$ (easy to adapt to other $p$) and, for $j\ge 1$, $k_n=4^j$ for exactly $2^j$ values of $n$ and no other value achieved.

In $\mathbf{Z}_2$ we can find pairwise disjoint closed balls $B_n$, $n\ge 1$, with $B_n$ of measure $2^{-n}$. Split $B_n$ into $2^n$ pairwise disjoint closed balls of measure $4^{-n}$. This realizes the given sequence $(k_n)$.

But the sum of square roots is clearly not convergent (equals $2^{-j}$ for exactly $2^j$ values of $n$).

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  • $\begingroup$ This shows that you cannot hope it to converge for all $\varepsilon >0$, e.g. $\varepsilon\leq 1/2$ won't work for $p=2$. I guess the question remains if there exists a $\varepsilon<1$ for which it converges. $\endgroup$
    – Chris Wuthrich
    Commented Jan 10, 2023 at 16:46
  • $\begingroup$ @ChrisWuthrich well, I guess not. I basically used that the convergent sum $\sum 1/n^2$ has diverging sum of square roots, adapting to the given setting. But probably starting with $\sum 1/n^s$ with $s>1$ can be used, with a little more work, to show that no $\varepsilon<1$ works. $\endgroup$
    – YCor
    Commented Jan 10, 2023 at 16:48
  • $\begingroup$ Very nice example, thanks. $\mathbb{Z}_p$ has the funny property that a disjoint unions of balls can sometimes be written a single ball. It may be the case that refining my balls to obtain a "minimal" representation of each ball, the result is still true. Any idea about this? $\endgroup$
    – Daniel Loughran
    Commented Jan 11, 2023 at 12:34

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