Confused by part of the proof of Stein's maximal principle

Question

I'm reading through the proof of Theorem 1 in Stein's "On limits of sequences of operators", and I'm confused at a step. I'll try to replicate all the parts I think are relevant. At the end of pg. 148 we assume for contradiction that there exists a sequence of functions $(f_n)_{n = 1}^\infty$ in $L^p(M, m)$ (where $M$ is a compact homogeneous space over a compact group $G$ with finite invariant measure $m$, and $1 \leq p \leq 2$) such that $$m(E_n) > n \|f_n\|_p^p ,$$ where $E_n = \left\{ x \in M : T_*f(x) > 1 \right\}$ and $T_* = \sup_n |T_n|$ is the relevant maximal operator. Stein then goes on to claim that we can extract a subsequence, possibly with repetition, for which $\sum_{n = 1}^\infty m(E_n) = \infty$ and $\sum_{n = 1}^\infty \|f_n\|_p^p < \infty$.

My confusion is how to do both of these simultaneously. On one hand, I understand how if I allow repetition, I can choose $n_1 \leq n_2 \leq n_3 \leq \cdots$ going to $\infty$ such that $\sum_{j = 1}^\infty m \left( E_{n_j} \right) = \infty$. Since each term is positive, and I'm allowed to repeat terms. For each $n$, just repeat the term $E_n$ for $\sim 1 / m(E_n)$ terms and I'll contribute $1$ to the sum, and repeat this process to get a divergent sum; in other words, I can make the sum diverge by repeating terms enough times. I also understand that because $\|f_n\|_p^p < m(E_n) / n \leq m(M) / n$, I have that $\|f_n\|_p^p \to 0$, so I can extract a summable subsequence. But by passing to a subsequence, couldn't I accidentally "break" the divergence of $\sum_{n = 1}^\infty m(E_n)$, accidentally passing to a subsequence where that series converges as well?

Would appreciate if anyone can explain this.

Answer 1

Lemma. Let $(c_n)$ be a sequence of numbers satisfying $0<c_n< C /n$ for some $C>0$. Then there exists a function $N(n):\{1,2,\dots\}\to \{0,1,2,\dots\}$ so that

$$(*)\quad \sum_{n=1}^\infty N(n) n c_n = \infty,$$ and $$(**)\quad \sum_{n=1}^\infty N(n) c_n < \infty.$$

Proof.

• If $\limsup n c_n =\rho >0$, pick a subsequence as follows: $n_1 = \min\{n: c_n <1 \mbox{ and }n c_n >\rho/2\}$ and continue inductively, $n_{k+1} = \min\{n> n_k: c_{n}<2^{-k-1} \mbox{ and }n c_n > \rho \frac{k}{k+1}\}$. Set $N(n)=1$ if $n=n_k$ for some $k$ and zero otherwise. The result immediately follows.

• Otherwise, $\lim n c_n = 0$. Let $n_k= 2^k$ for $k=1,2,\dots$, $N(n_k) = \lfloor\frac{1}{n_k c_{n_k}}\rfloor$ and for all other $n$, set $N(n)=0$. We have $\lim_{k\to\infty} N(n_k) n_k c_{n_k}= 1$, and therefore $(*)$ holds. We also have $N(n_k) c_{n_k} \le \frac{c_{n_k}}{n_k c_{n_k}}=2^{-k}$ and therefore $(**)$ holds $\Box$

Back to our problem. For each $n$, $m(E_n) /n > \|f_n\|_p^p$ and therefore we can pick $c_n$ such that $m(E_n)/n > c_n > \|f_n\|_p^p$. The sequence $(c_n)$ then satisfies the conditions with $C$ the measure of the space. The subsequence with repetitions is the set of $n$ such that $N(n)>0$, with each element $n$ in this set repeated $N(n)$ times.