I am having trouble verifying the following claim in Van Vu's 2000 paper "On a refinement of Waring's problem". First we define a few things.


Let $m \in \mathbb{N}\_0$ and $r \geq 2, r \in \mathbb{N}$ be fixed. Choose $P\_j \in $ {$2, 4, \cdots, 2^t$} where $t$ is chosen so that $2^t$ is the smallest integer power of 2 bigger than $m^{1/r}$. Suppose that $l \in \mathbb{N}$ is sufficiently large. Let $\mathcal{P}$ denote the set of $l$-tuples {$P\_1, \cdots, P\_l$} with $2 \leq P\_1 \leq \cdots \leq P\_l$. For each $A \in \mathcal{P}$, write $P\_A = \prod\_{P\_j \in A} P\_j$. Then verify the inequality
$$\displaystyle \sum\_{A \in \mathcal{P}} P\_A^{r/l} = O(m)$$


The idea here is that if $P\_A$ is as large as possible, then $P\_A^{r/l} = O(m^{l/r \cdot r/l}) = O(m)$, and the other terms are not so significant since the number of summands is small. However, how do I rigorously show that we indeed have the bound $O(m)$ instead of say, $O(m^{1 + \epsilon})$?



Sum over diadic boxes
As each $P\_A$ is a product of at most $l$ elements of {$2,4,\dots,2^t$} counted with multiplicity, we have for any $\lambda>0$,
$$\sum\_{A \in \mathcal{P}} P\_A^\lambda\leq(2^\lambda+2^{2\lambda}+\cdots+2^{t\lambda})^l<(2^{(t+1)\lambda}/(2^\lambda-1))^l.$$
Assuming $\lambda$ and $l$ are fixed, we obtain
$$\sum\_{A \in \mathcal{P}} P\_A^\lambda\ll(2^{t+1})^{\lambda l}<(4m^{1/r})^{\lambda l}\ll m^{\lambda l/r}.$$
Applying this for $\lambda:=r/l$ yields, for fixed $r$ and $l$,
$$\sum\_{A \in \mathcal{P}} P\_A^{r/l} \ll m.$$


EDIT: I corrected an oversight in my original message.


