I am reading a paper on PDEs and I has been struggling trying to understand some specific identities (that should be very easy). Let me introduce the main notation of the book, which says that each time a sum like $\sum\_\lambda f(\lambda)$ appears, then the $\lambda$-variable must to be understood as a dyadic variable, that is, the previous sum has to be understood as $$
\sum\_{k=0}^{\infty} f(2^k).
$$
This will also apply to $\mu$ below. Now, just after that, in the proof of a basic lemma the author writes the following sequence of equalities: $$
\sum\_\lambda \lambda^s\sum\_{\mu\geq \lambda/8} a\_{\mu}d\_{\lambda}=\sum\_\lambda \lambda^s\sum\_{\substack{k=-3,\\ 2^k\lambda\geq1}}^{\infty}a\_{2^k\lambda}d\_{\lambda}=\sum\_{k=-3}^\infty\sum\_{\lambda\geq 2^{-k}} a\_{2^k\lambda}d\_\lambda
$$
where $\{a\_\lambda\}$ and $\{d\_\mu\}$ are two sequences of real numbers indexed by dyadic integers and $s>0$. I don't really understand these identities for two reasons. First, I don't see how to change the $\mu$-variable to the index $2^k\lambda$ appearing in $a$ (I would just put $a\_{2^k}$ withouth the $\lambda$). But then, it seems confusing to me that the inequality in the inner sum becomes $2^k\lambda\geq 1$, particularly because in the left-hand side $\lambda$ is bounded from above, and in the other sums $\lambda$ seems to be bounded from below. Does anyone has any explanation?


**Edit:** The idea of the lemma is actually to prove the following inequality: $$
\sum\_\lambda \lambda^s\sum\_{\mu\geq \lambda/8} a\_{\mu}d\_{\lambda} \leq C \left(\sum\_\lambda \lambda^{2s}a\_\lambda^2\right)^{1/2}\left(\sum\_\lambda d\_\lambda^2\right)^{1/2},
$$
for some constant $C>0$ (whenever the right-hand side makes sense). Now I am feeling that the proof might be wrong, so if it is the case, I am wondering if anyone has a correct proof for it.



Understanding an identity for dyadic sums
We write $\mu=2^k \lambda$ with $k \ge -3$. The standing assumption $\mu \ge 1$ becomes $2^k \lambda \ge 1$. The RHS of the sequence of equalities you reproduce is missing a factor $\lambda^s$; is this typo in the original paper?


The sequence of equalities you included should thus be written (adding $L$ on the left) as
$$
L:= \sum\_\lambda \lambda^s\sum\_{\mu\geq \lambda/8} a\_{\mu}d\_{\lambda}=\sum\_\lambda \lambda^s\sum\_{\substack{k=-3,\\ 2^k\lambda\geq1}}^{\infty}a\_{2^k\lambda}d\_{\lambda}=\sum\_{k=-3}^\infty \sum\_{\lambda\geq 2^{-k}} \lambda^s a\_{2^k\lambda}d\_\lambda \,.
$$
Thus
$$
L= \sum\_{k=-3}^\infty 2^{-ks} \sum\_{\lambda\geq 2^{-k}} (2^k\lambda)^s a\_{2^k\lambda}\, d\_\lambda \,.
$$
Apply Cauchy-Schwarz to the inner sum:
$$
L \leq \sum\_{k=-3}^\infty 2^{-ks}\left(\sum\_{\lambda\geq 2^{-k}} (2^k \lambda)^{2s} a\_{2^k\lambda}^2\right)^{1/2}\left(\sum\_\lambda d\_\lambda^2\right)^{1/2} 
$$
Then substitute $\mu=2^k \lambda$:
$$
L \leq 
\sum\_{k=-3}^\infty 2^{-ks}\left(\sum\_{\mu} \mu^{2s} a\_{\mu}^2\right)^{1/2}\left(\sum\_\lambda d\_\lambda^2\right)^{1/2} \, ,
$$
which yields the inequality you asked about
$$
\sum\_\lambda \lambda^s\sum\_{\mu\geq \lambda/8} a\_{\mu}d\_{\lambda} \leq C \left(\sum\_\lambda \lambda^{2s}a\_\lambda^2\right)^{1/2}\left(\sum\_\lambda d\_\lambda^2\right)^{1/2} \, ,
$$
with $C:=\sum\_{k=-3}^\infty 2^{-ks}$.


