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I'm trying to get the basic idea behind the proof of Theorem 8.1 of this paper, but I'm having difficulty. Specifically, it says:

We shall show that there is a set $\Lambda_n\subset\mathcal{D}$ such that

(i) $|\Lambda_n|\leq6\cdot 2^n$, $n=1,2,\ldots$,

(ii) $|c_I|\leq C_1'2^{-n}$, $I\not\in\Lambda_n$,

where in (ii), $c_I$ is any of the three Haar coefficients associated to $I$. It is easy to see that this implies (8.8).

Here, (8.8) bounds the size of the $n$th largest Haar coefficient in terms of total variation: $$ \gamma_n(f)\leq 36C_1'\frac{V_Q(f)}{n}. $$ When they say (i) and (ii) implies (8.8), this is operating under the assumption that $V_Q(f)=1$. I assume the logic here is based on some well-known wavelet trick (like the Calderón–Zygmund decomposition), but I am not very familiar with this literature.

Why do (i) and (ii) together imply (8.8) when $V_Q(f)=1$?

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>Why do (i) and (ii) together imply (8.8)?< Erm... What is said is just that if there are at most $N$ elements in the sequence greater than $a$, then the $N+1$-st largest element is at most $a$. What is unclear about that? How to prove (i)&(ii) is another story but it is not what gives you trouble now, is it? – fedja May 29 '13 at 2:17
@fedja: It is not clear to me how (i) and (ii) are related to $V_Q(f)/n$. – Dustin G. Mixon May 29 '13 at 5:53
Apparently, (i) and (ii) are claimed under the normalization condition $V_Q(f)=1$. Also $n$ is (8.8) is $2^n$ in (i-ii) (give or take a factor of $2$) – fedja May 29 '13 at 6:20
I see. Yes, I forgot to mention the reduction to the case where $V_Q(f)=1$. I'll have to think about the fact that $n=2^{n\pm 1}$. – Dustin G. Mixon May 29 '13 at 6:26

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