# How to bound Haar coefficients in terms of total variation?

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