Inequality in the proof of the atomic decomposition for $L^p$ functions

Question

Given $f$ a nonnegative function in $L^p$, define $\lambda_k=\inf(\lambda: \mu(f>\lambda)<2^k)$, $c_k=2^{k/p}\lambda_k$ and $f_k=c_k^{-1}I_{(\lambda_{k+1}<f\le \lambda_k)}f \ $ with supports pairwise disjoint and $\mu(\mathrm{supp}(f_k))\le 2^{k+1}$. I know that $f=\sum_{k\in \mathbb{Z}}I_{(\lambda_{k+1}<f\le \lambda_k)}f$ and I want to prove that: $$\sum_{k:\lambda<\lambda_k}2^k\le\sum_{k: 2^k\le\mu(f>\lambda)}2^k \le 2\mu(f>\lambda)$$ The first inequality is ok because if $\lambda<\lambda_k$, hence $\mu(f>\lambda)\ge 2^k$ by the definition of $\lambda_k$, but why in the second inequality is there a 2?

Answer 1

This is due to the fact that for each positive number $A$ $$ \sum_{k: 2^k\leqslant A}2^k =\sum_{k: k\leqslant \log_2(A)}2^k=\sum_{k\in\mathbb Z,k\leqslant \lfloor \log_2(A)\rfloor}2^k $$ and an explicit computation of the previous sum: $$ \sum_{k\in\mathbb Z,k\leqslant \lfloor \log_2(A)\rfloor}2^k= \sum_{j\in\mathbb Z,j\leqslant 0 }2^{j+ \lfloor \log_2(A)\rfloor}=2^{1+\lfloor \log_2(A)\rfloor}\leqslant 2A. $$