The question may be very trivial. I am reading the proof of refined Young's inequality in page 8 of Bahouri,chemin and Danchin's famous book named Fourier Analysis and non-linear PDEs. The authors used the method of atomic decomposition (see page 7 of this book). Recall the atomic decomposition, let $(X,\mu)$ be a measure space and $p \geq 1,$ for a nonnegative function $f \in L^p,$ there is a sequence of positive real number $\{c_k\}_{k\in \mathbb{Z}}$ and a sequence of non-negative function $(f_k)_{k \in \mathbb{Z}}$ so that $f= \sum_{k \in \mathbb{Z}} c_kf_k,$ furthermore, the supports of $f_k$ are pairwise disjoint and satisfing $\mu(\text{Supp} f_k)\leq 2^{k+1}$ and $ \|f\|_{L^\infty} \leq 2^{-k/p},$ and $\frac{1}{2}\| f\|_{L^p}^p \leq \sum_{k \in \mathbb{Z}}c_k^p \leq 2\| f\|_{L^p}^p.$
In page 9, for $f \in L^p$ so that $\| f\|_{L^p}=1,$ the author said that $\| f_k\|_{L^a} \leq 2^{k(\frac{1}{a}-\frac{1}{p})},$ However, I think $$ \|f_k \|_{L^a}=(\int |f_k|^a d\mu)^{1/a} \leq \| f_k\|_{L^\infty}(\mu(f_k \neq 0))^{1/a} \leq 2^{1/a}\cdot 2^{k(\frac{1}{a}-\frac{1}{p})}. $$ Compared to the author's result, there is a factor $2^{1/a}$ above, and I haven't used the imformation that $\| f\|_{L^p}=1.$ What can I do?