
> 
> Let $ \{a\_k\}\_{i=1}^n $ is a positive sequence. For $ 0<p<\infty $, space $ L^{p,\infty} $ is defined by
> $$
> \left\{f:\|f\|\_{p,\infty}=\inf\left\{C>0:a\_f(\lambda)\leq C/\lambda^p\right\}\right\}
> $$
> where
> $$
> a\_f(\lambda)=|\{x\in\mathbb{R}^n:|f(x)|>\lambda\}|.
> $$
> Then
> $$
> \left\|\sum\_{k=1}^{n}f\_k\right\|\_{1,\infty}\leq \left(1+\sum\_{k=1}^{n}a\_k\right)\sum\_{k=1}^{n}\left(\ln(1+a\_k^{-1})\|f\_k\|\_{1,\infty}\right)
> $$
> 
> 
> 


I do not know why there are $ a\_k $ in the inequality. Can you give me some hints or references?



How to prove this inequality for the norm $ \|\cdot\|_{1,\infty} $?
For any $ \lambda>0 $, we have
\begin{align}
\left|\left\{x:\left|\sum\_{i=1}^nf\_k\right|>\left(1+\sum\_{i=1}^na\_k\right)\lambda\right\}\right|&\leq\left|\left\{x:\sum\_{i=1}^n|f\_k|>\left(1+\sum\_{i=1}^na\_k\right)\lambda\right\}\right|\\
&\leq\left|\left\{x:\sum\_{i=1}^n(|f\_k|-a\_k\lambda)>\lambda\right\}\right|\\
&\leq\left|\left\{x:\sum\_{i=1}^n(|f\_k|-a\_k\lambda)\_+>\lambda\right\}\right|.
\end{align}
For any $ \varepsilon>0 $, set
$$
g\_k(x)=\min\{(|f\_k|-a\_k\lambda)\_+,\lambda(1+\varepsilon)\}.
$$
Then
\begin{align}
\left|\left\{x:\left|\sum\_{i=1}^nf\_k\right|>\left(1+\sum\_{i=1}^na\_k\right)\lambda\right\}\right|&\leq\left|\left\{x:\sum\_{i=1}^ng\_k>\lambda\right\}\right|\\
&\leq\frac{\left\|\sum\_{i=1}^ng\_k\right\|\_{1,\infty}}{\lambda}\\
&\leq\frac{\sum\_{i=1}^n\|g\_k\|\_{1}}{\lambda}.
\end{align}
Finally, note that
\begin{align}
\|g\_k\|\_1&=\int\_{0}^{\lambda(1+\varepsilon)}|\{x:|g\_k|>\mu\}|d\mu\\
&\leq\int\_{0}^{\lambda(1+\varepsilon)}|\{x:|f\_k|>\mu+a\_k\lambda\}|d\mu\\
&\leq\int\_{0}^{\lambda(1+\varepsilon)}\frac{\|f\_k\|\_{1,\infty}}{\mu+a\_k\lambda}d\mu=\ln\left(\frac{a\_k\lambda+\lambda(1+\varepsilon)}{a\_k\lambda}\right).
\end{align}
Combining all above and letting $ \varepsilon\to 0 $, we can get the proof.


