In solving Brezis' exercise 6.23.2, I have to verify that
Let $(a_n)$ be a sub-additive sequence of real numbers, i.e., $a_{n+m} \le a_n + a_m$ for all $n,m \ge 1$. Then $\lim_n \frac{a_n}{n} = \inf_n \frac{a_n}{n}$.
There are possibly subtle mistakes that I could not recognize in below attempt. Could you please have a check on it?
It's clear that $\alpha := \inf_n \frac{a_n}{n} \le \liminf_n \frac{a_n}{n}$. It remains to prove that $\limsup_n \frac{a_n}{n} \le \alpha$. Fix $m \ge 1$. It suffices to prove that $\limsup_n \frac{a_n}{n} \le \frac{a_m}{m}$.
For each $n \ge 1$, we write $n=m q+r$ with $m$ being the divisor, $q$ the quotient and $r$ the remainder. Then $0 \le r <m$. By sub-additivity of $(a_n)$, we have $a_n \le q a_m + a_r$ and thus $$ \frac{a_n}{_n} \le \frac{mq}{n} \frac{a_m}{m} + \frac{a_r}{n}. $$
On the other hand, $\frac{mq}{n} = \frac{n-r}{n}$ and thus $\lim_n \frac{mq}{n} =1$. Then $$ \begin{align*} \limsup_n \frac{a_n}{n} &\le \limsup_n\frac{mq}{n} \frac{a_m}{m} + \lim_n \frac{a_r}{n} \\ &= \frac{a_m}{m}. \end{align*} $$
The claim then follows.