Proof of Fekete’s subadditive lemma

If there is a $m$ such that $a_{m}=-\\infty$ , then, by subadditivity, we have $a_{n}=-\\infty$ for all $n>m$ . Then, both sides of the equality are $-\\infty$ , and the theorem holds.So, we suppose that $a_{n}\\in\\textbf{R}$ for all $n$ . Let $L=\\inf_{n}\\frac{a_{n}}{n}$ and let $B$ be any number greater than $L$ . Choose $k\\geq 1$ such that

\\frac{a_{k}}{k}<B

For $n>k$ , we have, by the division algorithm there are integers $p_{n}$ and $q_{n}$ such that $n=p_{n}k+q_{n}$ , and $0\\leq q_{n}\\leq k-1$ .Applying the definition of subadditivity many times we obtain:

a_{n}=a_{p_{n}k+q_{n}}\\leq a_{p_{n}k}+a_{q_{n}}\\leq p_{n}a_{k}+a_{q_{n}}

So, dividing by $n$ we obtain:

\\frac{a_{n}}{n}\\leq\\frac{p_{n}k}{n}\\frac{a_{k}}{k}+\\frac{a_{q_{n}}}{n}

When $n$ goes to infinity, $\\frac{p_{n}k}{n}$ converges to $1$ and $\\frac{a_{q_{n}}}{n}$ converges to zero, because the numerator is bounded by the maximum of $a_{i}$ with $0\\leq i\\leq k-1$ . So, we have, for all $B>L$ :

L\\leq\\lim_{n}\\frac{a_{n}}{n}\\leq\\frac{a_{k}}{k}<B

Finally, let $B$ go to $L$ and we obtain

L=\\inf_{n}\\frac{a_{n}}{n}=\\lim_{n}\\frac{a_{n}}{n}