Given: A sequence ($a_n$) is called subadditive if $a_{m+n}$ ≤ $a_m$ + $a_n$ for all m, n ∈ N. Prove that if ($a_n$) is a subadditive sequence of positive real numbers, then $(\frac{a_n}{n})$ converges.
I am unaware of how to start this proof, but I know that it I need to show that $\frac{a_n}{n}$ --> inf{$\frac{a_k}{k}$: k $\geq$ 1}.
Any help would be appreciated.
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$\begingroup$Let $\epsilon >0$. Choose k such that $\alpha +\epsilon > \frac {a_k} k$ where $\alpha$ is the infimum of $\frac {a_n} n$. This is possible by definition of infimum. Now consider $\{1,k,2k,...\}$. Any positive integer $m$ lies between $nk$ and $(n+1)k$ for some $n$. Now $a_m \leq a_{nk} + a_{m-nk}$. Note that $a_{m-nk} \leq max\{a_j:1\leq j \leq k\}$. Call this maximum as M. Then $\frac {a_m} m \leq \frac {a_{nk}} m +\frac M m$. The last term approaches 0 as $m$ approaches $\infty$ and $ a_{nk} \leq {n a_k} $. Hence $a_{nk}<nk(\alpha +\epsilon)$. Divide by $m$ and note that $\frac {nk} m \leq 1$. Since $\epsilon$ is arbitrary we see that $limsup \frac {a_m} m$ does not exceed $\alpha$. Trivially, $\alpha$ does not exceed $\lim inf \frac {a_m} m$.
$\endgroup$ 2 $\begingroup$From $a_{n+m}\leq a_n+ a_m$, we have $(n+m)b_{n+m} \leq nb_n + mb_m$ for $b_n:=a_n/n$. Now, we show that the sequence $\{b_n\}_{n=1}^{\infty}$ is convergent. Indeed, by setting $m=1$, we have $(n+1) b_{n+1} \leq n b_n + b_1$. This is equivalent to $(n+1) (b_{n+1}-b_1) \leq n(b_n-b_1)$. It follows that $\{n(b_n-b_1)\}_{n=1}^{\infty}$ is a non-increasing function, so $\lim_{n\to \infty} n(b_n-b_1) =\beta$ for some $\beta \in R\cup \{-\infty\}$. Now, if $\beta =-\infty$, we have $b_n<0$ for $n$ sufficiently large, which contradicts to the assumption that $a_n \geq 0$. Hence, we must have $\beta \in R$ or $|\beta|<\infty$, so $b_n \to b_1$. In summary, we have $a_n/n \to a_1$ as $n\to \infty$.
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