Suppose that a sequence $\{a_n\}_{n\geqslant 0}$ satisfies $$\frac{a_{m+n}+a_{|m-n|}}{2}\leqslant a_m+a_n$$
for all non-negative integers $m,n$. Show that $\frac{a_n}{n^2}$ either converges or diverges to $-\infty$ as $n\to\infty$.
Solution:
For an arbitrarily fixed positive integer $k$ we put $n=qk+r$ with $0\leqslant r<k$. Put $c_l=a_{kl+r}$ for brevity. Adding $m-1$ inequalities
$c_{l+1}+c_{l-1}\leqslant 2(c_l+a_k)$
for $l=1,...,m-1$ one after another, we get
$c_0+c_m\leqslant c_1+c_{m-1}+2(m-1)a_k$.
Adding the above inequalities for $m=2,...,M$, we thus obtain
$$c_M\leqslant Mc_1-(M-1)c_0+M(M-1)a_k$$.
Therefore
$\limsup_{n\to\infty}\frac{a_n}{n^2}\leqslant\frac{a_k}{k^2}$
and the sequence $\frac{a_n}{n^2}$ is bounded above. Since $k$ is arbitrary, we conclude that $\limsup_{n\to\infty}\frac{a_n}{n^2}\leqslant inf_{k\leqslant 1}\frac{a_k}{k^2}\leqslant \liminf_{k\to\infty}\frac{a_k}{k^2}$
which completes the proof.
Questions:
I am still stuck at the beginning of the proof.
1) Considering the expression $c_{l+1}+c_{l-1}\leqslant 2(c_l+a_k)$. Since $c_l=a_{kl+r}$ and $l=1$, I think c_1=a_{k+r}, however on the expression the $r$ vanishes. How?
2) Considering this expression $c_0+c_m\leqslant c_1+c_{m-1}+2(m-1)a_k$. How did the author derive $c_1+c_{m-1}+2(m-1)a_k$?
Thanks in advance!
$c_\ell = a_{k\ell +r}$, so the $r$ "vanishes". In detail, WLOG we can assume that $m \geqslant n$ in the assumption, then $$ a_{m+n} + a_{m-n}\leqslant 2(a_m + a_n), $$ put $p =m+n, q =m-n$, then $$ a_p + a_q \leqslant 2(a_{(p+q)/2} + a_{(p-q)/2}). $$ Thus, $$ c_{\ell+1} + c_{\ell-1} = a_{k(\ell+1)+r} + a_{k(\ell-1)+r} \leqslant 2(a_{k\ell+r} + a_k)= 2(c_\ell + a_k) $$
List these inequalities and add them together, you get 1 $(c_0 + c_1)$, 1 $(c_m + c_{m-1})$, 2 $c_j$ for $j = 2, \ldots, m-2$ on the LHS, and 2 $c_j$ for $j = 1,\ldots, m-1 $ plus $2(m-1)a_k$ on the RHS. After cancellation you should get $$ c_0 + c_1 + c_{m-1} + c_m \leqslant 2c_1 + 2c _{m-1} \leqslant 2(m-1)a_k. $$ Then you get the inequality.