$\frac{a_{m+n}+a_{|m-n|}}{2}\leqslant a_m+a_n$ and the inequality $\limsup_{n\to\infty}\frac{a_n}{n^2}\leqslant\frac{a_k}{k^2}$

Question

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.

Question:

I have already asked a question about this proof. However as I went through I found myself with another doubt.

How does the author infers form this inequality $c_M\leqslant Mc_1-(M-1)c_0+M(M-1)a_k$ the following inequality $\limsup_{n\to\infty}\frac{a_n}{n^2}\leqslant\frac{a_k}{k^2}$?

Thanks in advance!

Answer 1

At this point, it is probably better to stick with write $q$ instead of $M$, so starting with $$a_n=c_q\leq qc_1 - (q-1)c_0 + q(q-1) a_k$$ dividing by $n^2$, you get $$ \frac{a_n}{n^2} \leq \frac{q}{n^2} c_1 - \frac{q-1}{n^2} c_0 + \frac{q(q-1)}{(kq+r)^2} a_k$$ So taking limsup, we get $$\limsup_{n=kq+r\to\infty} \frac{a_n}{n^2} \leq \frac{a_k}{k^2}$$ since the other two terms have order $n$ numerator but $n^2$ in denominator, so vanish in the limit. But this holds true for all possible remainder $r=0,1,\dots,k-1$, so we can finally drop the dependence on $r$ and just write $$\limsup_{n\to\infty} \frac{a_n}{n^2} \leq \frac{a_k}{k^2}$$

Answer 2

This is to provide an intuitive motivation for the proof.

The premise can be rewritten as $$a_{m+n}-2a_m+a_{m-n}\le 2a_n$$ letting $m>n$, or using the notation from the question $$c_{l+1}-2c_l+c_{l-1}\le 2a_k.$$ The left hand side is the second order difference operator operating on the sequence $c_l$ or $a_n$. The solution would naturally be to apply the summation operator twice. This method naturally generalizes to arbitrary difference orders.