Subfield: Algebra


Question: A sequence of real numbers $a_{1}, a_{2}, \ldots$ satisfies the relation

$$
a_{n}=-\max _{i+j=n}\left(a_{i}+a_{j}\right) \quad \text { for all } n>2017
$$

Prove that this sequence is bounded, i.e., there is a constant $M$ such that $\left|a_{n}\right| \leqslant M$ for all positive integers $n$.

Solution:
Set $D=2017$. Denote

$$
M_{n}=\max _{k<n} a_{k} \quad \text { and } \quad m_{n}=-\min _{k<n} a_{k}=\max _{k<n}\left(-a_{k}\right)
$$

Clearly, the sequences $\left(m_{n}\right)$ and $\left(M_{n}\right)$ are nondecreasing. We need to prove that both are bounded.

Consider an arbitrary $n>D$; our first aim is to bound $a_{n}$ in terms of $m_{n}$ and $M_{n}$.

(i) There exist indices $p$ and $q$ such that $a_{n}=-\left(a_{p}+a_{q}\right)$ and $p+q=n$. Since $a_{p}, a_{q} \leqslant M_{n}$, we have $a_{n} \geqslant-2 M_{n}$.

(ii) On the other hand, choose an index $k<n$ such that $a_{k}=M_{n}$. Then, we have

$$
a_{n}=-\max _{\ell<n}\left(a_{n-\ell}+a_{\ell}\right) \leqslant-\left(a_{n-k}+a_{k}\right)=-a_{n-k}-M_{n} \leqslant m_{n}-M_{n} .
$$

Summarizing (i) and (ii), we get

$$
-2 M_{n} \leqslant a_{n} \leqslant m_{n}-M_{n}
$$

whence

$$
m_{n} \leqslant m_{n+1} \leqslant \max \left\{m_{n}, 2 M_{n}\right\} \quad \text { and } \quad M_{n} \leqslant M_{n+1} \leqslant \max \left\{M_{n}, m_{n}-M_{n}\right\}
\tag{1}
$$

Now, say that an index $n>D$ is lucky if $m_{n} \leqslant 2 M_{n}$. Two cases are possible.

Case 1. Assume that there exists a lucky index $n$. In this case, (1) yields $m_{n+1} \leqslant 2 M_{n}$ and $M_{n} \leqslant M_{n+1} \leqslant M_{n}$. Therefore, $M_{n+1}=M_{n}$ and $m_{n+1} \leqslant 2 M_{n}=2 M_{n+1}$. So, the index $n+1$ is also lucky, and $M_{n+1}=M_{n}$. Applying the same arguments repeatedly, we obtain that all indices $k>n$ are lucky (i.e., $m_{k} \leqslant 2 M_{k}$ for all these indices), and $M_{k}=M_{n}$ for all such indices. Thus, all of the $m_{k}$ and $M_{k}$ are bounded by $2 M_{n}$.

Case 2. Assume now that there is no lucky index, i.e., $2 M_{n}<m_{n}$ for all $n>D$. Then (1) shows that for all $n>D$ we have $m_{n} \leqslant m_{n+1} \leqslant m_{n}$, so $m_{n}=m_{D+1}$ for all $n>D$. Since $M_{n}<m_{n} / 2$ for all such indices, all of the $m_{n}$ and $M_{n}$ are bounded by $m_{D+1}$.

Thus, in both cases the sequences $\left(m_{n}\right)$ and $\left(M_{n}\right)$ are bounded, as desired.
As in the previous solution, let $D=2017$. If the sequence is bounded above, say, by $Q$, then we have that $a_{n} \geqslant \min \left\{a_{1}, \ldots, a_{D},-2 Q\right\}$ for all $n$, so the sequence is bounded. Assume for sake of contradiction that the sequence is not bounded above. Let $\ell=\min \left\{a_{1}, \ldots, a_{D}\right\}$, and $L=\max \left\{a_{1}, \ldots, a_{D}\right\}$. Call an index $n$ good if the following criteria hold:

$$
a_{n}>a_{i} \text { for each } i<n, \quad a_{n}>-2 \ell, \quad \text { and } \quad n>D
\tag{2}
$$

We first show that there must be some good index $n$. By assumption, we may take an index $N$ such that $a_{N}>\max \{L,-2 \ell\}$. Choose $n$ minimally such that $a_{n}=\max \left\{a_{1}, a_{2}, \ldots, a_{N}\right\}$. Now, the first condition in (2) is satisfied because of the minimality of $n$, and the second and third conditions are satisfied because $a_{n} \geqslant a_{N}>L,-2 \ell$, and $L \geqslant a_{i}$ for every $i$ such that $1 \leqslant i \leqslant D$.



Let $n$ be a good index. We derive a contradiction. We have that

$$
a_{n}+a_{u}+a_{v} \leqslant 0\tag{3}
$$

whenever $u+v=n$.

We define the index $u$ to maximize $a_{u}$ over $1 \leqslant u \leqslant n-1$, and let $v=n-u$. Then, we note that $a_{u} \geqslant a_{v}$ by the maximality of $a_{u}$.

Assume first that $v \leqslant D$. Then, we have that

$$
a_{N}+2 \ell \leqslant 0
$$

because $a_{u} \geqslant a_{v} \geqslant \ell$. But this contradicts our assumption that $a_{n}>-2 \ell$ in the second criteria of (2).

Now assume that $v>D$. Then, there exist some indices $w_{1}, w_{2}$ summing up to $v$ such that

$$
a_{v}+a_{w_{1}}+a_{w_{2}}=0
$$

But combining this with (3), we have

$$
a_{n}+a_{u} \leqslant a_{w_{1}}+a_{w_{2}}
$$

Because $a_{n}>a_{u}$, we have that $\max \left\{a_{w_{1}}, a_{w_{2}}\right\}>a_{u}$. But since each of the $w_{i}$ is less than $v$, this contradicts the maximality of $a_{u}$.