Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 6.4 National Olympiad Prove it Ireland

Suppose a doubly infinite sequence of real numbers
$\dots, a_{-2}, a_{-1}, a_0, a_1, a_2, \dots$
has the following sub-Fibonacci property:
$a_{n+2} = \frac{a_n + a_{n+1}}{2}, \quad \text{for all integers } n.$
Show that if this sequence is bounded (i.e. if there exists a number $R$ such that $|a_n| \le R$ for all $n$ ), then $a_n$ has the same value for all $n$ .

Solution

For any $n \in \mathbb{Z}$ let $d_n = a_{n+1} - a_n$ . Then, for $n \in \mathbb{Z}$ ,
$2d_{n+1} = 2a_{n+2} - 2a_{n+1} = (a_{n+1} + a_n) - 2a_{n+1} = a_n - a_{n+1} = -d_n.$
This implies $d_n = (-2)^{-n}d_0$ for all $n \in \mathbb{Z}$ .
If $d_0 = 0$ , then $d_n = 0$ for all $n \in \mathbb{Z}$ , hence $a_n = a_0$ for all $n$ . If $d_0 \neq 0$ and $R > 0$ is any given number, there exists an integer $n > 0$ so that $d_{-n} = (-2)^n d_0 > 2R$ . If $|a_{-n}| < R$ and $|a_{-n+1}| < R$ , then
$|d_{-n}| = |a_{-n+1} - a_{-n}| \le |a_{-n+1}| + |a_{-n}| < 2R$

in contradiction to the choice of $n$ . This shows that the sequence cannot be bounded if it is not constant.

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