SHORTLISTED PROBLEMS FOR THE 62nd NMO

Let $L = \lim_{n \to \infty} x_n$. Since $x_n = \max\{a_n, a_{n+1}, a_{n+2}\}$, for every $n$, $a_n \le x_n$, $a_{n+1} \le x_n$, $a_{n+2} \le x_n$.

Let $\varepsilon > 0$. Since $x_n \to L$, there exists $N_1$ such that for all $n \ge N_1$, $|x_n - L| < \varepsilon/2$, i.e., $L - \varepsilon/2 < x_n < L + \varepsilon/2$.

Also, since $y_n = a_{n+1} - a_n \to 0$, there exists $N_2$ such that for all $n \ge N_2$, $|a_{n+1} - a_n| < \varepsilon/4$.

Let $N = \max\{N_1, N_2\}$. For $n \ge N$, $a_n, a_{n+1}, a_{n+2} \le x_n < L + \varepsilon/2$.

We claim that $a_n \to L$.

First, we show that $\limsup a_n \le L$.

Suppose not. Then there exists a subsequence $a_{n_k} \to L' > L$. But then for large $k$, $a_{n_k} > L + \varepsilon$ for some $\varepsilon > 0$. But $x_{n_k} \ge a_{n_k} > L + \varepsilon$, contradicting $x_{n_k} \to L$.

So $\limsup a_n \le L$.

Now, we show that $\liminf a_n \ge L$.

Suppose not. Then there exists a subsequence $a_{m_k} \to L'' < L$. For $k$ large, $a_{m_k} < L - \varepsilon$ for some $\varepsilon > 0$.

But $a_{m_k+1} = a_{m_k} + y_{m_k}$, $a_{m_k+2} = a_{m_k+1} + y_{m_k+1} = a_{m_k} + y_{m_k} + y_{m_k+1}$.

Since $y_n \to 0$, for large $k$, $|y_{m_k}| < \varepsilon/4$, $|y_{m_k+1}| < \varepsilon/4$.

So $a_{m_k+1} < L - \varepsilon + \varepsilon/4 = L - 3\varepsilon/4$,
$a_{m_k+2} < L - \varepsilon + \varepsilon/4 + \varepsilon/4 = L - \varepsilon/2$.

Therefore, $x_{m_k} = \max\{a_{m_k}, a_{m_k+1}, a_{m_k+2}\} < L - \varepsilon/2$, contradicting $x_{m_k} \to L$.

Therefore, $\liminf a_n \ge L$.

Thus, $a_n \to L$.