Convergence of a sequence defined by a recurrence inequality

Question

Let $(u_n)$ be a sequence of real numbers with the following property \begin{eqnarray*} \forall\,m,n\ge1,\quad u_{m+n}\le \frac{m}{m+n}u_m+\frac{n}{m+n}u_n \end{eqnarray*} Is it true that $(u_n)$ converges to $\inf u_n$ which is either $-\infty$ or a finite $u\in\mathbb{R}$? There's no obvious monotonicity property that I was able to use. It's not hard to see that $u_{2n}\le u_n$, but I was unable to make any use of it. My attempts to show that $(u_n)$ is Cauchy were not successful. Neither was my attempt to prove that $(u_{2n})$ and $(u_{2n+1})$ are convergent. Any hints would be appreciated.

Answer 1

I think I have another solution relying on proving the convergence of $(u_{2n+1}-u_1)$ and $(u_{2n}-u_1)$ to the same limit. For convenience, let $v_n=u_{2n+1}-u_1$ and $w_n=u_{2n}-u_1$ (set $v_0$ and $w_0$ to $0$). Using the inequality property of $(u_n)$, it is not hard to establish that \begin{eqnarray*} v_n\le\left(1-\frac{1}{2n+1}\right)w_n,\, w_n\le\left(1-\frac{1}{2n}\right)v_{n-1} \end{eqnarray*} Hence, \begin{eqnarray*} v_n\le\left(1-\frac{1}{n+1}\right)v_{n-1},\,w_n\le\left(1-\frac{1}{n}\right)w_{n-1} \end{eqnarray*} In other terms, $(v_n)$ and $(w_n)$ are decreasing sequences of real numbers.