# If $(t_n u_n-t_m u_m)(u_n-u_m) \le 0$ for all $m, n \in \mathbb N$, then $(u_n)$ is convergent

I'm trying to solve below exercise, i.e.,

Let $$(u_n) \subset \mathbb R$$ and $$(t_n) \subset \mathbb R_{>0}$$ such that $$(t_n)$$ is non-decreasing. Assume that $$(t_n u_n-t_m u_m)(u_n-u_m) \le 0 \quad \forall m, n \in \mathbb N.$$ Then $$(u_n)$$ is convergent.

1. Could you have a check on my below attempt?
2. Is there a more direct approach?

Thank you so much for your elaboration!

We have \begin{aligned} & (t_{n+1} u_{n+1}-t_n u_n)(u_{n+1}-u_n) \le 0 \\ \iff & (u_{n+1}-u_n)^2 \le \frac{t_{n} - t_{n+1}}{t_{n+1}} (u_{n+1}-u_n) u_n. \end{aligned}

If $$t_{n} - t_{n+1}=0$$ then $$u_{n+1} = u_n$$. Let $$t_{n} - t_{n+1}<0$$. Because $$(t_n)$$ is non-decreasing, $$\frac{|t_{n} - t_{n+1}|}{t_{n+1}} \le 1$$ and thus $$|u_{n+1}-u_n| \le |u_n|$$. We have three cases, i.e.,

1. If $$u_n=0$$ then $$u_{n+1}=0$$.
2. If $$u_n>0$$ then $$u_{n+1} \le u_n$$. On the other hand, $$|u_{n+1}-u_n| \le |u_n|$$. So $$0 \le u_{n+1} \le u_n$$.
3. If $$u_n<0$$ then $$u_{n+1} \ge u_n$$. On the other hand, $$|u_{n+1}-u_n| \le |u_n|$$. So $$u_n \le u_{n+1} \le 0$$.

It follows that

• if $$u_0=0$$, then $$u_n=0$$ for all $$n$$.
• if $$u_0>0$$, then $$(u_n) \subset \mathbb R_{>0}$$ is decreasing.
• if $$u_0<0$$, then $$(u_n) \subset \mathbb R_{<0}$$ is increasing.

Thus $$(u_n)$$ is convergent.

Your proof is correct. Instead of taking absolute values one can also rewrite the given inequality as $$t_n u_n^2 + t_m u_m^2 \le (t_n + t_m) u_n u_m \, ,$$ which implies that $$u_n$$ and $$u_m$$ are both zero or both positive or both negative, so that the $$u_n$$ are all zero or all positive or all negative.
Then $$0 \le t_{n+1}(u_{n+1}-u_n)^2 \le (t_{n+1}-t_n) u_n (u_n-u_{n+1})$$ implies that $$u_n$$ and $$u_n-u_{n+1}$$ have the same sign, so that (as you correctly concluded) the sequence is identically zero, or positive and decreasing, or negative and increasing.