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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.

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1 Answer 1

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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.

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