0
$\begingroup$

Let $\{x_n\}$ be a sequence in $\mathbb{R}_+^\mathbb{Z}$. Define
\begin{align} a_n&=\sup_{k \ge n} \{x_k\} \\ b_n&=a_n+\frac{1}{n} \\ c_1&=b_1 \text{ and } c_n=\max\left(c_{n-1}-\frac{1}{n},b_n\right) \end{align} Show $ \frac{1}{n} \ge c_n - c_{n+1} \ge \frac{1}{n+1} $

Writing \begin{align} c_n &\ge b_{n+1}, \\ c_n &\ge c_{n-1}-\frac{1}{n} \end{align} I get $ c_n-c_{n+1} \le c_n - c_ n+ \frac{1}{n+1} = \frac{1}{n+1} \le \frac{1}{n} $ I don't see how to get the lower bound.

  • 1
    A lower bound $1/(n+1)$ is most unlikely since you proved that $1/(n+1)$ is an **upper bound** of $c_n-c_{n+1}$.2012-10-15

1 Answers 1

1

The stated lower bound can fail. Take $x_n=1$ for all $n\in\Bbb N$. Then for $n\in\Bbb N$ we have $a_n=1$ and $b_n=1+\frac1n$. Finally, $c_1=b_1=2$, $c_2=\max\left\{2-\frac12,1+\frac12\right\}=\frac32\;,$ and $c_3=\max\left\{\frac32-\frac13,1+\frac13\right\}=\max\left\{\frac76,\frac43\right\}=\frac43\;,$ so

$c_2-c_3=\frac32-\frac43=\frac16<\frac13\;.$