How can we prove that $\displaystyle \limsup_{n \to \infty } b_n \le \limsup_{n \to \infty } a_n$

Question

Let $(a_n)_{n\ge1}$ be a bounded sequence in $\mathbb{R}$

Set $\displaystyle B_n=\left\{\sum_{j=n}^{\infty}(\theta_j\cdot a_j)\;:\theta_j\ge 0 \;,\;\sum_{j=n}^{\infty}\theta_j=1\right\}$ for $n\ge1$

Let $(b_n)_{n\ge 1}$ be a sequence such that $b_n \in B_n$ , $\forall \;n\ge1$

How can we prove that $\displaystyle \limsup_{n \to \infty } b_n \le \limsup_{n \to \infty } a_n$ ?

Any hints would be appreciated.

Answer 1

Let $n$ be fixed; then for all $j \ge n$, we have $a_j \le \sup\limits_{m \ge n} a_m$. Therefore:

$$b_n = \sum_{j=n}^\infty \theta_j a_j \le \sum_{j=n}^\infty \theta_j \sup_{m \ge n} a_m = \sup_{m \ge n} a_m$$

Moreover, since $\sup\limits_{m \ge n} a_m$ is decreasing as $n$ increases, we have for any $n' \ge n$ that: $$b_{n'} \le \sup\limits_{m\ge n'} a_m \le \sup\limits_{m \ge n} a_m$$

Can you take it from here?