$\limsup c_n\le \max(\limsup a_n,\limsup b_n)$

Question

have a question that im stuck on here

Let $a_n, b_n$ and $c_n$ be three sequences of real numbers.

Suppose $k_n \in [0,1]$ for all $n$. Let $$c_n = (k_n)(a_n) + (1-k_n)b_n\;.$$ Assuming that $\limsup_{n\to\infty}(a_n)$ and $\limsup_{n\to\infty}(b_n)$ are finite, prove the following inequality,

$$\limsup_{n\to\infty}(c_n) \le \max\big(\limsup_{n\to\infty}(a_n),\limsup_{n\to\infty}(b_n)\big)$$

Answer 1

Recall the definition: $\limsup_{n \to \infty} a_n = \lim_{n \to \infty} \sup_{m \geq n} a_m$.

Since $k_n \in [0,1]$, you have $c_n \leq \max(a_n,b_n)$, and so $\sup_{m \geq n} c_m \leq \sup_{m \geq n} \max(a_n,b_n) \leq \max(\sup_{m \geq n} a_n, \sup_{m \geq n} b_n)$. Now take limits to get the desired result.

Answer 2

HINT: For each $n$ let $I_n$ be the closed interval with endpoints $a_n$ and $b_n$. (If $a_n=b_n$, $I_n$ is the degenerate interval $[a_n,a_n]=\{a_n\}$.) Note that $c_n\in I_n$ for each $n$. Suppose that $$\limsup_nc_n>\max\{\limsup_na_n,\limsup_nb_n\}\;.$$ let $u$ be any real number satisfying $$\max\{\limsup_na_n,\limsup_nb_n\}<u<\limsup_nc_n\;.$$ There must be an $m\in\Bbb Z^+$ such that $$\sup_{n\ge m}a_n<u\quad\text{and}\quad\sup_{n\ge m}b_n<u\;;$$ why? This implies that $c_n<u$ for all $n\ge m$; why? What contradiction does this entail?