Prove the following properties of sequence

Question

Define $$L = \limsup_{k \rightarrow \infty}a_k =\inf_j(\sup_{k\geq j}a_k).$$

Prove that if $(a_k)$ and $(b_k)$ are sequence of real numbers then $$\limsup(a_k + b_k) \leq \limsup a_k + \limsup b_k.$$

I tried to use the very definition of sup and inf but my idea does not converge to the result. Any hint?

Answer 1

Let $\epsilon > 0$. First notice that since $\{ k \ge j\} \subset \{k \ge l\}$ whenever $j \ge l$, we get that $\sup_{k \ge j} a_k \le \sup_{k \ge l} a_k$, since we're supping over a smaller set; we're at best losing terms. From the definition of infimum, choose $j_1$ and $j_2$ so that $\sup_{k \ge j_1} a_k - \epsilon < \limsup_{k \to \infty} a_k$ and $\sup_{k \ge j_2} b_k - \epsilon < \limsup_{k \to \infty} b_k $. Choosing $j = \max \{j_1, j_2\}$, we have that $\sup_{k \ge j} a_k - \epsilon < \limsup a_k$ and $\sup_{k \ge j} b_k - \epsilon < \limsup b_k$/

So, $\limsup_{k \to \infty} (a_k + b_k) \le \sup_{k \ge j} (a_k + b_k) \le \sup_{k \ge j} a_k + \sup_{k \ge j} b_k < \limsup a_k + \limsup b_k + 2\epsilon$.

Letting $\epsilon \to 0$, we're done.

Answer 2

Hint. Let $(a_n)^\infty_{n=m}$ be a sequence of real number, and let $x\in\mathbf R$. Prove by contradiction and use

1. For every $x<\limsup_{n\to\infty}a_n$, and every $N\ge m$, there exists $n\ge N$ such that $a_n>x$. (In other words, for every $x<\limsup_{n\to\infty}a_n$, the elements of the sequence $(a_n)^\infty_{n=m}$ exceed $x$ infinitely often.)
2. For every $x>\limsup_{n\to\infty}a_n$, there exists an $N\ge m$ such that $a_n<x$ for all $n\ge N$. (In other words, for every $x>\limsup_{n\to\infty}a_n$, the elements of the sequence $(a_n)^\infty_{n=m}$ are eventually less than $x$.)