Let $a_{n,k}\in \mathbb{R}$ with:
- $0\leq a_{n,k}\leq M$
- $\displaystyle a_{n,k} \rightarrow_k a_n$, uniformly with $n$
Prove: $$\displaystyle \sup_n\lim_ka_{n,k}\geq\limsup_k\sup_na_{n,k}$$
Does anybody know of a result that would help me out?
If I knew $\displaystyle a_{n,k} \rightarrow_n a_k$ I could conclude the existence of the double limits, and their equality, however here I cannot suppose this.
Take $\epsilon \gt 0$. By hypothesis on the uniform convergence of $\{a_{n,k}\}$ it exists $K \in \mathbb N$ such that for $k \ge K$ and $n \in \mathbb N$
$$ a_n - \epsilon \lt a_{n,k} \lt a_n + \epsilon.$$
As $0\leq a_{n,k}\leq M$ for all $n,k \in \mathbb N$ $\sup_n a_n$ exists and for $k \ge K$ we have
$$\sup_n a_{n,k} \le \sup_n a_n + \epsilon$$ which means that
$$\sup_n a_{n,k} \le \sup_n \lim_k a_{n,k} + \epsilon$$ as $a_n$ is defined by $a_n = \lim_k a_{n,k}$. As the sequence in $k$ $\{\sup_n a_{n,k}\}$ is less than the real number $\sup_n \lim_k a_{n,k} + \epsilon$ for $k \ge K$ we obtain
$$\limsup_k\sup_n a_{n,k} \le \sup_n \lim_k a_{n,k} + \epsilon.$$ As this is true for any $\epsilon \gt 0$ we finally get the desired result
$$\limsup_k\sup_n a_{n,k} \le \sup_n \lim_k a_{n,k}.$$