Let $f_n: \mathcal{X} \to [0,1]$ be a sequence of functions and $\alpha \in (0,1)$.
I want to show that
$$ \limsup_{n \to \infty} \sup_{x \in \mathcal{X}} f_n(x) \leq \alpha $$
implies
$$ \sup_{x \in \mathcal{X}} \limsup_{n \to \infty} f_n(x) \leq \alpha $$
but my $\limsup$'s are a little rusty. Any tips?
Suppose $$\sup_{x\in X}\limsup_{n\to \infty }f_n(x)>\alpha .$$ Then, there is $y\in X$ s.t. $$\limsup_{n\to \infty }f_n(y)>\alpha ,$$ i.e. there is $N$ s.t. $$\sup_{k\geq n}f_k(y)>\alpha.$$ for all $n\geq N$.
But $$f_k(y)\leq \sup_{x\in X}f_k(y),$$ i.e., if $n\geq N$ $$\alpha <\sup_{k\geq n}f_k(y)\leq \sup_{k\geq n}\sup_{x\in X}f_k(x).$$ Therefore $$\limsup_{n\to \infty }\sup_{x\in X}f_n(x)>\alpha .$$ Contradiction.
Consider, for each $n$, $$m_n=\sup_{x \in \mathcal{X}} f_n(x)$$ Then, for every $x$ in $\mathcal X$ and every $n$, $$f_n(x)\leqslant m_n$$ hence, for each given $x$ in $\mathcal X$, $$\limsup\limits_{n\to\infty}f_n(x)\leqslant\limsup\limits_{n\to\infty}m_n$$ This inequality holds for every $x$ in $\mathcal X$, hence, as desired, $$\sup_{x\in\mathcal X}\limsup\limits_{n\to\infty}f_n(x)\leqslant\limsup\limits_{n\to\infty}m_n$$
