Interchanging limit and supremum Sequence of functions is bounded $f_k(x) \le c\ \forall\ k, \forall\ x \ge 0,\ c \in \mathbb{R}$ and decreasing $\forall\ x \ge 0$. Is it possible to show such inequality $$\lim_{x \to \infty} \sup_{k \ge 1} f_k(x) \le \sup_{k \ge 1} \lim_{x \to \infty} f_k(x) \text{ ?}$$

Here is a counter-example: let $$f_k(x) = 1 - \left( \frac {x}{1+x}\right)^k, \quad x\ge 0$$
Then:


*

*since $g : [0,\infty) \to [0,1) : x \mapsto \frac {x}{1+x}$ is increasing, and $h : [0,1) \to [0,1) : x \mapsto x^k$ is also increasing, so is their composition $h\circ g$. Therefore $f_k : [0,\infty) \to (0,1] : x \mapsto 1 - h\circ g(x)$ is decreasing and bounded below by $0$.

*For all $k, \lim_{x \to \infty} f_k(x) = 0$, so $\sup_k \lim_{x \to \infty} f_k(x) = 0$.

*For all $x, \sup_k f_k(x) = 1$, so $\lim_{x \to \infty} \sup_k f_k(x) = 1$.



Original post (where I thought the sequence was decreasing, not the functions):
If the sequence of functions is decreasing, then for all $k, x, f_1(x) \ge f_k(x)$. Therefore $\sup_{k \ge 1} f_k(x) = f_1(x)$. And $$\lim_{x\to\infty} f_1(x) \ge \lim_{x\to\infty} f_k(x)$$ when they converge (or diverge to $\pm\infty$). So again $$\lim_{x\to\infty} f_1(x) \ge \sup_{k \ge 1} \lim_{x\to\infty} f_k(x)$$
But since the LHS is also one of the values the supremum is being taken over, we must have 
$$\lim_{x\to\infty} f_1(x) = \sup_{k \ge 1} \lim_{x\to\infty} f_k(x)$$
Thus 
$$\lim_{x\to\infty} \sup_{k \ge 1} f_k(x) = \sup_{k \ge 1} \lim_{x\to\infty} f_k(x)$$
Being bounded below by a constant is not necessary, but convergence of the limits is.
