I'd appreciate hints on proving the following theorem:
If $f(x) = \lim_{n\to\infty} f_n(x)$ for each $x \in [0,1]$ and $M = \sup_{x\in[0,1]} f(x)$, then there is $t \in [0,1]$ such that $f(t) = M$. We are also given (from a previous part to this problem) that $f_n : [0,1] \to [0,\infty)$ is continuous for each $n$, and $f_n$ is a decreasing sequence of functions.
Thus far, I've tried following another fellow student's work, which follows in abbreviated format:
Set $M_n = \sup_{x\in[0,1]} f_n(x_n)$. Note $M_n$ is decreasing and bounded below by $M$. Thus, $M_n$ converges to some $N$, where $N \geq M$. Choose $t$ such that there is a subsequence of $x_n$, $t_k$, where $t_k \to t$. (Such a subsequence exists since $\{x_n\} \subseteq [0,1]$, which is compact.) We want to manipulate the following string of inequalities:
$$M \leq N = \lim_{k\to\infty} M_k = \lim_{k\to\infty} \sup_{x\in[0,1]} f_k(t_k) = \sup_{x\in[0,1]} f_k(t)$$ where the last step theoretically uses the continuity of the $f_k$s, but we can't take the limit of each $k$ individually.... Ideas? I have also tried using contradiction, but that hasn't gotten me anywhere so far.
Thank you in advance.
$f(x)=\inf_n{f_n(x)}$ and $f_n$ continuous, means $f$ upper continuous as {$f(x)<\alpha$} is the union of the open sets {$f_n(x)<\alpha$}, hence open; but then using the above definition of upper continuity, it is easy to see that for any $x_0$ in the domain of $f$, $\limsup f(x) \leq f(x_0)$ as $x$ goes to $x_0$, so in particular if $x_0 \in [0,1]$ is a limit point of $x_n$ for which $f(x_n)$ converges to $M$, $f(x_0) \geq M$, hence $f(x_0)=M$ by the definition of $M$.