Proof that the pointwise limit of this sequence of functions attains its supremum For each $n\in\mathbb{N}$, let $f_n : [0,1]\rightarrow [0,\infty)$ be a continuous function, and assume that the sequence $\{f_n\}$ satisfies the property that for all $x\in[0,1]$ and for all $n\in\mathbb{N}$, $f_n(x) \geq f_{n+1}(x)$.
Prove that the sequence $\{f_n\}$ is pointwise convergent:
Fix an $x \in \mathbb{R}$. Then the sequence $\{f_n(x)\}$ is decreasing and bounded below by $0$. Therefore the sequence $\{f_n(x)\}$ converges for every fixed $x \in \mathbb{R}$, and so $\{f_n\}$ is pointwise convergent on $\mathbb{R}$.
Set $f(x) = \lim_{n\rightarrow\infty}f_n(x)$ for each $x\in[0,1]$ and set $M = \sup_{x\in[0,1]}f(x)$. Prove that there exists $t\in[0,1]$ such that $f(t) = M$.
For each $n$, set $M_n = \sup_{x\in[0,1]} f_n(x)$, and let $x_n$ be a sequence such that $f_n(x_n) = M_n$ (as each $f_n$ is continuous on a compact set and so attains its supremum. We have that for all $n\in\mathbb{N}$, $M_n = f_n(x_n) \geq f_n(x_{n+1}) \geq f_{n+1}(x_{n+1}) = M_{n+1}$, as $x_n$ is the max of $f_n(x)$ and $f_n(x_{n+1}) \geq f_{n+1}(x_{n+1})$ as given. Also, for each $n\in \mathbb{}N$ and $x\in[0,1]$, $M_n = f_n(x_n) \geq f_n(x) \geq f(x)$ for the same reason. Therefore $M_n$ is a decreasing sequence which is bounded below, and so converges to some $N\geq M$. Now, since $[0,1]$ is compact, we can find a convergent subsequence of $x_n$, say $t_k$.
What I want to do is say that $M \leq N = \lim_{k\rightarrow\infty}f_k(t_k) = f(t) \leq M$ so that $M\leq f(t)\leq M$ but I believe I would need uniform convergence for that. Am I on the right track?
 A: Let $x \in [0,1]$ and $\epsilon >0$. There exists $n_0$ such that for all $n \geqslant n_0$, $f_n(x)<f(x)+\epsilon$. Since $f_{n_0}$ is continuous, there is a neighborhood $V$ of $x$ such that $f_{n_0} < f(x)+\epsilon$ on $V$. Because of the hypothesis on $\{f_n\}$, we find $f_n(y)<f(x)+\epsilon$ for all $n \geqslant n_0$ and $y \in V$.
Thus for $t=\lim t_k$ from your text, for all $\epsilon >0$ and all $k \geqslant k_0$ (for some $k_0$), we have $f_k(t_k) < f(t)+\epsilon$. So $\lim f_k(t_k) < f(t)+\epsilon$ for all $\epsilon >0$, whence $\lim f_k(t_k) \leqslant f(t)$. Here because $f_k(t_k)$ is the max of $f_k$ we have also $f(t)=\lim f_k(t) \leqslant \lim f_k(t_k)$ but actually it is not necessary for your problem.
A: Compactness gives us a sequence $(t_k)$ such that $t_k\to t\in [0,1]$ and $f(t_k)\to M$ and of course
$\tag 1f(t)\le M.$
But, $f(x)$ is the infimum of $(f_n(x))_{n\in\mathbb N}$ and each $f_n$ is continuous, so $f$ is upper semicontinuous, which means that $\underset{x\to t}\limsup f(x) \leq f(t).$ Now, $\underset {x\in U}\sup f(x)=M$ on any neighborhood $U$ of $t$ so
$\tag2 M=\underset{x\to t}\limsup f(x) \leq f(t).$
