# 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?

• I don't think we have that the limit function is necessarily continuous. – Aphyd Sep 4 '20 at 0:33
• oops my bad, you're right ! – user598294 Sep 4 '20 at 0:38

Let $$x \in [0,1]$$ and $$\epsilon >0$$. There exists $$n_0$$ such that for all $$n \geqslant n_0$$, $$f_n(x). Since $$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) 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.
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
$$\tag2 M=\underset{x\to t}\limsup f(x) \leq f(t).$$