Showing that the pointwise limit of continuous functions equals its supremum somewhere on compact domain

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.

$$f(x)=\inf_n{f_n(x)}$$ 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 $$M$$.
• I don't think I understand why we can say "if $x_0 \in [0,1]$ is a limit point of $x_n$ for which $f(x_n)$ converges to $M$, $f(x_0) \geq M$". Can you elaborate a bit? Maybe I'm tired, but how do we know there is a sequence $x_n$ converging to $x_0$ such that $f(x_n) \to M$? I understand that there is a sequence of $x$s such that $f(x_n) \to M$. But does that mean that $x_n$ has to converge to $x_0$? – walter595738 Mar 29 at 4:01
• $[0,1]$ is compact so any sequence in it like $x_n$ must have a subsequence converging to some $x_0$ in $[0,1]$ so by passing to such and renaming if you wish, we can assume $x_n$ converges; note that it is quite possible for $x_n$ to be a constant sequence as the characteristic function of a closed set like a point can be obtained as above and in that case obviously $f$ is 0 except at a point where it is 1 – Conrad Mar 29 at 11:34
• $x^n$ on $[0,1]$ satisfies the requirements and converges to the characteristic function of {$1$} and with more care you can get example where the minimum is not attained as any upper semicontinuos function is an infimum as here, $x+(1-x)^n$ for example – Conrad Mar 29 at 11:43