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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.

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$f(x)=\inf_n{f_n(x)}$ 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 $M$.

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  • $\begingroup$ 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 $x_0$? $\endgroup$ – walter595738 Mar 29 at 4:01
  • $\begingroup$ $[0,1]$ 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 $\endgroup$ – Conrad Mar 29 at 11:34
  • $\begingroup$ $x^n$ 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 $\endgroup$ – Conrad Mar 29 at 11:43

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