TITLE: Series of minimum values of uniformly convergent series of functions
QUESTION [0 upvotes]: Let $(f_n(x))$ be a series of continuous functions in $[0,1]$, that uniformly converges to $f(x)$ in $[0,1]$. Let $m=min\{f(x) | x\in [0,1]\}, c_n=min\{f_n(x)|x\in [0,1]\}$ Prove that $c_n \rightarrow m$

I tried directly by definition:
Let $\epsilon \gt 0$, so we know that there is $N$ such that for every $n \gt N$: $|f_n(x)-f(x)|\lt \epsilon$, specifically it's correct for the minimum values, and thus $|min f_n(x)-min f(x)|=|c_n -m| \lt \epsilon$ but obviously it isn't formal...
How can I formalize it? If my solution is not correct, any hint please?

REPLY [1 votes]: Let $y \in [0,1]$ such that $f(y) \leq m + \epsilon$ then
$$m + \epsilon \geq f(y) = \limsup_{n \to \infty}f_n(y) \geq \limsup_{n \to \infty} c_n$$
the arbitarity of $\epsilon$ gives
$$m \geq \limsup_{n \to \infty} c_n$$
Now let $N$ such that $|f_n(x) - f(x)| < \frac{\epsilon}{2}$ for every $n > N$ and $x \in [0,1]$ and let $y_n \in [0,1]$ such that $f_n(y_n) \leq c_n + \frac{\epsilon}{2}$ for every $n$. (Here you need the uniform convergence)
$$-\frac{\epsilon}{2} < f_n(y_n) - f(y_n) < \frac{\epsilon}{2}$$
So $m \leq f(y_n) < f_n(y_n) + \frac{\epsilon}{2} \leq c_n + \epsilon$ for every $n$. The arbitarity of $\epsilon$ and passing to the $\liminf$ we get
$$m \leq \liminf_{n \to \infty} c_n$$
But now since
$$m \leq \liminf_{n \to \infty} c_n \leq \limsup_{n \to \infty} c_n \leq m$$
we have that
$$\exists \lim_{n \to \infty} c_n = m$$
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