For $f_n,f$ continuous on $[a,b]$, and $f_n \to f$ uniformly, define $M_n=\max_{[a,b]}f_n$ and $M=\max_{[a,b]}f$, then it was asked to show that $M_n \to M$. And it was also asked is it also true for sequence of minima. I did the first part as follows:
$f_n(x)-f(x)\leq \max_{[a,b]}\lvert f_n(x)-f(x) \rvert$,
$\Rightarrow f_n(x)\leq \max_{[a,b]}\lvert f_n(x)-f(x) \rvert + f(x)$,
Then first taking maximum on right side and then on left side, we will get,
$M_n-M\leq \max_{[a,b]}\lvert f_n(x)-f(x) \rvert$.
Similarly, we can also show that,
$M-M_n\leq \max_{[a,b]}\lvert f_n(x)-f(x) \rvert$, which gives,
$\lvert M_n-M \rvert \leq \max_{[a,b]}\lvert f_n(x)-f(x) \rvert$.
By uniform continuity, it is evident that right hand side of above equation goes to zero, which implies $M_n \to M$.
But I am having difficulty in showing the second part of question. I guess it is true and I thought proof will be similar but the same idea doesn't seem to work. Any hint. Thanks.