Can someone please verify this?
Let $f_n$ be a sequence of continuous functions on a closed interval $I$ converging uniformly to $f$. Is it true that max $\{f_n(x):x\in I\}$ converges to max $\{f(x):x \in I\}$?
Yes. Let $\alpha = $ max $\{f(x):x \in I\}$
Assume, for the sake of contradiction, that max $\{f_n(x):x \in I\}$ does not converge to $\alpha$.
Then, $$\forall \epsilon > 0, \forall N \in \mathbb{N}, \exists n > N : \left|\operatorname{max} \{ f_n(x)|x \in I\} – \alpha\right| > \epsilon$$
However, this contradicts the fact that $f_n$ converges uniformly to $f$. Note that
$$\forall \epsilon > 0, \forall N \in \mathbb{N}, \exists n > N:|f_n(y) – \alpha| > \epsilon$$
Here, $y$ is the value at which $f$ attains its maximum (it must attain its maximum at some point within $I$, since $I$ is closed). Therefore, it must be the case that max $\{f_n(x):x\in I\}$ converges to max $\{f(x):x \in I\}$
Best Answer
I assume $I$ is closed and bounded, so $f_n$ is bounded.
Uniform convegence implies for every $\epsilon>0$ there exists $N \in \mathbb{N}$ such that for every $x \in I$ and $n \geq N$
$$ f(x) - \epsilon < f_n(x) < f(x) + \epsilon.$$
Hence, for every $\epsilon > 0,$
$$f_n(x) < \max(f)+ \epsilon, \\ f(x) - \epsilon < \max(f_n).$$
It follows that,
$$\max(f_n) \leq \max(f)+ \epsilon, \\ \max(f) - \epsilon \leq \max(f_n).$$
Therefore for every $n \geq N,$
$$|\max(f_n)-\max(f)| \leq \epsilon$$