Let {$f_n$} be a sequence of continuous functions on $[a,b]$ that converges uniformly to the function $f$ on $[a,b]$. Show that there is some $K ∈ R$ such that $|f_n(x)|≤ K$ for all $n ≥ 1$ and all $x∈[a,b]$.
Give an example of a sequence of continuous functions {$g_n$} on some interval $[a,b]$ such that {$g_n$} converges pointwise to the function $g$ on $[a,b]$ for which there is no number $K ∈R$ with $|g_n(x)|≤ K$ for all $n$ and all $x∈[a,b]$.
I'm not sure how to go about the first part, but I know that since f is continuous, by the extreme value theorem, it must obtain a max and min on its closed interval. If we let the max=K, I believe that satisfies the first part of the question. It just seems a little to simple.
As $\{f_n\}$ converges uniformly to $f$, one can find $N \in \mathbb N$ such that $\vert f_n(x)- f(x) \vert \le 1$ for all $n > N$ and all $x \in [a,b]$. Hence $\vert f_n(x) \vert \le 1 + \vert f(x) \vert$ again for all $n > N$ and all $x \in [a,b]$.
As $f$ is continuous on the compact $[a,b]$, $f$ is bounded on that interval. Let's say by $k$. We get that $\vert f_n(x) \vert \le 1 +k$ again for all $n > N$ and all $x \in [a,b]$. With a similar argument, the functions $f_1, f_2, \dots ,f_{n}$ are bounded respectively by $k_1, \dots, k_n$. Finally we get that $\vert f(x) \vert \le \sup(1+k, k_1, \dots, k_n)$ for all $x \in [a,b]$.
Now for the second part of your question. Take for $g$ the always vanishing function and for $g_n$ the piecewise continuous map defined by $g_n(0)=g_n(1)=g_n(\frac{1}{n})=0 \text{ and } g_n(\frac{1}{2n})=n$. $\{g_n\}$ converges pointwise to $g$ but cannot be uniformly bounded on $[a,b]$.