Rudin's PMA Theorem 7.25 Rudin's theorem 7.25:

7.25     Theorem      If $K$ is compact, if $f_n \in \mathscr{C}(K)$ for $n=1,2,3,\dots,$ and if $\{f_n\}$ is pointwise bounded and equicontinuous on $K$, then

*

*$\{f_n\}$ is uniformly bounded on $K$,

*$\{f_n\}$ contains a uniformly convergent subsequence.


The proof is the following:

Proof
(1) Let $\varepsilon > 0$ be given and choose $\delta > 0$, in accordance with Definition 7.22, so that
\begin{equation}
|f_n(x) - f_n(y)| < \varepsilon \tag{44}
\end{equation}
for all $n$, provided that $d(x,y) < \delta$.
Since $K$ is compact, there are finitely many points $p_1, \dots, p_r$ in $K$ such that to every $x \in K$ corresponds at least one $p_i$ with $d(x, p_i) < \delta$. Since $\{f_n\}$ is pointwise bounded, there exist $M_i < \infty$ such that $|f_n(p_i)| < M_i$ for all $n$. If $M = \max(M_1, \dots, M_r)$, then $|f_n(x)| < M + \varepsilon$ for every $x \in K$. This proves (1).
(2) Let $E$ be a countable dense subset of $K$. (For the existence of such a set $E$, see Exercise 25, Chap. 2.) Theorem 7.23 shows that $\{f_n\}$ has a subsequence $\{f_{n_i}\}$ such that $\{f_{n_i}(x)\}$ converges for every $x \in E$.
Put $f_{n_i} = g_i$, to simplify the notation. We shall prove that $\{g_i\}$ converrges uniformly on $K$.
Let $\varepsilon > 0$, and pick $\delta > 0$ as in the beginning of this proof. Let $V(x,\delta)$ be the set of all $y \in K$ with $d(x,y) < \delta$. Since $E$ is dense in $K$, and $K$ is compact, there are finitely many points $x_1, \dots, x_m$ in $E$ such that
$$
K \subset V(x_1, \delta) \cup \dots \cup V(x_m, \delta). \tag{45}
\label{45}
$$
Since $\{g_i(x)\}$ converges for every $x \in E$, there is an integer $N$ such that
$$
|g_i(x_s) - g_j(x_s)| < \varepsilon \tag{46}
\label{46}
$$
whenever $i \geq N$, $j \geq N$, $1 \leq s \leq m$.
If $x \in K$, $(\ref{45})$ shows that $x \in V(x_s, \delta)$ for some $s$, so that
$$
|g_i(x) - g_i(x_s) | < \varepsilon
$$
for every $i$. If $i \geq N$ and $j \geq N$, it follows from $(\ref{46})$ that
\begin{align}
|g_i(x) - g_j(x)| &\leq |g_i(x) - g_i(x_s)| + |g_i(x_s) - g_j(x_s)| + |g_j(x_s) - g_j(x)| \\
&< 3\varepsilon.
\end{align}

My question about the proof is that why is equicontinuity needed in part (1)?
 A: Hint:
Take a sequence of piecewise linear continuous functions $f_n$ that are $0$ on most of the domain ,$[0,1]$, but peak sharply at $1/n$. These functions are continuous and pointwise bounded because the all functions other than $f_n$ are zero at $1/n$. This sequence isn't uniformly bounded.
Drawing a picture would help. The peaks have to lie in the $(\frac{1}{n+1},\frac{1}{n-1})$ range to avoid interference.
Answer:$\\$
Define $f_n(x) =
\left\{
 \begin{array}{ll}
  0  & \mbox{if } x \in (\frac{1}{n+1},\frac{1}{n-1})^C\\
  n^2[(n+1)x-1] & \mbox{if } x \in (\frac{1}{n+1},\frac{1}{n}]\\
       -n^2[(n-1)x-1] & \mbox{if } x\in (\frac{1}{n},\frac{1}{n-1})
 \end{array}
\right.$
This function looks like a single triangular peak of height $n$ at the point $1/n$. By how it is defined, the function is zero outside of the interval  $(\frac{1}{n+1},\frac{1}{n-1})$. This function is continuous.
Let us see why this function is pointwise bounded. Pick a point $x$ in the interval $[\frac{1}{k},\frac{1}{k+1})$. Then, we can say that $f_n(x)=0$ for all $n\geq k+2$. Therefore, $f_n(x) < Max(f_1(x),\cdots,f_{k+1}(x))$ for each $x\in  [\frac{1}{k},\frac{1}{k+1})$  and all $n$. This is the pointwise bound.
However, the $f_n$ are not uniformly bounded because $f_n(\frac{1}{n})=n$.
A: We get (44) from the definition of equicontinuity. So, we get $\delta$ from (44). We used $\delta$ and the compactness of $K$ to get the points $p_1,\ldots,p_r$. In the last, we used $p_1,\dots,p_r$ to get $M$.
We have $|f_n(x)-f_n(y)|<\epsilon$ for all n, provided $d(x,y)<\delta$.
We know that there exists $p_i$ for each $x$ such that $d(x,p_i)<\delta$.
We get $|f_n(x)-f_n(p_i)|<\epsilon$ for all $n$.
This implies $|f_n(x)|<|f_n(p_i)|+\epsilon$ for all $n$.
For each x, there exists $M_i$ such that $|f_n(x)|<M_i+\epsilon$ for all $n$.
This implies $|f_n(x)|<M+\epsilon$ for all $n$ and $x\in K$.
Therefore, $M+\epsilon$ is a uniform bound for $\{f_n\}$ and we got $\epsilon$ from (44), $M$ from $\delta$ and $\delta$ from (44).
A: The following is a detail proof of part (a) from which you can see what you wanted to know.
(a) Let $\varepsilon >0$ be given and choose $\delta>0$, in accordance with Definition $7.22$ (of equicontinuous), so that
$$\tag{44} \left|f_n(x)-f_n(y)\right|<\varepsilon$$
for all $n$, provided that $d(x,y)<δ$.
Since $K$ is compact, there are finitely many points $p_1, \dots , p_r$ in $K$ such that
$$K\subset V(p_1,\delta)\cup\ V(p_2,\delta),\ \cup \ \dots \ \cup \ V(p_r,δ).$$
That is, to every $x∈ K$  corresponds at least one $p_i$ with $d(x,p_i)<\delta$.
Since $\{f_n\}$ is pointwise bounded, there exist $M_i <\infty$  such that $\left|f_n(p_i)\right|<M_i$ for all $n$.
If $M=\max (M_1,\ …\ , M_r)$, then (44) implies $\left|f_n(x)-f_n(p_i)\right|<\varepsilon$.
This implies that
$$\left|f_n(x)\right|\le \left|f_n(p_i)\right|+ \left|f_n(x)-f_n(p_i)\right|\le \left|f_n(p_i)\right|+ \varepsilon <M_i+\varepsilon\le M+\varepsilon. $$
That is,
$\left|f_n(x)\right|<M+\varepsilon $ for every $x∈ K$. This proves (a).
