About 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 equicontinuity is needed for proving (1). But I don't see how it is needed for proving (2).
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}
This completes the proof.
So my question is: if I remove the equicontinuity from the assumptions, part (1) would not hold anymore, but would the part (2) of the theorem still hold?
Best Answer
No, it would not hold anymore. Think of $K=[0,1]$ and
$$ f_n(x)= \begin{cases} 1-nx, & x \in [0,1/n), \\0, & \text{else.} \end{cases}$$
The way $\delta$ is chosen in the proof above makes use of equicontinuity.