Dini's Theorem states that
Given a sequence of real-valued continuous functions $(f_n)$ on a compact set $E\subseteq \mathbb{R},$ if $(f_n)$ decreases to a continuous function $f$ pointwise on $E$, then $(f_n)$ converges to $f$ uniformly on $E$.
Egoroff's Theorem states that
Given a measure space $(E,\mathcal{A},\mu)$ where $E \subseteq \mathbb{R}$ has finite measure. $\mathcal{A}$ is an $\sigma$-algebra and $\mu$ is a measure on $E.$
If a sequence of measurable functions $(f_n)$ converges pointwise to $f$ almost everywhere, then for every $\varepsilon>0,$ there exists a measurable subset $F\subseteq E$ with $\mu(E\setminus F) <\varepsilon$ such that $(f_n)$ converges uniformly to $f$ on $F.$
Question: Do there exist theorems, other than Dini and Egoroff, which give sufficient conditions for pointwise convergence to be uniform convergence?
I would like to see techniques involved in proving those theorems.
Dini and Egoroff used similar technique, which is to define a set containing elements such that $f_n$ and $f$ are 'closed to each other'. If possible, I would like to know other technique to show pointwise convergence implies uniform convergence.
Best Answer
Here's one that I rather like that uses equicontinuity and compactness as the engine for moving from pointwise to uniform convergence. For the sake of clarity, let's recall what equicontinuity means.
With this definition in hand, we can state the result.
Proof:
Let $\varepsilon >0$. By the equicontinuity property, for each $x \in X$ we can find $\delta_x >0$ such that $$ y \in B_X(x,\delta_x) \Rightarrow d_Y(f_n(x),f_n(y)) < \frac{\varepsilon}{4} \text{ for all } n \in \mathbb{N}. $$ The collection of open balls $\{B_X(x,\delta_x)\}_{x \in X}$ is an open cover of $X$. The compactness of $X$ allows us to find a finite subcover, namely $x_1,\dotsc,x_k \in X$ such that $X = \bigcup_{i=1}^k B_X(x_i,\delta_{x_i})$. Due to the pointwise convergence $f_n \to f$ (or really the pointwise Cauchy property), for each $i=1,\dotsc,k$ we can choose $N_i \in \mathbb{N}$ such that $$ m,n \ge N_i \Rightarrow d_Y(f_n(x_i),f_m(x_i)) < \frac{\varepsilon}{4}. $$
Set $N = \max \{N_1,\dotsc,N_m\} \in \mathbb{N}$. For each $x \in X$ there exists $i\in \{1,\dotsc,k\}$ such that $x \in B(x_i,\delta_{x_i})$. Then for $m \ge n \ge N$ we have the estimate $$ d_Y(f_m(x),f_n(x)) \le d_Y(f_m(x),f_m(x_i)) + d_Y(f_m(x_i),f_n(x_i)) + d_Y(f_n(x_i),f_n(x)) \\ < \frac{\varepsilon}{4} + \frac{\varepsilon}{4} + \frac{\varepsilon}{4} = \frac{3\varepsilon}{4}. $$ Hence for $n \ge N$, $$ d_Y(f(x),f_n(x)) = \lim_{m \to \infty} d_Y(f_m(x),f_n(x)) \le \frac{3\varepsilon}{4}. $$ This holds for all $x \in X$, so $$ n \ge N \Rightarrow \sup_{x \in X} d_Y(f(x),f_n(x)) \le \frac{3\varepsilon}{4} < \varepsilon, $$ and we deduce that $f_n \to f$ uniformly as $n \to \infty$.