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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.

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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.

Definition: Let $X$ and $Y$ be metric spaces. A set $E$ of continuous functions from $X$ to $Y$ (i.e. $E \subseteq C(X,Y)$ ) is equicontinuous if for every $x \in X$ and every $\varepsilon>0$ there exists $\delta >0$ such that $$ y \in X \text{ and } d_X(x,y) < \delta \Rightarrow d_Y(f(x),f(y)) < \varepsilon \text{ for all } f \in E. $$

With this definition in hand, we can state the result.

Theorem: Let $X$ and $Y$ be metric spaces with $X$ compact. Suppose that for $n \in \mathbb{N}$ the function $f_n: X \to Y$ is continuous and that $\{f_n | n \in \mathbb{N}\}$ is equicontinuous. Let $f : X \to Y$ and suppose that $f_n \to f$ pointwise as $n \to \infty$. Then $f_n \to f$ uniformly as $n \to \infty$. In particular, this means that $f$ is continuous.

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 $n \to \infty$.

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  • $\begingroup$ Oops, how can i forgot compactness and equicontonuity. $\endgroup$
    – Idonknow
    Commented Nov 6, 2017 at 23:34
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    $\begingroup$ So your equicontinuity refers to pointwise equicontinuity? $\endgroup$
    – Idonknow
    Commented Nov 7, 2017 at 6:00
  • $\begingroup$ Yes, this is really pointwise equicontinuity, as opposed to uniform equicontinuity. For this result only the pointwise version is needed. $\endgroup$
    – Glitch
    Commented Nov 7, 2017 at 11:55
  • $\begingroup$ Can this be considered as a corollary to the generalized Arzela-Ascoli theorem? $\endgroup$
    – Hashimoto
    Commented Aug 2, 2018 at 20:21

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