# Theorems similar to Dini's Theorem and Egoroff's Theorem

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.

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 $n \to \infty$.

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