Dini’s Theorem states that

Given a sequence of real-valued continuous functions (fn) on a compact set E⊆R, if (fn) decreases to a continuous function f pointwise on E, then (fn) converges to f uniformly on E.

Egoroff’s Theorem states that

Given a measure space (E,A,μ) where E⊆R has finite measure. A is an σ-algebra and μ is a measure on E.

If a sequence of measurable functions (fn) converges pointwise to f almost everywhere, then for every ε>0, there exists a measurable subset F⊆E with μ(E∖F)<ε such that (fn) 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 fn and f are 'closed to each other'. If possible, I would like to know other technique to show pointwise convergence implies uniform convergence.

**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.

Definition:Let X and Y be metric spaces. A set E of continuous functions from X to Y (i.e. E⊆C(X,Y) ) is equicontinuous if for every x∈X and every ε>0 there exists δ>0 such that

y∈X and dX(x,y)<δ⇒dY(f(x),f(y))<ε for all f∈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∈N the function fn:X→Y is continuous and that {fn|n∈N} is equicontinuous. Let f:X→Y and suppose that fn→f pointwise as n→∞. Then fn→f uniformly as n→∞. In particular, this means that f is continuous.

**Proof:**

Let ε>0. By the equicontinuity property, for each x∈X we can find δx>0 such that

y∈BX(x,δx)⇒dY(fn(x),fn(y))<ε4 for all n∈N.

The collection of open balls {BX(x,δx)}x∈X is an open cover of X. The compactness of X allows us to find a finite subcover, namely x1,…,xk∈X such that X=⋃ki=1BX(xi,δxi). Due to the pointwise convergence fn→f (or really the pointwise Cauchy property), for each i=1,…,k we can choose Ni∈N such that

m,n≥Ni⇒dY(fn(xi),fm(xi))<ε4.

Set N=max. 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.

**Attribution***Source : Link , Question Author : Idonknow , Answer Author : Glitch*