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A sequence of real-valued functions converges uniformly if and only if it is uniformly Cauchy

Statement

Let $X$ be a set and let $f_{k} : X \to R$ for every $k \in N$ . Then $(f_{k})$ converges uniformly on $X$ to some $f : X \to R$ if and only if $(f_{k})$ is uniformly Cauchy on $X$ (Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions).

Facts & Assumptions

Given: A set $X$ and a sequence of functions $f_{k} : X \to R$ .

[A1]

Uniform convergence to $f$ means that for every real $ε > 0$ there is $N$ such that $∣ f_{k} (x) - f (x) ∣ < ε$ for every $k \geq N$ and every $x \in X$ (Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions).

[A2]

Uniform Cauchyness means that for every real $ε > 0$ there is $N$ such that $∣ f_{m} (x) - f_{n} (x) ∣ < ε$ for every $m, n \geq N$ and every $x \in X$ (Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions).

[A3]

Pointwise convergence as defined through real sequences can equivalently be tested with every positive real error (Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions).

[L1]

For reals $u, v$ , $∣ u + v ∣ \leq ∣ u ∣ + ∣ v ∣$ (The triangle inequality).

[L2]

Every Cauchy sequence of reals converges to a real (The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges).

Proof

technique · direct

1.1

Suppose first that $f_{k} \to f$ uniformly on $X$ , and let $ε > 0$ be real. By [A1] choose $N$ with $∣ f_{j} (x) - f (x) ∣ < ε / 2$ for every $j \geq N$ and $x \in X$ . Thus, for $m, n \geq N$ and $x \in X$ , $∣ f_{m} (x) - f_{n} (x) ∣ \leq ∣ f_{m} (x) - f (x) ∣ + ∣ f (x) - f_{n} (x) ∣ < ε$ . Since $ε$ was arbitrary, $(f_{k})$ is uniformly Cauchy.

A1A2L1choosealgebra

1.2

Conversely, suppose that $(f_{k})$ is uniformly Cauchy on $X$ . For each $x \in X$ , [A2] makes $(f_{k} (x))$ a Cauchy real sequence; by [L2] it has a real limit $f (x)$ . These values define a function $f : X \to R$ .

A2L2construct

1.3

Under this converse assumption, let $ε > 0$ be real and choose $N$ such that $∣ f_{m} (x) - f_{n} (x) ∣ < ε / 2$ for every $m, n \geq N$ and every $x \in X$ .

A2choose

2.1

Fix $k \geq N$ and $x \in X$ . Pointwise convergence at $x$ gives a threshold $M$ such that $∣ f_{n} (x) - f (x) ∣ < ε / 2$ for $n \geq M$ . Choose $n \geq N, M$ . Then $∣ f_{k} (x) - f (x) ∣ \leq ∣ f_{k} (x) - f_{n} (x) ∣ + ∣ f_{n} (x) - f (x) ∣ < ε$ .

step 1.2step 1.3A3L1choose

3.1

The index $N$ in step 1.3 is independent of $k$ and $x$ , so step 2.1 proves $f_{k} \to f$ uniformly. Together with step 1.1 this proves both directions.

step 1.1step 2.1A1∎

Depends on

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Sources

Stanford Math 63CM, Additional Lecture Notes, Theorem 1.12 (standard reference, not scraped)
W. Trench, Introduction to Real Analysis (standard reference, not scraped)