Uniformly Cauchy Sequences of Functions

# Uniformly Cauchy Sequences of Functions

Recall from the Pointwise Cauchy Sequences of Functions page that a sequence of functions $(f_n(x))_{n=1}^{\infty}$ with common domain $X$ is said to be pointwise Cauchy if for all $\epsilon > 0$ and for all $x \in X$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then:

(1)
\begin{align} \quad \mid f_m(x) - f_n(x) \mid < \epsilon \end{align}

In other words, the sequence of functions $(f_n(x))_{n=1}^{\infty}$ is pointwise Cauchy if for each $x_0 \in X$ the numerical sequence $(f_n(x_0))_{n=1}^{\infty}$ is Cauchy.

We can also similarly define a sequence of functions that is uniformly Cauchy.

 Definition: A sequence of real-valued functions $(f_n(x))_{n=1}^{\infty}$ with common domain $X$ is said to be uniformly Cauchy if for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then for all $x \in X$ we have that $\mid f_m(x) - f_n(x) \mid < \epsilon$.

We now look at a very important theorem which states that a sequence of real-valued functions is uniformly Cauchy if and only if it is uniformly convergent.

 Theorem 1: A sequence of real-valued functions $(f_n(x))_{n=1}^{\infty}$ with common domain $X$ is uniformly Cauchy on $X$ if and only if it is uniformly convergent on $X$.
• Proof: Let $\epsilon > 0$ be given.
• $\Rightarrow$ Suppose that $(f_n(x))_{n=1}^{\infty}$ is uniformly Cauchy. Then for all $\epsilon_1 = \frac{\epsilon}{2} > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ and for all $x \in X$ we have that:
(2)
\begin{align} \quad \mid f_m(x) - f_n(x) \mid < \epsilon_1 = \frac{\epsilon}{2} \quad (*) \end{align}
• Let $f(x) = \lim_{n \to \infty} f_n(x)$. We will show that $(f_n(x))_{n=1}^{\infty}$ converges uniformly to $f$. For $m \geq N$ and for $n = m + k \geq N$ we have that $(*)$ holds for all $x \in X$ and so:
(3)
\begin{align} \quad \mid f_m(x) - f(x) \mid = \lim_{k \to \infty} \mid f_m(x) - f_{m+k}(x) \mid \leq \frac{\epsilon}{2} < \epsilon \end{align}
• So for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m \geq N$ and for all $x \in X$ we have that $\mid f_m(x) - f(x) \mid < \epsilon$, so $(f_n(x))_{n=1}^{\infty}$ is uniformly convergent.
• $\Leftarrow$ Suppose that $(f_n(x))_{n=1}^{\infty}$ is uniformly convergent to the limit function $f(x)$. Then for $\epsilon_1 = \frac{\epsilon}{2} > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ we have that for all $x \in X$ that:
(4)
\begin{align} \quad \mid f_n(x) - f(x) \mid < \epsilon_1 = \frac{\epsilon}{2} \quad (**) \end{align}
• Take this $N$ and let $m, n \geq N$. Then by the triangle inequality we have that for all $x \in X$:
(5)
\begin{align} \quad \quad \mid f_m(x) - f_n(x) \mid = \mid f_m(x) - f(x) + f(x) - f_n(x) \mid \leq \mid f_m(x) - f(x) \mid + \mid f(x) - f_n(x) \mid < \epsilon_1 + \epsilon_1 = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}
• So for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ and for all $x \in X$ we have that $\mid f_m(x) - f_n(x) \mid < \epsilon$, so $(f_n(x))_{n=1}^{\infty}$ is uniformly Cauchy. $\blacksquare$