Pointwise Cauchy Sequences of Functions

# Pointwise Cauchy Sequences of Functions

 Definition: A sequence of functions $(f_n(x))_{n=1}^{\infty}$ with common domain $X$ is said to be Pointwise Cauchy on $X$ 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 $\mid f_m(x) - f_n(x) \mid < \epsilon$.

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

For example, consider the following sequence of functions with common domain $X = [0, 1]$:

(1)
\begin{align} \quad (f_n(x))_{n=1}^{\infty} = \left ( \frac{1}{n} x^2 \right )_{n=1}^{\infty} \end{align} We claim that this sequence of functions is pointwise Cauchy. To prove this, let $\epsilon > 0$ and let $x_0 \in X$. If $x_0 = 0$ then $(f_n(x_0))_{n=1}^{\infty} = (0, 0, ... )$ which is clearly Cauchy, so assume that $x_0 \neq 0$ and consider:

(2)
\begin{align} \quad \mid f_m(x) - f_n(x) \mid = \biggr \lvert \frac{1}{m} x^2 - \frac{1}{n}x^2 \biggr \rvert \leq \biggr \lvert \frac{1}{m} \biggr \rvert x_0^2 + \biggr \lvert \frac{1}{n} \biggr \rvert x_0^2 = \frac{x_0^2}{m} + \frac{x_0^2}{n} \end{align}

Let $N \in \mathbb{N}$ be such that $N > \frac{2x_0^2}{\epsilon}$. Then if $m, n \geq N$ we have that:

(3)

So then:

(4)
\begin{align} \quad \mid f_m(x_0) - f_n(x_0) \mid < \frac{\epsilon}{2x_0^2} \cdot x_0^2 + \frac{\epsilon}{2x_0^2} \cdot x_0 = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}

So 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 $\mid f_m(x) - f_n(x) \mid < \epsilon$, so the sequence of functions $(f_n(x))_{n=1}^{\infty}$ is pointwise Cauchy.

We now look at a theorem which says that a sequence of real-valued functions is pointwise Cauchy if and only if it is pointwise convergent.

 Theorem 1: Every sequence of real-valued functions $(f_n(x))_{n=1}^{\infty}$, $f_n : X \to \mathbb{R}$ for all $n \in \mathbb{N}$, is pointwise Cauchy on $X$ if and only if it is pointwise convergent on $X$.

Note that Theorem 1 relies on the fact that $\mathbb{R}$ is a complete metric space.

• Proof: $\Rightarrow$ Suppose that the sequence of functions $(f_n(x))_{n=1}^{\infty}$ is pointwise Cauchy. Then each of the numerical sequences $(f_n(x_0))_{n=1}^{\infty}$ for $x_0 \in X$ is a Cauchy sequence. Since this sequence is a Cauchy sequence in the complete metric space $\mathbb{R}$ we have that every Cauchy sequence converges in $\mathbb{R}$, so each numerical sequence $(f_n(x_0))_{n=1}^{\infty}$ converges. Therefore $(f_n(x))_{n=1}^{\infty}$ is pointwise convergent on $X$.
• $\Leftarrow$ Suppose that the sequence of functions $(f_n(x))_{n=1}^{\infty}$ is pointwise convergent. Then each of the numerical sequnces $(f_n(x_0))_{n=1}^{\infty}$ for $x_0 \in X$ is a convergent sequence. But we know that every numerical convergent sequence is Cauchy, so each of the numerical sequences $(f_n(x_0))_{n=1}^{\infty}$ is a Cauchy sequence. Therefore, $(f_n(x))_{n=1}^{\infty}$ is pointwise Cauchy on $X$. $\blacksquare$