Cauchy's Uniform Convergence Criterion for Series of Functions

Recall from the Pointwise Convergent and Uniformly Convergent Series of Functions page that if we have a sequence of functions $(f_n(x))_{n=1}^{\infty}$ with common domain $X$ then the corresponding series of functions $\displaystyle{\sum_{n=1}^{\infty} f_n(x)}$ is said to be uniformly convergent if the corresponding sequence of partial sums $(s_n(x))_{n=1}^{\infty}$ is a uniformly convergent sequence of functions.

We will now look at a nice theorem known as Cauchy's uniform convergence criterion for series of functions which gives us a nice criterion for when a series of functi

Theorem 1: Let $(f_n(x))_{n=1}^{\infty}$ be a sequence of real-valued functions with common domain $X$. Then $\displaystyle{\sum_{n=1}^{\infty} f_n(x)}$ is uniformly convergent on $X$ if and only if for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ and for all $x \in X$ we have that $\displaystyle{\biggr \lvert \sum_{k=n+1}^{n+p} f_k(x) \biggr \rvert < \epsilon}$ for all $p \in \mathbb{N}$.

• Proof: $\Rightarrow$ Suppose that $\displaystyle{\sum_{n=1}^{\infty} f_n(x)}$ is uniformly convergent to some limit function $f(x)$ on $X$. Let $(s_n(x))_{n=1}^{\infty}$ denote the sequence of partial sums for this series. Then we must have that $\displaystyle{\lim_{n \to \infty} s_n(x) = f(x)}$ uniformly on $X$. So, for $\epsilon_1 = \frac{\epsilon}{2}$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ and for all $x \in X$ we have that:

(1)

\begin{align} \quad \mid s_n(x) - f(x) \mid < \epsilon \end{align}

• For any $p \in \mathbb{N}$ let $m = n + p$. Then $m \geq N$ and so:

(2)

\begin{align} \quad \quad \biggr \lvert \sum_{k=1}^{m} f_k(x) - \sum_{k=1}^{n} f_k(x) \biggr \rvert = \biggr \lvert \sum_{k=1}^{n+p} f_k(x) - \sum_{k=1}^{n} f_k(x) \biggr \rvert = \biggr \lvert \sum_{k=n+1}^{n+p} f_k(x) \biggr \rvert = \mid s_m(x) - s_n(x) \mid \leq \mid s_m(x) - f(x) \mid + \mid f(x) - s_n(x) \mid < \epsilon_1 + \epsilon_1 = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}

• $\Leftarrow$ Suppose that for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ and for all $x \in X$ we have that:

(3)

\begin{align} \quad \biggr \lvert \sum_{k=n+1}^{n+p} f_k(x) \biggr \rvert < \epsilon \end{align}

• Let $m, n \geq N$. Assume without loss of generality that $m > n$ and that $m = n + p$ for some $p \in \mathbb{N}$. Then from above we see that for all $x \in X$:

(4)

\begin{align} \quad \mid s_m(x) - s_n(x) \mid = \biggr \lvert \sum_{k=1}^{n+p} f_k(x) - \sum_{k=1}^{n} f_k(x) \biggr \rvert = \biggr \lvert \sum_{k=n+1}^{n+p} f_k(x) \biggr \rvert < \epsilon \end{align}

• So $(s_n(x))_{n=1}^{\infty}$ converges uniformly by the Cauchy uniform convergence criterion for sequences of functions. So $\displaystyle{\sum_{n=1}^{\infty} f_n(x)}$ converges uniformly on $X$. $\blacksquare$