Sequences and Series of Complex Functions

Often times one first looks at sequences and series of real numbers and then look at sequences and series of real functions. We have already glanced over Sequences and Series of Complex Numbers so it is only natural to begin to look over sequences and series of complex functions. The notions are much the same with real functions so once again, the material is only briefly mentioned and more details can be found on the Real Analysis section for reference.

Definition: A Sequence of Complex Functions defined on $A \subseteq \mathbb{C}$ is an infinite ordered list of complex functions, $(f_n(z))_{n=1}^{\infty} = (f_1(z), f_2(z), ..., f_n(z), ...)$ where $f_n$ is defined on $A$ for all $n \in \mathbb{N}$.

For example, the sequence of complex functions $\left ( \frac{z^n}{n!} \right )_{n=1}^{\infty}$ when expanded is:

(1)

\begin{align} \quad \left ( \frac{z^n}{n!} \right )_{n=1}^{\infty} = \left ( \frac{z}{1!}, \frac{z^2}{2!}, ..., \frac{z^n}{n!}, ... \right ) \end{align}

Definition: A sequence of complex functions $(f_n(z))_{n=1}^{\infty}$ on $A \subseteq \mathbb{C}$ is said to be Pointwise Convergent on $A$ if for each $z_0 \in A$, $(f_n(z_0))_{n=1}^{\infty}$ converges as a sequence of complex numbers. The sequence of complex functions $(f_n(z))_{n=1}^{\infty}$ is said to be Uniformly Convergent on $A$ if for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $\mid f_n(z) - f(z) \mid < \epsilon$ for all $z \in A$.

Definition: If $(f_n)_{n=1}^{\infty}$ is a sequence of complex functions on $A$ then the corresponding Series of Complex Functions is the infinite formal sum $\displaystyle{\sum_{n=1}^{\infty} f_n(z)}$. The corresponding Sequence of Partial Sums is the sequence of complex functions $(s_n(z))_{n=1}^{\infty}$ where $\displaystyle{s_n(z) = \sum_{k=1}^{n} f_n(z)}$ for each $n \in \{ 1, 2, ... \}$. The complex series $\displaystyle{\sum_{n=1}^{\infty} f_n(z)}$ is said to Converge Pointwise on $A$ if for each $z_0 \in A$, $\displaystyle{\sum_{n=1}^{\infty} f_n(z_0)}$ converges as a series of complex numbers. The series of complex functions $\displaystyle{\sum_{n=1}^{\infty} f_n(z)}$ is said to Converge Uniformly on $A$ if the sequence of partial sums, $(s_n(z))_{n=1}^{\infty}$ converges uniformly on $A$.

We will now state some extremely important results regarding uniform convergent sequences and series of complex functions. The first result below tells us that the notion of uniform convergence is stronger than pointwise convergence.

Theorem 1: a) If $(f_n(z))_{n=1}^{\infty}$ is a uniformly convergent sequence of complex functions then $(f_n(z))_{n=1}^{\infty}$ is also a pointwise convergent sequence of complex functions. b) If $\displaystyle{\sum_{n=1}^{\infty} f_n(z)}$ is a uniformly convergent series of complex functions then $\displaystyle{\sum_{n=1}^{\infty} f_n(z)}$ is a pointwise convergent series of complex functions.

The next result gives us a nice criterion for determining when a sequence of series of complex functions uniformly converges on a set $A$.

Theorem 2 (Cauchy's Uniform Convergence Criterion): a) A sequence of complex functions $(f_n(z))_{n=1}^{\infty}$ converges uniformly on $A$ if and only if for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $\mid f_n(z) - f_{n+p}(z) \mid < \epsilon$ for all $z \in A$ and for all $p \in \{ 1, 2, ... \}$. b) A series of complex functions $\displaystyle{\sum_{n=1}^{\infty} f_n(z)}$ converges uniformly on $A$ if and only if for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $\displaystyle{\biggr \lvert \sum_{k=n+1}^{n+p} f_k(z)} \biggr \rvert < \epsilon$ for all $z \in A$ and for all $p \in \{ 1, 2, ... \}$.

The final result - perhaps the most important result on this page, gives us a relatively simple test to check whether a sequence of complex functions converges uniformly on a set $A$ by looking at relative series of positive real numbers. This is useful since there are many easy to apply convergence tests for series of positive real numbers.

Theorem 3 (The Weierstrass M-Test): If $(f_n(z))_{n=1}^{\infty}$ is a sequence of complex functions on $A$ and $(M_n)_{n=1}^{\infty}$ is a sequence of nonnegative real-numbers such that $\mid f_n(z) \mid \leq M_n$ for all $z \in A$ and for all $n \in \mathbb{N}$ and if $\displaystyle{\sum_{n=1}^{\infty} M_n}$ converges then $\displaystyle{\sum_{n=1}^{\infty} f_n(z)}$ converges absolutely and uniformly on $A$.

• Proof: Let $(f_n(z))_{n=1}^{\infty}$ be a sequence of complex functions on $A$ and $(M_n)_{n=1}^{\infty}$ be a sequence of nonnegative real numbers such that $\mid f_n(z) \mid \leq M_n$ for all $z \in A$ and for all $n \in \mathbb{N}$ with $\displaystyle{\sum_{n=1}^{\infty} M_n}$ being convergent. Since $\displaystyle{\sum_{n=1}^{\infty} M_n}$ converges, for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then for all $p \in \{ 1, 2, ... \}$:

(2)

\begin{align} \quad \sum_{k = n+1}^{n+p} M_k < \epsilon \end{align}

• So:

(3)

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

• By Theorem 2 this means that $\displaystyle{\sum_{n=1}^{\infty} f_n(z)}$ converges uniformly on $A$. The inequality above also shows that $\displaystyle{\sum_{n=1}^{\infty} \mid f_n(z) \mid}$ converges, so $\displaystyle{\sum_{n=1}^{\infty} f_n(z)}$ converges absolutely as well. $\blacksquare$