Complex Analysis/Complex Functions/Continuous Functions

In this section, we

• introduce a 'broader class of limits' than known from real analysis (namely limits with respect to a subset of $\mathbb {C}$) and
• characterise continuity of functions mapping from a subset of the complex numbers to the complex numbers using this 'class of limits'.

Limits of complex functions with respect to subsets of the preimage

We shall now define and deal with statements of the form

$\lim _{z\to z_{0} \atop z\in B'}f(z)=w$

for $B\subseteq \mathbb {C} ,f:B\to \mathbb {C} ,B'\subseteq B,w\in \mathbb {C}$ , and prove two lemmas about these statements.

Definition 2.2.1:

Let $B\subseteq \mathbb {C}$ be a set, let $f:B\to \mathbb {C}$ be a function, let $B'\subseteq B$ , let $z_{0}\in B'$ and let $w\in \mathbb {C}$ . If

$\forall \varepsilon >0:\exists \delta >0:{\bigl (}z\in B'\cap B(z_{0},\delta )\Rightarrow |f(z)-w|<\varepsilon {\bigr )}$

we define:

$\lim _{z\to z_{0} \atop z\in B'}f(z):=w$

Lemma 2.2.2:

Let $B\subseteq \mathbb {C}$ be a set, let $f:B\to \mathbb {C}$ be a function, let $B''\subseteq B'\subseteq B$ , let $z_{0}\in B''$ and $w\in \mathbb {C}$ . If

$\lim _{z\to z_{0} \atop z\in B'}f(z)=w$

then

$\lim _{z\to z_{0} \atop z\in B''}f(z)=w$

Proof: Let $\varepsilon >0$ be arbitrary. Since

$\lim _{z\to z_{0} \atop z\in B'}f(z)=w$

there exists a $\delta >0$ such that

$z\in B'\cap B(z_{0},\delta )\Rightarrow |f(z)-w|<\varepsilon$

But since $B''\subseteq B'$ , we also have $B''\cap B(z_{0},\delta )\subseteq B'\cap B(z_{0},\delta )$ , and thus

$z\in B''\cap B(z_{0},\delta )\Rightarrow z\in B'\cap B(z_{0},\delta )\Rightarrow |f(z)-w|<\varepsilon$

and therefore

$\lim _{z\to z_{0} \atop z\in B''}f(z)=w$$\Box$

Lemma 2.2.3:

Let $B\subseteq \mathbb {C}$ , $f:B\to \mathbb {C}$ be a function, $O\subseteq B$ be open, $z_{0}\in O$ and $w\in \mathbb {C}$ . If

$\lim _{z\to z_{0} \atop z\in O}f(z)=w$

then for all $B'\subseteq B$ such that $z_{0}\in B'$ :

$\lim _{z\to z_{0} \atop z\in B'}f(z)=w$

Proof

Let $B'\subseteq B$ such that $z_{0}\in B'$ .

First, since $O$ is open, we may choose $\delta _{1}>0$ such that $B(z_{0},\delta _{1})\subseteq O$ .

Let now $\varepsilon >0$ be arbitrary. As

$\lim _{z\to z_{0} \atop z\in O}f(z)=w$

there exists a $\delta _{2}>0$ such that:

$z\in B(z_{0},\delta _{2})\cap U\Rightarrow |f(z)-f(z_{0})|<\varepsilon$

We define $\delta :=\min\{\delta _{1},\delta _{2}\}$ and obtain:

$z\in B(z_{0},\delta )\cap B'\Rightarrow z\in B(z_{0},\delta )\Rightarrow z\in B(z_{0},\delta _{2})\cap U\Rightarrow |f(z)-f(z_{0})|<\varepsilon$$\Box$

Continuity of complex functions

We recall that a function

$f:M\to M'$

where $M,M'$ are metric spaces, is continuous if and only if

$x_{l}\to x,l\to \infty \Rightarrow f(x_{l})\to f(x)$

for all convergent sequences $(x_{l})_{l\in \mathbb {N} }$ in $M$ .

Theorem 2.2.4:

Let $B\subseteq \mathbb {C}$ and $f:B\to \mathbb {C}$ be a function. Then $f$ is continuous if and only if

$\forall z_{0}\in B:\lim _{z\to z_{0} \atop z\in B}f(z)=f(z_{0})$

Proof

Exercises

1. Prove that if we define

   $f:\mathbb {C} \to \mathbb {C} ,f(z)={\begin{cases}{\dfrac {z^{2}}{|z|^{2}}}&:z\neq 0\\1&:z=0\end{cases}}$

   then $f$ is not continuous at $0$ . Hint: Consider the limit with respect to different lines through $0$ and use theorem 2.2.4.

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