Complex Analysis/Limits and continuity of complex 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'.

Complex functions

Definition 2.1:

Let $S_{1},S_{2}$ be sets and $f:S_{1}\to S_{2}$ be a function. $f$ is a complex function if and only if $S_{1},S_{2}\subseteq \mathbb {C}$.

Example 2.2:

The function

$f:\mathbb {C} \to \mathbb {C} ,f(z):=z^{2}$

is a complex function.

Limits of complex functions with respect to subsets of the preimage

We shall now define and deal with statements of the form

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

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

Definition 2.3:

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

$\forall \epsilon >0:\exists \delta >0:\left(z\in A\cap B(z_{0},\delta )\Rightarrow |f(z)-w|<\epsilon \right)$,

we define

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

Lemma 2.4:

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

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

then

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

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

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

there exists a $\delta >0$ such that

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

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

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

and therefore

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

Lemma 2.5:

Let $S\subseteq \mathbb {C}$, $f:S\to \mathbb {C}$ be a function, $O\subseteq S$ 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 $A\subseteq S$ such that $z_{0}\in A$

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

Proof:

Let $A\subseteq S$ such that $z_{0}\in A$.

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

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

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

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

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

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

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

Continuity of complex functions

Definition 2.6:

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

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

Exercises

1. Prove that if we define

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

   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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