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

* 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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