Lecture 4: Continuity of Functions

Limits of functions

We now introduce some fundamental concepts in differential calculus for complex functions. We start with the concept of limit and continuity, two of the most fundamental concepts in analysis. The concept of limit can be introduced the same way as how it was introduced in calculus.

Definition 1 (Limit). For a given function $f:\mathbb C\to \mathbb C$ , we say the function $f$ has the limit $f_z$ at $z\in \mathbb C$ if: $\forall \varepsilon>0$ , $\exists \delta(\varepsilon, z)>0$ such that $|\widetilde z-z|< \delta \implies |f(\widetilde z)-f_z|<\varepsilon,\ \ \forall\ Dom(f)\ni \widetilde z\neq z\,.$

Notation 1. We write $\lim_{\widetilde z \to z} f(\widetilde z) = f_z,$ when $f(z)$ has the limit $f_z$ at $z$ .

Example 2. Let $f: \mathbb C\to \mathbb C$ be $f(z)=2z$ . Then $\lim_{z\to i} f(z) = 2i.$ To see this, we compute $|f(z)-2i|=2|z-i|$ . For any given $\varepsilon>0$ , our objective is to find a $\delta(\varepsilon, z_0=i)$ such that $|z-i|<\delta \implies |f(z)-2i|<\varepsilon$ . It is clear this can be done by selecting $\delta =\varepsilon/2$ .

Example 3. Let $f: \mathbb C\to \mathbb C$ be $f(z)=z^2$ . Then $\lim_{z\to -i} z^2= -1.$

Proof. We compute $|f(z)-(-1)|=|z^2+1|=|(z+(-i))(z-(-i))|=|(z-(-i))(z-(-i)-2i)|=|z-(-i)||z-(-i)-2i| \le |z-(-i)|(|z-(-i)|+2|-i|)=|z-(-i)|(|z-(-i)|+2)$ . For any given $\varepsilon>0$ , we can take $\delta$ such that $\delta(\delta+2)<\varepsilon$ (that is, $\delta\le -1+\sqrt{\varepsilon+1^2}$ ). This way, we have that $|z-(-i)|<\delta \implies |f(z)-(-1)|<\varepsilon$ . ◻

From the definition, it is easy to verify the following properties of limits of real and complex functions.

Theorem 4 (Properties of Limits). Let $f(z)$ and $g(z)$ be functions defined in a region such that $\displaystyle\lim_{z\to z_0} f(z) = w_0$ and $\displaystyle\lim_{z\to z_0} g(z) = u_0$ . Then we have $\lim_{z\to z_0}(f+g)(z) = w_0+u_0$ $\lim_{z\to z_0}(fg)(z) = w_0u_0$ and $\lim_{z\to z_0}(f/g)(z) = w_0/u_0,\ \ \mbox{provided that} \ u_0\neq 0.$

Proof. We prove the second property. (i) $\displaystyle\lim_{z\to z_0} f(z) = w_0$ means $\forall \varepsilon>0$ , $\exists \delta_1$ such that $|z-z_0|<\delta_1 \implies |f(z)-w_0|<\varepsilon$ ; (ii) $\displaystyle\lim_{z\to z_0} f(z) = w_0$ means $\exists \delta_2$ such that $|z-z_0|<\delta_2 \implies |f(z)-w_0|<1 \implies |f(z)|<|w_0|+1$ ; (iii) $\displaystyle\lim_{z\to z_0} g(z) = u_0$ means $\forall \varepsilon>0$ , $\exists \delta_3$ such that $|z-z_0|<\delta_3 \implies |g(z)-u_0|<\varepsilon$ ; (iii) $|f(z)g(z)-w_0u_0|=|f(z)g(z)-f(z)u_0+f(z)u_0-w_0u_0|=|f(z)(g(z)-u_0)+(f(z)-w_0)u_0|\le |f(z)||g(z)-u_0|+|f(z)-w_0||u_0|$ . Therefore, $\forall \varepsilon$ , take $\delta=\min(\delta_1, \delta_2, \delta_3)$ , then $|z-z_0|<\delta \implies |f(z)g(z)-w_0u_0|<(|w_0|+1)\varepsilon+|u_0|\varepsilon=(|w_0|+|u_0|+1)\varepsilon$ . ◻

Exercise 5. Prove the first and the third properties of limit in Theorem Theorem 4.

Continuity of functions

Intuitively, continuous functions defined on a set take nearby elements to nearby elements.

Definition 6 (Continuity). Let $f:\mathbb C\to \mathbb C$ be a given function with $Dom(f)\subseteq \mathbb C$ . We say that $f$ is continuous at $z\in \mathbb C$ if $\forall\varepsilon>0$ , $\exists \delta(\varepsilon, z)>0$ such that $|\widetilde z-z|\le \delta \implies |f(\widetilde z)-f(z)|\le \varepsilon,\ \ \forall \widetilde z.$ If $f$ is continuous at any point $z\in Dom(f)$ , we say $f$ is continuous.

Example 7. The function $f:\mathbb C\to \mathbb C$ and $f(z)=5 z$ is continuous on $Dom(f)=\mathbb C$ .

Example 8. The function $f:\mathbb C\to \mathbb C$ and $f(z)=\overline z$ is continuous on $Dom(f)=\mathbb C$ .

Example 9. The function $f(z)=z^2$ is continuous on $Dom(f)=\mathbb C$ .

Proof. Let $z_0$ be arbitrary. $\forall \varepsilon>0$ , we need to find $\delta(\varepsilon, z_0)$ such that $|z-z_0|\le \delta(\varepsilon, z_0)$ implies $|f(z)-f(z_0)|\le\varepsilon$ . We observe that $|f(z)-f(z_0)|=|z^2-z_0^2|=|z+z_0||z-z_0|=|z-z_0+z_0||z-z_0|\le |z-z_0|^2+|z_0||z-z_0|$ . To get $|f(x)-f(z_0)|\le\varepsilon$ , we can take $\delta(\varepsilon,z_0)$ such that $\delta^2+|z_0|\delta\le \varepsilon$ , that is $\Big(\delta\le \sqrt{|z_0|^2+4\varepsilon}-|z_0|\Big)/2$ . ◻

Remark 10. It is clear from the definition of continuity and the definition of limit that $f(x)$ is continuous at $z_0$ if $\lim_{z\to z_0}f(z)=f(z_0)$ .

Remark 11. Let $f: \mathbb C\to \mathbb C$ be a given function. For $f$ to have a limit at a point $z_0\in \mathbb C$ , the function needs not be defined at $z_0$ . However, for a function to be continuous at $z_0$ , the function has to be defined at $z_0$ (that is, $z_0$ has to be in the domain of the function). For instance, if we define a function $f(z)=e^z$ on $\Omega=\{z\ |\ |z|>0\}$ . Then $\displaystyle\lim_{z\to 0} f(z)=1$ . However, $f(z)$ is not defined at $z=0$ , so we could not talk about continuity of $f(z)$ at $z=0$ in this case.

The following properties of continuity are straightforward to verify.

Theorem 12 (Properties of Continuity). Let $f: \mathbb C\to \mathbb C$ and $g: \mathbb C\to \mathbb C$ be continuous functions on $Dom(f)$ and $Dom(g)$ respectively. Then
(a) $kf$ is continuous on $Dom(f)$ , $\forall k\in\mathbb R$ .
(b) $f+g$ is continuous on $Dom(f)\cap Dom(g)$ .
(c) $fg$ is continuous on $Dom(f)\cap Dom(g)$ .
(d) $f/g$ is continuous on $\{z\in Dom(f)\cap Dom(g)|g(z)\neq 0\}$ .
(e) $f\circ g$ is continuous on $\{z\in Dom(g)| g(z)\in Dom(f)$ .

Proof. The fact that $f$ is continuous means that at any $z\in Dom(f)$ , for any $\varepsilon$ , we have $\delta_f(z, \varepsilon)$ such that $|\widetilde z-z|\le \delta_f(z, \varepsilon)$ implies $|f(\widetilde z)-f(z)|\le \varepsilon$ . In the same manner, at any $z\in Dom(g)$ , for any $\varepsilon$ , we have $\delta_g(z, \varepsilon)$ such that $|\widetilde z-z|\le \delta_g(z, \varepsilon)$ implies $|g(\widetilde z)-g(z)|\le \varepsilon$ .
(a) At any $z\in Dom(f)\cap Dom(g)$ , for any $\varepsilon$ , we take $\delta_f(z, \varepsilon)$ such that $|\widetilde z-z|\le \delta_f(z, \varepsilon)$ implies $|f(\widetilde z)-f(z)|\le \varepsilon/|k|$ (assuming that $k\neq 0$ ). Then $|\widetilde z-z|\le \delta_f$ implies $|kf(\widetilde z)-kf(z)|=k |f(\widetilde z)-f(z)| \le |k| \varepsilon/|k|=\varepsilon$ .
(b) At any $z\in Dom(f)\cap Dom(g)$ , for any $\varepsilon$ , we take $\delta_f(z, \varepsilon)$ such that $|\widetilde z-z|\le \delta_f(z, \varepsilon)$ implies $|f(\widetilde z)-f(z)|\le \varepsilon/2$ , and $\delta_g(z, \varepsilon)$ such that $|\widetilde z-z|\le \delta_g(z, \varepsilon)$ implies $|g(\widetilde z)-g(z)|\le \varepsilon/2$ . Take $\delta=\min(\delta_f, \delta_g)$ , then $|\widetilde z-z|\le \delta$ implies $|(f+g)(\widetilde z)-(f+g)(z)|\le |f(\widetilde z)-f(z)|+|g(\widetilde z)-g(z)|\le \varepsilon/2+\varepsilon/2=\varepsilon$ .
(c) At any $z\in Dom(f)\cap Dom(g)$ , for any $\varepsilon$ , we take $\delta_g(z, \varepsilon)$ such that $|\widetilde z-z|\le \delta_g(z, \varepsilon)$ implies $|g(\widetilde z)-g(z)|\le \varepsilon/(2|f(z)|)$ , and $\delta_f(z, \varepsilon)$ such that $|\widetilde z-z|\le \delta_f(z, \varepsilon)$ implies $|f(\widetilde z)-f(z)|\le \varepsilon/(2(|g(z)|+\varepsilon/(2|f(z)|)))$ . Take $\delta=\min(\delta_f, \delta_g)$ , then $|\widetilde z-z|\le \delta$ implies $\begin{gathered} |(fg)(\widetilde z)-(fg)(z)|=|(f(\widetilde z)-f(z))g(\widetilde z)+f(z)(g(\widetilde z)-g(z)| \\ \le |f(\widetilde z)-f(z)||g(\widetilde z)|+|f(z)||g(\widetilde z)-g(z)| \\ \le \frac{\varepsilon}{2(|g(z)|+\varepsilon/(2|f(z)|))}|g(\widetilde z)|+|f(z)|\frac{\varepsilon}{2|f(z)|} \le \varepsilon\,, \end{gathered}$ where we used the fact that $|g(\widetilde z)|\le |g(z)|+\varepsilon/(2|f(z)|)$ . ◻

Exercise 13. Prove (d) and (e) of Theorem Theorem 12.

Guided review and additional examples

Learning goals

Interpret limits using both $\varepsilon$ - $\delta$ language and sequences.
Prove continuity by estimating the modulus of a difference.
Apply algebraic and composition rules for continuous functions.
Separate continuity from complex differentiability.

Concept connection: one modulus controls every direction

A complex limit requires the same value along every path in the plane. The modulus packages all directions into one distance, so familiar real-variable estimates often transfer directly. To disprove a limit, it is enough to find two paths or two sequences that give different limiting values.

Worked example 1: an $\varepsilon$ - $\delta$ proof for $f(z)=z^2$

At $z_0$ ,

$|z^2-z_0^2|=|z-z_0|\,|z+z_0|.$

If $|z-z_0|<1$ , then $|z+z_0|\le |z-z_0|+2|z_0|<1+2|z_0|$ . Given $\varepsilon>0$ , choose

$\delta=\min\left\{1,\frac{\varepsilon}{1+2|z_0|}\right\}.$

Then $|z-z_0|<\delta$ implies $|z^2-z_0^2|<\varepsilon$ .

Worked example 2: conjugation is continuous

For $f(z)=\overline z$ ,

$|f(z)-f(z_0)|=|\overline z-\overline{z_0}| =|\overline{z-z_0}|=|z-z_0|.$

Thus $\delta=\varepsilon$ works at every point. This example is also a useful warning: continuity alone does not imply complex differentiability.

Check your understanding

Suppose $z_n\to z_0$ and $f$ is continuous at $z_0$ . What must happen to $f(z_n)$ ?

Show the answer

We must have $f(z_n)\to f(z_0)$ . Conversely, in metric spaces such as $\mathbb C$ , this sequential property for every sequence $z_n\to z_0$ is equivalent to continuity at $z_0$ .