Last modified on 18 May 2015, at 19:37

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 preimageEdit

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 and 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 \epsilon > 0 : \exists \delta > 0 : \left( z \in B' \cap B(z_0, \delta) \Rightarrow |f(z) - w| < \epsilon \right)

, 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 \epsilon > 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| < \epsilon

. 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| < \epsilon

, and therefore

\lim_{z \to z_0 \atop z \in B''} f(z) = w
////

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 \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 B' \Rightarrow z \in B(z_0, \delta) \Rightarrow z \in B(z_0, \delta_2) \cap U \Rightarrow |f(z) - f(z_0)| < \epsilon
////

Continuity of complex functionsEdit

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:

ExercisesEdit

  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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