Complex Analysis/Complex Functions/Continuous Functions

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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[edit]

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 functions[edit]

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[edit]

  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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