Continuity

A function $w = f (z)$ , defined over a domain $D$ , is said to be continuous in $D$ if for every point $ζ$ of $D$ we have \tag{2.00} \lim _{z \rightarrow \zeta} f(z)=f(\zeta). The limit of a complex function is defined here as in the real case. If for any $ε > 0$ there is a $δ (ε, ζ) > 0$ such that $| f (z) - a | < ε whenever 0 < | z - ζ | < δ (ε, ζ)$ we say that $f (z)$ approaches the limit $a$ as $z$ tends to $ζ$ and write $lim_{z \to ζ} f (z) = a .$ Thus, the definition of continuity amounts to the requirement that given any preassigned $ε > 0$ there is a $δ (ε, ζ)$ such that $| f (z) - f (ζ) | < ε$ for all $z$ for which $| z - ζ | < δ (ε, ζ)$ . Geometrically this means that no matter what circle we draw with $f (ζ)$ as center, it is always possible to find a neighborhood of $ζ$ which maps completely inside the given circle.

The notion of limit for a function is only a little more general than that of continuity. A function which possesses a limit at a point can be made continuous simply by altering its value at the point to coincide with the limiting value.

The definition of continuity may be given in other ways. A function $w = f (z)$ is continuous at the point $ζ$ in its domain of definition if for every sequence ${z_{n}}$ in $D$ with $z_{n} \to ζ$ we have \tag{2.01} \lim _{n \rightarrow \infty} f(z_{n})=f(\zeta). Again, $w = f (z) = u (x, y) + i v (x, y)$ is continuous in $D$ if the real and imaginary parts, $u$ and $v$ , separately are continuous functions of $x$ and $y$ in $D$ . The reader may demonstrate the equivalence of these three definitions for himself.

If two functions $f (z)$ and $g (z)$ are both continuous in $D$ then evidently $f (z) \pm g (z)$ and $f (z) . g (z)$ are also continuous in $D$ . Furthermore, the quotient $\frac{f (z)}{g (z)}$ is continuous in any subdomain of $D$ in which $g (z) \neq 0$ .

Theorem 2.1 . A continuous function of a continuous function is continuous. More precisely, if $g (z)$ is continuous and maps a domain $D$ onto a point set D' and if $f (z)$ is continuous in a domain containing D' then the function $F (z) = f (g (z))$ is continuous in $D$ .

The proof is a direct application of (2.01) . It is now easy to find large classes of continuous functions. From the fact that $f (z) = constant$ and $f (z) = z$ are both continuous it follows that any polynomial $p (z) = a_{0} + a_{1} z + \dots + a_{n} z^{n}$ is continuous in the entire $z$ -plane and moreover, that any rational function $g (z) = \frac{a_{0} + a_{1} z + \dots + a_{m} z^{m}}{b_{0} + b_{1} z + \dots + b_{n} z^{n}}$ is continuous in any domain in which the denominator is not zero.

For a continuous function, the function values remain within an $ε$ -neighborhood of $f (ζ)$ for all values in a sufficiently small $δ$ -neighborhood of $ζ$ where $δ$ depends on both $ε$ and $ζ, δ = δ (ε, ζ)$ . In general, it will not be possible to pick $δ$ completely independently of the point $ζ$ . The function $\frac{1}{z}$ , for example, is continuous in the domain $0 < | z | < 1$ . But, given any $ε$ , there is no fixed value of $δ$ which can be employed for the entire domain; for, clearly, as $ζ$ approaches the origin we must let $δ \to 0$ . In contrast, we say that a function is uniformly continuous if for every positive $ε$ there is a $δ (ε) > 0$ such that $| f (z) - f (ζ) | < ε$ for all points $z$ , $ζ$ in $D$ satisfying $| z - ζ | < δ (ε) .$ For a finite region continuity implies uniform continuity. In fact we may state more generally:

Theorem 2.2 . If a function $f (z)$ is continuous in a closed bounded point set $\overset{―}{D}$ then it is uniformly continuous.

Setting $f (z) = u (x, y) + i v (x, y)$ we see that this result is a corollary of the theorem for real functions. ¹

There are numerous powerful techniques for representing a function by a convergent series of functions. It is therefore essential to have a method of determining whether a function given as the sum of a convergent series is continuous. Such a criterion is provided by the theorem:

Theorem 2.3 . A function $f (z)$ which is defined in a domain $D$ as the sum of a uniformly convergent series of functions continuous in $z$ must be continuous.

Proof. For suppose we have $f (z) = \sum_{n = 1}^{\infty} f_{n} (z)$ where the $f_{n} (z)$ are all continuous in $D$ . Then, denoting the remainder after $n$ terms by $R_{n} (z)$ , we may write $f (z) = \sum_{ν = 1}^{n} f_{ν} (z) + R_{n} (z) .$ Hence, for any pair of points $z$ and $ζ$ in $D$ we have \begin{align} |f(z)-f(\zeta)| &=\left|\sum_{\nu = 1}^{n} f_{\nu}(z)-\sum_{\nu=1}^{n} f_{\nu}(\zeta)+R_{n}(z)-R_{n}(\zeta)\right|\\ &\leq\left|\sum_{\nu=1}^{n} f_{\nu}(z)-\sum_{\nu=1}^{n} f_{\nu}(\zeta)\right|+\left|R_{n}(z)-R_{n}(\zeta)\right|. \end{align}Since the series converges uniformly in $D$ it follows for any positive $ε$ that there is an $N (ε)$ for which $| R_{n} (z) | < \frac{ε}{3}$ whenever $n > N (ε)$ and for all $z$ in $D$ . Furthermore, a finite sum of continuous functions is continuous and therefore we can determine a $δ (ε, ζ)$ such that $| \sum_{ν = 1}^{n} f_{ν} (z) - \sum_{ν = 1}^{n} f_{ν} (ζ) | < \frac{ε}{3} for | z - ζ | < 0.$ Thus given an $ε > 0$ we can find a $δ > 0$ such that $| f (z) - f (ζ) | < ε whenever | z - ζ | < δ (ε, ζ)$ and so we have proved $f (z)$ continuous. ◻

As a corollary of this theorem and the first theorem from Section on Power Series we observe that a power series represents a continuous function in the interior of its convergence .

Quick Links

Social Media