A function \(w=f(z)\) , defined over a domain \(D\) , is said to be continuous in \(D\) if for every point \(\zeta\) 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 \(\varepsilon>0\) there is a \(\delta(\varepsilon, \zeta)>0\) such that \[|f(z)-a|<\varepsilon\ \text { whenever }\ 0<|z-\zeta|<\delta(\varepsilon, \zeta)\] we say that \(f(z)\) approaches the limit \(a\) as \(z\) tends to \(\zeta\) and write \[\lim _{z \rightarrow \zeta} f(z)=a.\] Thus, the definition of continuity amounts to the requirement that given any preassigned \(\varepsilon>0\) there is a \(\delta(\varepsilon, \zeta)\) such that \(|f(z)-f(\zeta)|<\varepsilon\) for all \(z\) for which \(|z-\zeta|<\delta(\varepsilon, \zeta)\) . Geometrically this means that no matter what circle we draw with \(f(\zeta)\) as center, it is always possible to find a neighborhood of \(\zeta\) 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 \(\zeta\) in its domain of definition if for every sequence \(\left\{z_{n}\right\}\) in \(D\) with \(z_{n} \rightarrow \zeta\) 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 \(f(z)= \textit{constant}\) and \(f(z)=z\) are both continuous it follows that any polynomial \[p(z)=a_{0}+a_{1} z+\cdots+a_{n} z^{n}\] is continuous in the entire \(z\) -plane and moreover, that any rational function \[g(z)=\frac{a_{0}+a_{1} z+\cdots+a_{m} z^{m}}{b_{0}+b_{1} z+\cdots+b_{n} z^{n}}\] is continuous in any domain in which the denominator is not zero.
For a continuous function, the function values remain within an \(\varepsilon\) -neighborhood of \(f(\zeta)\) for all values in a sufficiently small \(\delta\) -neighborhood of \(\zeta\) where \(\delta\) depends on both \(\varepsilon\) and \(\zeta, \delta=\delta(\varepsilon, \zeta)\) . In general, it will not be possible to pick \(\delta\) completely independently of the point \(\zeta\) . The function \(\frac{1}{z}\) , for example, is continuous in the domain \(0<|z|<1\) . But, given any \(\varepsilon\) , there is no fixed value of \(\delta\) which can be employed for the entire domain; for, clearly, as \(\zeta\) approaches the origin we must let \(\delta \rightarrow 0\) . In contrast, we say that a function is uniformly continuous if for every positive \(\varepsilon\) there is a \(\delta(\varepsilon)>0\) such that \[|f(z)-f(\zeta)|<\varepsilon\] for all points \(z\) , \(\zeta\) in \(D\) satisfying \[|z-\zeta|<\delta(\varepsilon).\] 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 \(\overline{D}\) then it is uniformly continuous.
Setting \(f(z)=u(x, y)+ iv(x, y)\) 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 \[f(z)=\sum_{\nu = 1}^{n} f_{\nu}(z)+R_{n}(z).\] Hence, for any pair of points \(z\) and \(\zeta\) 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 \(\varepsilon\) that there is an \(N(\varepsilon)\) for which \(\left|R_{n}(z)\right|<\frac{\varepsilon}{3}\) whenever \(n>N(\varepsilon)\) and for all \(z\) in \(D\) . Furthermore, a finite sum of continuous functions is continuous and therefore we can determine a \(\delta(\varepsilon, \zeta)\) such that \[\left|\sum_{\nu = 1}^{n} f_{\nu}(z)-\sum_{\nu = 1}^{n} f_{\nu}(\zeta)\right|<\frac{\varepsilon}{3}\quad \text{for }|z-\zeta|<0.\] Thus given an \(\varepsilon>0\) we can find a \(\delta>0\) such that \[|f(z)-f(\zeta)|<\varepsilon\quad \text{whenever } |z-\zeta|<\delta(\varepsilon, \zeta)\] 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 .