2.3 Uniform convergence

Definition 2.15

Suppose \(E\subseteq \mathbb {C}\) and \(\{ f_n\} \) is a sequence of complex-valued functions on \(E\). We say that \(\{ f_n\} \) converges uniformly to a function \(f:E\to \mathbb {C}\) if, for every positive real number \(\epsilon \), there exists a nonnegative integer \(N\) such that, for all \(z\in E\),

\begin{equation} \label{conv} n\geq N\Rightarrow |f_n(z) - f(z)| < \epsilon . \end{equation}

2.16

We say that \(\{ f_n\} \) converges to \(f\) pointwise on \(E\) if, for every \(z\in E\), \(\lim _{n\to \infty } f_n(z) = f(z)\). So, \(f_n\to f\) pointwise if, for every \(z\in E\) and every \(\epsilon {\gt} 0\), there exists an \(N\) such that 2.16 holds. The difference between pointwise and uniform convergence is that, for pointwise convergence, the \(N\) can depend on \(\epsilon \) and on \(z\), while, for uniform convergence, \(N\) can only depend on \(\epsilon \).

Example 2.17

Let

\[ f_n(z) = \begin{cases} 1 - n|z|, & |z| \leq 1/n;\\ 0, & |z| {\gt} 1/n, \end{cases} \]

and let

\[ f(z) = \begin{cases} 1, & z = 0;\\ 0, & z \neq 0. \end{cases} \]

Then \(f_n\to f\) pointwise on \(\mathbb {C}\), but not uniformly.

Theorem 2.22

Suppose \(E\subseteq \mathbb {C}\) and \(\{ f_n\} \) is a sequence of continuous complex-valued functions on \(E\) converging uniformly to a function \(f:E\to \mathbb {C}\). Then \(f\) is continuous.

Proof ▼

Pick \(\epsilon {\gt} 0\). Since \(f_n\to f\) uniformly on \(E\), we can an \(N\in \mathbb {N}\) such that, for all \(s\in E\) and \(n\geq N\), \(|f_n(s) - f(s)| {\lt} \epsilon /3\). Now suppose \(z\in E\). Since \(f_{N}\) is continuous at \(z\), we can find \(\delta {\gt} 0\) such that \(|s-z| {\lt} \delta \Rightarrow |f_N(s) - f_N(z) | {\lt} \epsilon /3\).

So now, if \(|s-z| {\lt}\delta \), then

\begin{align*} |f(s) - f(z)| & \leq |f(s) - f_N(s)| + |f_N(s) - f_N(z)| + |f_N(z) - f(z)|\\ & {\lt} \epsilon /3 + \epsilon /3 + \epsilon /3 = \epsilon . \end{align*}

And this proves that \(f\) is continuous at \(z\).

Theorem 2.23

Suppose \(\Omega \subseteq \mathbb {C}\) is an open subset, and \(\{ f_n\} \) is a sequence of continuous complex-valued functions on \(\Omega \) converging uniformly to a function \(f:\Omega \to \mathbb {C}\). Let \(g\in C(\Omega )\) be another continuous function on \(\Omega \). Then \(f_ng\to fg\) pointwise on \(\Omega \), and for any piecewise smooth contour \(C\subset \Omega \),

\[ \int _C f_n(z)g(z)\, dz \to \int _C f(z)g(z)\, dz. \]

Proof ▼

Since \(f_n\to f\) uniformly on \(\Omega \), for any \(z\in \Omega \), \(f_n(z)\to f(z)\). So \(f_n(z)g(z)\to f(z)g(z)\).

Now, for the integral, pick contour \(C\) and let \(L\) denote the length of \(C\). If \(L = 0\), then all the integrals in the statement of the theorem are \(0\). So we can assume \(L {\gt} 0\).

If \(\gamma :[a,b]\to \Omega \) is a parametrization of the contour \(C\), then the function \(t\mapsto |g(\gamma (t))|\) is continuous on \([a,b]\). Therefore, it has a maximum value \(M\). So, for all \(z\in C\), \(|g(z)|\leq M\).

Let \(\epsilon {\gt} 0\) be given. Then, since \(f_n\to f\) uniformly on \(\Omega \), we can find \(N\in \mathbb {N}\) such that \(n\geq N\Rightarrow |f_n(z) - f(z)| {\lt} \epsilon /LM\) for all \(z\in \Omega \). Therefore, if \(n\geq N\), we have

\begin{align*} |\int _C f_ng\, dz - \int _C fg\, dz| & \leq \int _C |f_n(z) - f(z)||g(z)|\, dz\\ & \leq LM(\epsilon /LM) = \epsilon . \end{align*}

So the theorem is proved.

Theorem 2.24

Suppose \(\Omega \subseteq \mathbb {C}\) is an open subset, and \(\{ f_n\} \) is a sequence of analytic functions on \(\Omega \) converging uniformly to a function \(f:\Omega \to \mathbb {C}\). Then \(f\in A(\Omega )\). Moreover, for every nonnegative integer \(k\), \(f_n^{(k)} \to f^{(k)}\).

Proof ▼

Pick \(z\in \Omega \). Since \(\Omega \) is open, we can find a \(r {\gt} 0\), such that the disk \(D = D(z,r) = \{ s\in \mathbb {C}: |s - z| {\lt} r\} \) is contained in \(\Omega \). We’ll use Morera’s theorem to show that \(f\) is analytic on \(D\). This will show that \(f\) is analytic at \(z\), and, since \(z\) was an arbitrary point in \(\Omega \), this will show that \(f\) is analytic on \(\Omega \).

Pick a simple, closed, piecewise-smooth curve \(C\) in \(D\). Then, since \(D\) is simply-connected and each \(f_n\) is analytic, \(\int _C f_n\, dz = 0\). Therefore, since \(f_n\to f\) uniformly on \(\Omega \), \(f\) is continuous, and \(\int _C f\, dz = 0\), by Theorem 2.23. But then, by Morera, \(f\) is analytic on \(D\).

To prove the last assertion, suppose \(k\) is a nonnegative integer, and set

\[ g(s) = \frac{k!}{2\pi i (s - z)^{k+1}} \]

for \(s\in D\). Let \(C\) be a circle of radius \(r/2\) centered at \(z\). Then, by Cauchy, \(f_n^{(k)}(z) = \int _C f_n(s) g(s)\, ds\), and, since \(f\) is analytic, \(f(z) = \int _C f(s) g(s)\, ds\). So, by Theorem 2.23, \(f_n^{(k)}(z)\to f^{(k)}(z)\).