Uniform convergence.

Definition 7.3.1   The series $\underset{n \geq 0}{\sum} f_k(z)$ is uniformly convergent to $S(z)$ in a domain $D$ if

$$\forall \varepsilon >0, \exists N_0 \in \mathbb{N} s.t. n > N_0 \Longrightarrow \vert S(z)-S_n(z)\vert< \varepsilon$$

where $S_n(z)$ denotes the $n^{th}$ partial sum of the series.

Note that here $N_0$ depends on $\varepsilon$ only, not on the point $z$ .

Theorem 7.3.2 (Weierstrass)   We use the notations of the definition. Let $\underset{n \geq 0}{\sum} M_k$ be a convergent series of positive real numbers. Suppose that for any $k \in \mathbb{N}$ , $M_k$ is an upper bound for $\vert f_k(z)\vert$ on $D$ . Then the series $\underset{n \geq 0}{\sum} f_k(z)$ converges absolutely and uniformely on $D$ .

Theorem 7.3.3   Suppose that the series $\underset{n \geq 0}{\sum} f_k(z)$ converges uniformely to $F(z)$ on $D$ and that the function $g$ is bounded on $D$ . Then the series $\underset{n \geq 0}{\sum} g(z)f_k(z)$ is uniformly convergent on $D$ and its sum is equal to $g(z)F(z)$ .

Theorem 7.3.4   Suppose that the series $\underset{n \geq 0}{\sum} f_k(z)$ converges uniformely to $F(z)$ on $D$ . If every function $F_k$ is continuous on $D$ , then the sum $F$ is continuous on $D$ .

A famous example of a non-uniformly convergent sequence of functions is given in Calculus: take $f_k(x)=x^k$ oj the interval $[0,1]$ . The limit of this sequence is the function $F$ such that $F(x)$ for $0 \leq x < 1$ and $F(1)=1$ . As this limit is non continuous at 1, the sequence is not uniformly convergent (see Fig. 1).

Figure 1: Non uniform convergence of a sequence of functions.

For the series $\underset{n \geq 0}{\sum} x^k$ , it is defined on $[-1,1]$ , but convergent only on $(-1,1)$ .

Theorem 7.3.5 (term-by-term integration)   Suppose that the series $\underset{n \geq 0}{\sum} f_k(z)$ converges uniformly to $F(z)$ on the domain $D$ and that every $f_k$ is continuous on $D$ . Denote by $\mathcal{C}$ a Jordan curve in $D$ . Then, we have:

$$\int_{\mathcal{C}}F(z) \; dz = \underset{n \geq 0}{\sum} \int_{\mathcal{C}} f_k(z) \; dz.$$

Corollary 7.3.6 (analyticity of the sum)   Suppose that the series $\underset{n \geq 0}{\sum} f_k(z)$ converges uniformely on the domain $D$ and that every $f_k$ is analytic on $D$ . Then the sum $F(z)$ of the series is analytic on $D$ .

As a consequence, we have that power series converge uniformly on their open domain of convergence (v.i. 0.8).

Theorem 7.3.7 (term-by-term differentiation)   Suppose that the series $\underset{n \geq 0}{\sum} f_k(z)$ converges uniformely on the domain $D$ and denote its sum by $F(z)$ . If all the functions $f_k$ are analytic on $D$ , then at any interior point of $D$ , we have:

$$\frac {dF}{dz} = \underset{n \geq 0}{\sum} \frac {f_k(z)}{dz}$$