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

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

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.
\begin{figure}\mbox{\epsfig{file=NonUniformCv.eps,height=4cm}}\end{figure}

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:

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

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:

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

Noah Dana-Picard 2007-12-24