Weierstrass Convergence Theorem
Domain Boundary Convergence Corollary
Theorem: let \{u_n(z)\}_{n=1}^\infty be a sequence of functions, each analytic in an open domain \mathcal{D}. If the series \sum_{n=1}^\infty u_n(z) converges uniformly to a function f(z) on every compact subset of \mathcal{D}, then:
f^{(k)}(z) = \sum_{n=1}^\infty u_n^{(k)}(z)
Proof: the proof is structured in three parts, corresponding to the three conclusions of the theorem.
We will establish the analyticity of f(z) by employing Morera’s theorem (here), which states that a continuous function in a domain \mathcal{D} whose integral over every simple closed contour within \mathcal{D} is zero must be analytic in \mathcal{D}.
By hypothesis, each function u_n(z) is analytic and therefore continuous in \mathcal{D}. We have previously established that the uniform limit of a series of continuous functions is continuous. Since the series \sum u_n(z) converges uniformly to f(z) on compact subsets of \mathcal{D}, the function f(z) is continuous in \mathcal{D}.
Next, let C be any simple closed contour contained within \mathcal{D}. Since C is a compact set, the series converges uniformly on C. We can therefore apply the theorem on term-by-term integration:
\oint_C f(z) \, \mathrm dz = \oint_C \left( \sum_{n=1}^\infty u_n(z) \right) \, \mathrm dz = \sum_{n=1}^\infty \oint_C u_n(z) \, \mathrm dz
By hypothesis, each function u_n(z) is analytic in \mathcal{D}. According to Cauchy’s integral theorem, the integral of an analytic function over a simple closed contour within its domain of analyticity is zero. Therefore, for every n:
\oint_C u_n(z) \, \mathrm dz = 0
Substituting this result into the sum, we find:
\oint_C f(z) \, \mathrm dz = \sum_{n=1}^\infty 0 = 0
The conditions for Morera’s theorem are satisfied: f(z) is continuous in \mathcal{D}, and its integral along any simple closed contour is zero. We conclude that f(z) is analytic in \mathcal{D}.
To derive the formula for the derivatives of f(z), we use Cauchy’s Integral Formula for derivatives. Let z be an arbitrary point in \mathcal{D}, and let C be a simple closed contour in \mathcal{D} that encloses z. Consider the series:
\sum_{n=1}^\infty \frac{u_n(\zeta)}{(\zeta - z)^{k+1}}
where \zeta is a point on the contour C. The original series \sum u_n(\zeta) converges uniformly on C. The term |\zeta - z| has a minimum positive value, d = \min_{\zeta \in C}|\zeta - z| > 0. The modulus of each term in the new series is bounded:
\left| \frac{u_n(\zeta)}{(\zeta - z)^{k+1}} \right| \le \frac{|u_n(\zeta)|}{d^{k+1}}
By the Weierstrass M-test, the uniform convergence of \sum u_n(\zeta) implies the uniform convergence of this new series on C. This permits us to integrate term-by-term along the contour C:
\oint_C \frac{f(\zeta)}{(\zeta - z)^{k+1}} \, \mathrm d\zeta = \oint_C \left( \sum_{n=1}^\infty \frac{u_n(\zeta)}{(\zeta - z)^{k+1}} \right) \, \mathrm d\zeta = \sum_{n=1}^\infty \oint_C \frac{u_n(\zeta)}{(\zeta - z)^{k+1}} \, \mathrm d\zeta
We now apply Cauchy’s Integral Formula for derivatives:
g^{(k)}(z) = \frac{k!}{2\pi i} \oint_C \frac{g(\zeta)}{(\zeta - z)^{k+1}} \, \mathrm d\zeta
to the entire expression. The left-hand side becomes:
\oint_C \frac{f(\zeta)}{(\zeta - z)^{k+1}} \, \mathrm d\zeta = \frac{2\pi i}{k!} f^{(k)}(z)
Each term on the right-hand side becomes:
\oint_C \frac{u_n(\zeta)}{(\zeta - z)^{k+1}} \, \mathrm d\zeta = \frac{2\pi i}{k!} u_n^{(k)}(z)
Substituting these back into the integrated equation gives:
\frac{2\pi i}{k!} f^{(k)}(z) = \sum_{n=1}^\infty \frac{2\pi i}{k!} u_n^{(k)}(z)
Canceling the constant factor \frac{2\pi i}{k!} yields the desired formula. Since z was an arbitrary point in \mathcal{D}, the result is valid throughout the domain:
f^{(k)}(z) = \sum_{n=1}^\infty u_n^{(k)}(z)
Finally, we must show that the series of derivatives, \sum u_n^{(k)}(z), converges uniformly on any compact subset of \mathcal{D}.
Let K be a compact subset of \mathcal{D}. We can choose a simple closed contour C that lies entirely in \mathcal{D} and encloses K.
Because K is compact and C is closed, there is a minimum distance d > 0 between any point z \in K and any point \zeta \in C, so |\zeta - z| \ge d > 0.
The remainder term of the derivative series is:
r_n^{(k)}(z) = f^{(k)}(z) - \sum_{j=1}^n u_j^{(k)}(z) = \sum_{j=n+1}^\infty u_j^{(k)}(z)
The remainder of the original series is R_n(\zeta) = f(\zeta) - \sum_{j=1}^n u_j(\zeta). From the previous part, we can write the derivative of the remainder using the integral formula:
r_n^{(k)}(z) = \frac{k!}{2\pi i} \oint_C \frac{R_n(\zeta)}{(\zeta - z)^{k+1}} \, \mathrm d\zeta
We bound the modulus of r_n^{(k)}(z) using the ML-inequality. Let L be the length of C.
\begin{aligned} |r_n^{(k)}(z)| &= \left| \frac{k!}{2\pi i} \oint_C \frac{R_n(\zeta)}{(\zeta - z)^{k+1}} \, \mathrm d\zeta \right| \\ &\le \frac{k!}{2\pi} \oint_C \frac{|R_n(\zeta)|}{|\zeta - z|^{k+1}} \,|\mathrm d\zeta| \\ &\le \frac{k!}{2\pi} \frac{\sup_{\zeta \in C}|R_n(\zeta)|}{d^{k+1}} L \end{aligned}
The original series \sum u_n(\zeta) converges uniformly on the compact set C. This means that \sup_{\zeta \in C}|R_n(\zeta)| \to 0 as n \to \infty. For any \varepsilon > 0, we can find an integer N such that for all n > N, \sup_{\zeta \in C}|R_n(\zeta)| is smaller than any positive number we choose.
Let’s make it smaller than \varepsilon \frac{2\pi d^{k+1}}{k! L}. Then for all n > N and for all z \in K:
|r_n^{(k)}(z)| \le \frac{k! L}{2\pi d^{k+1}} \left(\varepsilon \frac{2\pi d^{k+1}}{k! L}\right) = \varepsilon
Since N depends only on \varepsilon (and the fixed geometry of K and C) but not on the specific point z \in K, we have shown that the remainder term r_n^{(k)}(z) converges uniformly to zero on K. This establishes the uniform convergence of the derivative series on any compact subset of \mathcal{D}.
Corollary: let \{u_n(z)\} be a sequence of functions that are analytic within a bounded domain \mathcal{D} and continuous on its closure \overline{\mathcal{D}}. If the series \sum_{n=1}^\infty u_n(z) converges uniformly on the boundary of \mathcal{D}, then the series converges uniformly on the entirety of the closed domain \overline{\mathcal{D}}.
Proof: the difference of the partial sum of the considered function, S_{n+p}(z) - S_n(z) as it is a sum of analytical function is itself analytical in \mathcal D and continuous in \overline{\mathcal D}. From the uniformly convergence on the boundary it follows that, for n \ge N:
\left| S_{n+p}(z) - S_n(z) \right| = \left| u_{n+p}(z) + \cdots + S_{n+1}(z) \right| < \varepsilon
for every p \in \mathbb N and z \in \partial \mathcal D. The maximum modulus theorem (here) implies that this inequality is true for every p \in \mathbb N and z \in \overline{\mathcal D}.