2.3 Uniform Convergence
The notes go on to differentiate and integrate series term by term. That step is not free, and uniform convergence is the hypothesis that pays for it. The distinction below is the whole of the matter.
Definition 2.9 (pointwise and uniform convergence). Let \(f_n\) and \(f\) be functions on a set \(S \subseteq \mathbb {C}\).
- (i).
- \(f_n \rightarrow f\) pointwise on \(S\) if for every \(z \in S\) and every \(\varepsilon > 0\) there is \(N\), possibly depending on both \(z\) and \(\varepsilon \), with \(\left |f_n(z) - f(z)\right | < \varepsilon \) for all \(n \geq N\).
- (ii).
- \(f_n \rightarrow f\) uniformly on \(S\) if for every \(\varepsilon > 0\) there is \(N\), depending on \(\varepsilon \) alone, with \[\left |f_n(z) - f(z)\right | < \varepsilon \quad \text {for all } n \geq N \,\text {and all } z \in S .\]
The single word that separates them is alone. Under pointwise convergence each point may demand its own \(N\), and there need be no bound on how large those \(N\) have to be as \(z\) moves about \(S\); under uniform convergence one \(N\) serves the whole set at once.
Example 2.10. On \(S = \{\left |z\right | < 1\}\) take \(f_n(z) = z^n\). Then \(f_n \rightarrow 0\) pointwise, since \(\left |z\right | < 1\) gives \(\left |z\right |^n \rightarrow 0\). The convergence is not uniform on \(S\): taking \(z\) with \(\left |z\right | = 1 - \tfrac {1}{n}\) gives \(\left |f_n(z)\right | = \left (1 - \tfrac {1}{n}\right )^n \rightarrow e^{-1}\), which does not go to \(0\), so no single \(N\) can work everywhere. On the smaller disc \(\left |z\right | \leq r\) with \(r < 1\) the convergence is uniform, because \(\left |f_n(z)\right | \leq r^n \rightarrow 0\) with a bound independent of \(z\).
That example is the pattern for power series generally: uniform on every closed disc strictly inside the circle of convergence, and typically not on the open disc itself.
Theorem 2.11 (the Weierstrass \(M\)-Test). Let \(\displaystyle \sum ^{\infty }_{n = 1} u_n(z)\) be a series of functions on \(S\), and suppose there are constants \(M_n\) with \[\left |u_n(z)\right | \leq M_n \quad \text {for all } z \in S, \quad \text {and}\quad \sum ^{\infty }_{n=1} M_n < \infty .\] Then \(\displaystyle \sum u_n\) converges absolutely and uniformly on \(S\).
Proof. Absolute convergence at each point is immediate by comparison with \(\sum M_n\). Write \(s(z)\) for the sum and \(s_N(z)\) for the \(N\)-th partial sum. Then for every \(z \in S\) \[\left |s(z) - s_N(z)\right | = \left |\sum ^{\infty }_{n = N+1} u_n(z)\right | \leq \sum ^{\infty }_{n = N+1}\left |u_n(z)\right | \leq \sum ^{\infty }_{n = N+1} M_n .\] The right-hand side is the tail of a convergent series of constants, so it can be made smaller than any \(\varepsilon > 0\) by choosing \(N\) large — and that choice does not involve \(z\). \(\blacksquare \)
The \(M\)-test is the workhorse: it converts a question about functions into a question about a series of numbers. □
Theorem 2.12. A uniform limit of continuous functions is continuous. That is, if each \(f_n\) is continuous on \(S\) and \(f_n \rightarrow f\) uniformly on \(S\), then \(f\) is continuous on \(S\).
Proof. Fix \(z_0 \in S\) and \(\varepsilon > 0\). Choose \(N\) with \(\left |f_N(z) - f(z)\right | < \tfrac {\varepsilon }{3}\) for all \(z \in S\), which uniformity permits. Since \(f_N\) is continuous at \(z_0\) there is \(\delta > 0\) such that \(\left |z - z_0\right | < \delta \) gives \(\left |f_N(z) - f_N(z_0)\right | < \tfrac {\varepsilon }{3}\). Then for such \(z\) \[\left |f(z) - f(z_0)\right | \leq \left |f(z) - f_N(z)\right | + \left |f_N(z) - f_N(z_0)\right | + \left |f_N(z_0) - f(z_0)\right | < \tfrac {\varepsilon }{3} + \tfrac {\varepsilon }{3} + \tfrac {\varepsilon }{3} = \varepsilon . \quad \blacksquare \]
The three-way split above is worth remembering; it recurs whenever a uniform estimate is traded for a local one. □
Theorem 2.13 (term-by-term integration). If \(f_n \rightarrow f\) uniformly on a contour \(C\) of finite length \(L\), and each \(f_n\) is continuous on \(C\), then \[\lim _{n \rightarrow \infty }\int _C f_n(z)\,dz = \int _C f(z)\,dz .\]
Proof. Given \(\varepsilon > 0\) pick \(N\) so that \(\left |f_n(z) - f(z)\right | < \frac {\varepsilon }{L}\) on \(C\) for \(n \geq N\). By the estimation (ML) inequality, \[\left |\int _C f_n(z)\,dz - \int _C f(z)\,dz\right | = \left |\int _C \big (f_n(z) - f(z)\big )dz\right | \leq \frac {\varepsilon }{L}\cdot L = \varepsilon . \quad \blacksquare \]
Note where uniformity was used: the bound \(\varepsilon / L\) had to hold at every point of \(C\) at once. Pointwise convergence gives no such bound and the theorem is false without it. □
Theorem 2.14 (Weierstrass’ theorem on analytic limits). Let \(f_n\) be analytic on a domain \(D\) and suppose \(f_n \rightarrow f\) uniformly on every compact subset of \(D\). Then \(f\) is analytic on \(D\), and moreover \(f_n' \rightarrow f'\) uniformly on every compact subset of \(D\).
Proof. \(f\) is continuous by the theorem above. For any triangle \(\Delta \) with its interior contained in \(D\), Cauchy’s theorem gives \(\int _{\partial \Delta } f_n = 0\), and term-by-term integration passes to the limit to give \(\int _{\partial \Delta } f = 0\). Morera’s theorem then makes \(f\) analytic. The statement about derivatives follows from the Cauchy integral formula for \(f'\) applied to \(f_n - f\) on a circle inside \(D\). \(\blacksquare \)
This is the result that licenses differentiating a power series term by term, and it has no analogue for real functions: a uniform limit of real polynomials need not be differentiable even once, whereas here the limit is infinitely differentiable. □
Corollary 2.15 (power series converge uniformly on compact subdiscs). If \(\displaystyle \sum a_n (z - z_0)^n\) has radius of convergence \(R > 0\), then for every \(r\) with \(0 < r < R\) the series converges uniformly on \(\left |z - z_0\right | \leq r\).
Proof. Pick \(\rho \) with \(r < \rho < R\). Since \(\rho < R\) the series \(\sum \left |a_n\right | \rho ^n\) converges. On \(\left |z - z_0\right | \leq r\), \[\left |a_n (z - z_0)^n\right | \leq \left |a_n\right | r^n = \left |a_n\right |\rho ^n \left (\frac {r}{\rho }\right )^n \leq \left |a_n\right | \rho ^n =: M_n ,\] and \(\sum M_n < \infty \). The Weierstrass \(M\)-test applies. \(\blacksquare \)
Consequently a power series is analytic inside its circle of convergence and may be differentiated and integrated term by term there — which is what the rest of this course assumes throughout. □
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