Section 7.3 Sequences and Series of Functions
The fun starts when we study sequences of functions.
Definition 7.3.1.
Let \(G \subseteq \C\) and \(f_n : G \to \C\) for \(n
\ge 1\text{.}\) We say that \((f_n)\) converges pointwise to \(f: G \to \C\) if for each \(z \in G\text{,}\)
\begin{equation*}
\lim_{ n \to \infty } f_n(z) \ = \ f(z) \,\text{.}
\end{equation*}
We say that \((f_n)\) converges uniformly to \(f: G \to \C\) if for all \(\epsilon>0\) there is an \(N\) such that for all \(z
\in G\) and for all \(n \geq N\)
\begin{equation*}
\left| f_n(z) - f(z) \right| \ \lt \ \epsilon\,\text{.}
\end{equation*}
Sometimes we want to express that either notion of convergence holds only on a subset
\(H\) of
\(G\text{,}\) in which case we say that
\((f_n)\) converges pointwise/uniformly on
\(H\text{.}\)
It should be clear that uniform convergence on a set implies pointwise convergence on that set; but the converse is not generally true.
Let’s digest these two notions of convergence of a function sequence by describing them using quantifiers; as usual, \(\forall\) denotes for all and \(\exists\) means there exists. Pointwise convergence on \(G\) says
\begin{equation*}
\forall \, \epsilon > 0 \ \ \forall \, z \in G \ \ \exists \, N
\ \ \forall \, n \geq N \ \ \left| f_n(z) - f(z) \right| \ \lt
\ \epsilon \, \text{,}
\end{equation*}
whereas uniform convergence on \(G\) translates into
\begin{equation*}
\forall \, \epsilon > 0 \ \ \exists \, N \ \ \forall \, z \in G
\ \ \forall \, n \geq N \ \ \left| f_n(z) - f(z) \right| \ \lt
\ \epsilon \, \text{.}
\end{equation*}
No big deal — we only exchanged two of the quantifiers. In the first case,
\(N\) may well depend on
\(z\text{,}\) in the second case we need to find an
\(N\) that works for all
\(z \in G\text{.}\) And this can make all the difference …
Example 7.3.2.
Example 7.3.3.
The differences between
Example 7.3.2 and
Example 7.3.3 are subtle, and we suggest you meditate over them for a while with a good cup of coffee. You might already suspect that the function sequence in
Example 7.3.2 does
not converge uniformly, as we will see in a moment.
The first application illustrating the difference between pointwise and uniform convergence says, in essence, that if we have a sequence of functions \(\left( f_n \right)\) that converges uniformly on \(G\) then for all \(z_0 \in G\)
\begin{equation*}
\lim_{n \to \infty} \ \lim_{z \to z_0} f_n(z) \ = \ \lim_{z \to
z_0} \ \lim_{n \to \infty} f_n(z) \, \text{.}
\end{equation*}
We will need similar interchanges of limits frequently.
Proposition 7.3.4.
Suppose
\(G\subset\C\) and
\(f_n: G \to \C\) is continuous, for each
\(n \ge 1\text{.}\) If
\((f_n)\) converges uniformly to
\(f: G \to \C\) then
\(f\) is continuous.
Proof.
Let \(z_0 \in G\text{;}\) we will prove that \(f\) is continuous at \(z_0\text{.}\) By uniform convergence, given \(\epsilon>0\text{,}\) there is an \(N\) such that for all \(z \in G\) and all \(n \geq N\)
\begin{equation*}
\left| f_n(z) - f(z) \right| \ \lt \ \frac \epsilon 3 \,\text{.}
\end{equation*}
Now we make use of the continuity of the \(f_n\)’s. This means that given (the same) \(\epsilon>0\text{,}\) there is a \(\delta>0\) such that whenever \(|z-z_0| \lt \delta\text{,}\)
\begin{equation*}
\left| f_n(z) - f_n(z_0) \right| \ \lt \ \frac \epsilon 3 \,\text{.}
\end{equation*}
All that’s left is putting those two inequalities together: by the triangle inequality (
Corollary 1.3.5 c)),
\begin{align*}
\left| f(z) - f(z_0) \right| \ \amp = \ \left| f(z) -
f_n(z) + f_n(z) - f_n(z_0) + f_n(z_0) - f(z_0) \right|\\
\amp \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|\\
\amp \lt \ \epsilon \, \text{.}
\end{align*}
This proves that
\(f\) is continuous at
\(z_0\text{.}\)
Proposition 7.3.4 can sometimes give a hint that a function sequence does not converge uniformly.
Example 7.3.5.
We modify
Example 7.3.2 and consider the real function sequence
\(f_n: [0,1] \to \R\) given by
\(f_n(x) = x^n\text{.}\) It converges pointwise to
\(f: [0,1] \to \R\) given by
\begin{equation*}
f(x) = \begin{cases}0 \amp \text{ if } 0 \le x \lt 1 \, ,
\\ 1 \amp \text{ if } x = 1 \, . \end{cases}
\end{equation*}
As this limiting function is not continuous, the above convergence cannot be uniform. This gives a strong indication that the convergence in
Example 7.3.2 is not uniform either, though this needs a separate proof, as the domain of the functions in
Example 7.3.2 is the unit disk (
Exercise 7.5.20(b)).
Now that we have established
Proposition 7.3.4 about continuity, we can ask about integration of sequences or series of functions. The next theorem should come as no surprise; however, its consequences (which we will see shortly) are wide ranging.
Proposition 7.3.6.
Suppose \(f_n: G \to \C\) is continuous, for \(n \ge 1\text{,}\) \((f_n)\) converges uniformly to \(f: G \to \C\text{,}\) and \(\gg \subseteq G\) is a piecewise smooth path. Then
\begin{equation*}
\lim_{n \to \infty} \int_\gamma f_n \ = \ \int_\gamma f \,\text{.}
\end{equation*}
Proof.
We may assume that \(\gg\) is not just a point, in which case the proposition holds trivially. Given \(\epsilon > 0\text{,}\) there exists \(N\) such that for all \(z \in G\) and all \(n \ge N\text{,}\)
\begin{equation*}
\left| f_n(z) - f(z) \right| \ \lt \ \frac{ \epsilon }{
\length(\gg) } \, \text{.}
\end{equation*}
\begin{equation*}
\left| \int_\gamma f_n - \int_\gamma f \right| \ = \ \left|
\int_\gamma f_n - f \right| \ \leq \ \max_{z \in \gamma}
\left| f_n(z) - f(z) \right| \cdot \length (\gamma) \ \lt \
\epsilon \, \text{.}
\end{equation*}
All of these notions for sequences of functions hold verbatim for series of functions. For example, if \(\sum_{ k \ge 1 } f_k(z)\) converges uniformly on \(G\) and \(\gg \subseteq G\) is a piecewise smooth path, then
\begin{equation*}
\int_\gg \sum_{ k \ge 1 } f_k(z) \, \diff{z} \ = \ \sum_{ k \ge 1 }
\int_\gg f_k(z) \, \diff{z} \, \text{.}
\end{equation*}
In some sense, the above identity is
the reason we care about uniform convergence.
There are several criteria for uniform convergence; see, e.g.,
Exercise 7.5.19 and
Exercise 7.5.20, and the following result, sometimes called the
Weierstraß \(M\)-test.
Proposition 7.3.7.
Suppose
\(f_k : G \to \C\) for
\(k \ge 1\text{,}\) and
\(|f_k(z)| \le M_k\) for all
\(z \in G\text{,}\) where
\(\sum_{ k \ge 1 } M_k\) converges. Then
\(\sum_{k \geq 1} \left|f_k\right|\) and
\(\sum_{k \geq 1} f_k\) converge uniformly in
\(G\text{.}\) (We say the series
\(\sum_{k \geq 1} f_k\) converges absolutely and uniformly.)
Proof.
For each fixed
\(z\text{,}\) the series
\(\sum_{k \geq 1} f_k(z)\) converges absolutely by
Corollary 7.2.6. To show that the convergence is uniform, let
\(\epsilon > 0\text{.}\) Then there exists
\(N\) such that for all
\(n \ge N\text{,}\)
\begin{equation*}
\sum_{k \ge 1} M_k - \sum_{k=1}^nM_k \ = \ \sum_{k >n}M_k \ \lt
\ \epsilon \, \text{.}
\end{equation*}
Thus for all \(z \in G\) and \(n \ge N\text{,}\)
\begin{equation*}
\abs{\sum_{k \geq 1} f_k(z)-\sum_{k=1}^nf_k(z)} \ = \
\abs{\sum_{k >n}f_k(z)} \ \le \ \sum_{k >n}\abs{f_k(z)} \ \le
\ \sum_{k >n}M_k \ \lt \ \epsilon \, \text{,}
\end{equation*}
which proves uniform convergence. Replace \(f_k\) with \(\left|f_k\right|\) in this argument to see that \(\sum_{k \geq 1} \left|f_k\right|\) also converges uniformly.
Example 7.3.8.
We revisit
Example 7.2.2 and consider the geometric series
\(\sum_{ k \ge 1 } z^k\) as a series of functions in
\(z\text{.}\) We know from
Example 7.2.2 that this function series converges pointwise for
\(|z| \lt 1\text{:}\)
\begin{equation*}
\sum_{ k \ge 1 } z^k \ = \ \frac z {1-z} \,\text{.}
\end{equation*}
To study uniform convergence, we apply
Proposition 7.3.7 with
\(f_k(z) = z^k\text{.}\) We need a series of upper bounds that converges, so fix a real number
\(0 \lt r \lt 1\) and let
\(M_k = r^k\text{.}\) Then
\begin{equation*}
|f_k(z)| \ = \ |z|^k \ \le \ r^k \qquad \text{ for } \ |z| \le
r \, \text{,}
\end{equation*}
and
\(\sum_{ k \ge 1 } r^k\) converges by
Example 7.2.2. Thus,
Proposition 7.3.7 says that
\(\sum_{ k \ge 1 } z^k\) converges uniformly for
\(|z| \le r\text{.}\)
We note the subtle distinction of domains for pointwise/uniform convergence:
\(\sum_{ k \ge 1 } z^k\) converges (absolutely) for
\(|z|\lt 1\text{,}\) but to force
uniform convergence, we need to shrink the domain to
\(|z| \le r\) for some (arbitrary but fixed)
\(r\lt 1\text{.}\)
