Lecture. Sequences and series of functions [l1]

Apr 2, 2026

Theorem 1. uniform convergence allows switching limits [br711]

Suppose $f_{n} \to f$ uniformly on a set $E$ in a metric space. Let $x$ be a limit point of $E$, and suppose that $$ \lim_{ t \to x } f_{n}(t) = A_{n} \quad (n = 1, 2, 3, \dots). $$ Then $\{ A_{n} \}$ converges, and $$ \lim_{ t \to x } f(t) = \lim_{ n \to \infty } A_{n}. $$ In other words, the conclusion is that $$ \lim_{ t \to x } \lim_{ n \to \infty } f_{n}(t) = \lim_{ n \to \infty } \lim_{ t \to x } f_{n}(t). $$

Proof. sketch [br711A]

Omitted.

An immediate corollary follows:

Theorem 2. uniform convergence of continuous fns implies continuity of the limit fn [br712]

If $\{ f_{n} \}$ is a sequence of continuous functions on $E$, and if $f_{n} \to f$ uniformly on $E$, then $f$ is continuous on $E$.

Definition 3. equicontinuous [br722]

A family $\mathscr{F}$ of complex functions $f$ defined on a set $E$ in a metric space $X$ is said to be equicontinuous on $E$ if for every $\epsilon > 0$, there exists a $\delta > 0$ such that $$ \lvert f(x) - f(y) \rvert < \epsilon $$ whenever $d(x, y) < \delta, x \in E, y \in E$, and $f \in \mathscr{F}$. Here $d$ denotes the metric of $X$.

It is clear that every member of an equicontinuous family is uniformly continuous.

Theorem 4. pointwise boundedness of $\mathscr{C}$ in a countable set [br723]

If $\{ f_{n} \}$ is a pointwise bounded sequence of complex functions on a countable set $E$, then $\{ f_{n} \}$ has a subsequence $\{ f_{n_{k}} \}$ such that $\{ f_{n_{k}}(x) \}$ converges for every $x \in E$.

Proof. sketch [br723A]

Use diagonal magic.

Let $E = \{ x_{i} \}$. Since $\{ f_{n}(x_{1}) \}$ is a bounded sequence on $\mathbb{C}$, we can find a subsequence that converges to a point of $\mathbb{C}$, denote it by $S_{1} = \{f_{1, k}(x_{1})\}$.

Now, given $\{ f_{1, k}(x_{2}) \}$, a bounded sequence on $\mathbb{C}$, we can again construct a subsequence that converges, denote it by $S_{2} = \{ f_{2, k}(x_{2}) \}$.

Similarly, we do the same for every $x_{i}$, and thus obtain sequences $S_{1}, S_{2}, S_{3}, \dots$, which we represent by the array $$ \begin{align*} &S_{1}: \ f_{1,1} \; f_{1, 2} \; f_{1, 3} \; f_{1, 4} \; \dots \\ &S_{2}: \ f_{2,1} \; f_{2, 2} \; f_{2, 3} \; f_{2, 4} \; \dots \\ &S_{3}: \ f_{3,1} \; f_{3, 2} \; f_{3, 3} \; f_{3, 4} \; \dots \\ &\dots \dots \dots \dots \dots \dots \dots \dots \end{align*} $$ Then, $\{ f_{n,n} \}$ is the desired subsequence of $\{ f_{n} \}$.

Theorem 5. compactness and uniform convergence implies equicontinuity [br724]

If $K$ is a compact metric space, if $f_{n} \in \mathscr{C}(K)$ for $n = 1, 2, 3, \dots$, and if $\{ f_{n} \}$ converges uniformly on $K$, then $\{ f_{n} \}$ is equicontinuous on $K$.

Proof. sketch [br724A]

~~搬砖题~~

$\{ f_{n} \}$ converges uniformly on $K$ $\implies$ for all $\epsilon > 0$, there exists an integer $N > 0$ such that for any $n, m > N$ we have $\lVert f_{n} - f_{m} \rVert < \epsilon$.

Pick any $k > N$. $f_{k} \in \mathscr{C}(K) \implies$ $f_{k}$ continuous on $K$. Now $K$ is compact $\implies$ $f_{k}$ is uniformly continuous on $K$. So there exists $\delta > 0$ such that for any $x, y \in K$ with $d(x, y) < \delta$, we have $\lvert f_{k}(x) - f_{k}(y) \rvert < \epsilon$. Then for any $n > N$, $$ \lvert f_{n}(x) - f_{n}(y) \rvert = \lvert (f_{n}(x) - f_{k}(x)) + (f_{k}(x) - f_{k}(y)) + (f_{k}(y) - f_{n}(y)) \rvert < 3\epsilon. $$ For $1 \leq n \leq N$, $f_{n}$ is uniformly continuous, just take $\delta' = \min_{1 \leq i \leq N}\delta_{i} > 0$, and our final choice of $\delta^*$ is $\min \{ \delta, \delta' \}$, which is positive.

Theorem 6. [br725]

If $K$ is compact, if $f_{n} \in \mathscr{C}(K)$ for $n = 1, 2, 3, \dots$, and if $\{ f_{n} \}$ is pointwise bounded and equicontinuous on $K$, then

$\{ f_{n} \}$ is uniformly bounded on $K$,
$\{ f_{n} \}$ contains a uniformly convergent subsequence.

Proof. sketch [br725A]

$\{ f_{n} \}$ is equicontinuous on $K$ $\implies$ for all $\epsilon > 0$, there exists $\delta > 0$ such that any $x, y \in K$ with $d(x, y) < \delta$ implies $\lvert f_{n}(x) - f_{n}(y) \rvert < \epsilon$ for all $n$.

For all $x \in K$, let $B_{x}$ denote the neighborhood of $x$ with radius $\delta$. Then $\{ B_{x} \}$ is an open cover of $K$. Since $K$ is compact, there exists a finite set of points $\{ p_{1}, \dots, p_{m} \}$ such that $\{ B_{p_{i}} \}$ is a finite subcover of $K$.

Now consider $B_{p_{i}}$. Let $M_{i}$ be an upper bound of $\{ \lvert f_{n}(p_{i}) \rvert \}$. Then for all $x \in B_{p_{i}}$, $|f_{n}(x)| < |f_{n}(p_{i})| + \epsilon \leq M_{i} + \epsilon$ for all $n$. Hence $M_{i} + \epsilon$ is a global upper bound of $\{ f_{n} \}$ on $B_{p_{i}}$.

Take $M = \max_{i} M_{i} + \epsilon < \infty$, then for all $x \in K$, $\lvert f_{n}(x) \rvert < M$ for all $n$. So $\{ f_{n} \}$ is uniformly bounded on $K$.
Choose any countable dense set $E$ on $K$. Then there exists a subsequence $\{ f_{n_{k}} \}$ of $\{ f_{n} \}$ such that $\{ f_{n_{k}} \}$ converges at every point of $E$. Let $\{ g_{k} \}$ denote $\{ f_{n_{k}} \}$.

Use the same $\{ B_{p_{i}} \}$ as in the first part. Pick any $x \in E$, then $x \in B_{p_{i}}$ for some $i$. Then $\{ g_{k}(x) \}$ converges, i.e., there exists a positive integer $N_{i}$ such that any $n, m > N_{i}$ means $\lvert g_{n}(x) - g_{m}(x) \rvert < \epsilon / 3$. Now for any $y \in B_{p_{i}} \cap E$, we have that for any $n, m > N_{i}$, $$\lvert g_{n}(y) - g_{m}(y) \rvert = \lvert \left( g_{n}(y) - g_{n}(x) \right) + \left( g_{n}(x) - g_{m}(x) \right) + \left( g_{m}(x) - g_{m}(y) \right) \rvert < \epsilon$$ Now choose $N = \max_{i} N_{i} < \infty$. For any $x \in E$, $n, m > N$ implies $\lvert g_{n}(x) - g_{m}(x) \rvert < \epsilon$. Hence $\{ g_{k} \}$ is uniformly convergent. Therefore, $\{ f_{n} \}$ contains a uniformly convergent subsequence.