Dini's Theorem

On a compact set, a monotone sequence of continuous functions converging pointwise converges uniformly — a finite subcover of the open sets

E_n

grabs them all at once. The twin tool of the M-test.

Examples & Exercises

Let $K$ be a compact metric space. If each $f_n : K \to \mathbb{R}$ is continuous and $f_n(x)$ increases monotonically to a continuous limit $f : K \to \mathbb{R}$ at each $x \in K$ , prove that $f_n \to f$ uniformly on $K$ .

E_n = \{x \in K : f(x) - f_n(x) < \varepsilon\}, \qquad E_n \subseteq E_{n+1}, \qquad K = \bigcup_{n=1}^{\infty} E_n

n=8

: worst gap

1/9 \approx 0.1111

— the gap shrinks across the whole interval at once as

n

grows.

Show proof

Fix $\varepsilon > 0$ and set $E_n = \{x \in K : f(x) - f_n(x) < \varepsilon\}$ . Each $E_n$ is open in $K$ because $f - f_n$ is continuous. Since $f_n(x) \le f_{n+1}(x)$ monotonically, $f(x) - f_n(x) \ge f(x) - f_{n+1}(x)$ , so $E_n \subseteq E_{n+1}$ .
Since $f_n(x) \to f(x)$ at each $x$ , there is $n$ with $f(x) - f_n(x) < \varepsilon$ , so $x \in E_n$ . Hence $\{E_n\}$ is an open cover of $K$ .
Compactness of $K$ gives a finite subcover $E_{n_1}, \dots, E_{n_k}$ . With $N = \max\{n_1, \dots, n_k\}$ , monotonicity yields $K \subseteq E_N$ . For $n \ge N$ , $E_N \subseteq E_n$ , so $f(x) - f_n(x) < \varepsilon$ for every $x \in K$ — one $N$ works for all of $K$ , which is uniform convergence. (For the decreasing case, apply the argument to $-f_n$ .)

Show that $g_n(x) = x^n$ converges pointwise on $(0,1]$ to $g(x) = 0$ (on $(0,1)$ , with $g(1) = 1$ ), yet, despite being monotone decreasing, does not converge uniformly.

\sup_{x \in (0,1]} |g(x) - g_n(x)| = 1 \quad \text{for every } n

n=20

: the sup gap is stuck at 1 — the curves pile toward 0 but never hug it uniformly.

Show proof

Each $g_n$ is continuous, and on $0 < x < 1$ we have $g_{n+1}(x) = x^{n+1} \le x^n = g_n(x)$ , so the sequence decreases monotonically. Pointwise: $x^n \to 0$ for $0 < x < 1$ , while $g_n(1) = 1$ at $x = 1$ , so the limit $g$ is discontinuous at $x = 1$ .
On $(0,1)$ , $\sup_{x \in (0,1)} |g(x) - g_n(x)| = \sup_{x \in (0,1)} x^n = 1$ (take $x \to 1^-$ ). So for $\varepsilon = \tfrac{1}{2}$ no $n$ works for all $x$ at once: the convergence is not uniform.
Monotonicity and continuity match Dini's setup, yet it fails — what is missing is compactness of the domain. $(0,1]$ is not closed, and the finite-subcover step of the proof breaks down here. The fact that a uniform limit of continuous functions must be continuous gives the same conclusion: the limit $g$ is discontinuous at $x = 1$ .

Show that $\sum_{n=1}^{\infty} \frac{x^n}{n^2}$ converges uniformly on $[0,1]$ using Dini's theorem.

S_N(x) = \sum_{k=1}^{N} \frac{x^k}{k^2} \uparrow S(x), \qquad 0 \le \frac{x^k}{k^2} \le \frac{1}{k^2}

n=12

, compact mode: a monotone sequence of continuous functions on

[0,1]

— the worst gap

1/13 \approx 0.0769

shrinks uniformly.

Show proof

The partial sums $S_N(x) = \sum_{k=1}^{N} \frac{x^k}{k^2}$ are continuous (finite sums of continuous functions) and increasing on $[0,1]$ , since $S_{N+1}(x) - S_N(x) = \frac{x^{N+1}}{(N+1)^2} \ge 0$ .
For each $x \in [0,1]$ , $0 \le \frac{x^k}{k^2} \le \frac{1}{k^2}$ and $\sum \frac{1}{k^2} = \frac{\pi^2}{6} < \infty$ , so the pointwise limit $S(x) = \sum_{k=1}^{\infty} \frac{x^k}{k^2}$ exists.
$S$ is continuous: the M-test with $M_k = \frac{1}{k^2}$ gives uniform convergence of the series on $[0,1]$ , making $S$ the uniform limit of continuous functions, hence continuous. All of Dini's hypotheses hold (compact $[0,1]$ , continuous monotone sequence, continuous limit), so $S_N \to S$ uniformly on $[0,1]$ .

Part of these learning paths

Examples & Exercises