Dini's Theorem

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 $x \in K$.

Explicitly, increasing convergence gives $f_n(x) \le f(x)$, hence $0 \le f(x) - f_n(x)$, so $|f(x) - f_n(x)| = f(x) - f_n(x) < \varepsilon$ — one $N$ works for all of $K$, which is uniform convergence.

(For the decreasing case, apply the argument to $-f_n$.)

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.

Separate the two cases.

(i) On $(0,1)$ the limit $g \equiv 0$ is continuous, each $g_n$ is continuous and monotone decreasing, so the only failed Dini hypothesis is compactness ($(0,1)$ is not compact) — the $\sup = 1$ computation above is exactly this example.

(ii) On $(0,1]$ compactness also fails, but the limit $g$ is discontinuous at $x = 1$, so the continuous-limit hypothesis fails as well.

Hence the $(0,1]$ example alone does not isolate compactness.

The fact that a uniform limit of continuous functions must be continuous also gives non-uniformity in case (ii).

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.

Show $S$ is continuous without the M-test (the M-test already gives uniform convergence, so citing it would be circular).

For $x, x_0 \in [0,1]$, the tail estimate $0 \le S(x) - S_N(x) = \sum_{k>N} \frac{x^k}{k^2} \le \sum_{k>N} \frac{1}{k^2}$ gives $|S(x) - S(x_0)| \le |S_N(x) - S_N(x_0)| + 2\sum_{k>N} \frac{1}{k^2}$.

Given $\varepsilon > 0$, choose $N$ so the tail is below $\varepsilon/3$, then use continuity of $S_N$ to make $|S_N(x) - S_N(x_0)| < \varepsilon/3$ for $|x - x_0|$ small enough; then $|S(x) - S(x_0)| < \varepsilon$.

Hence $S$ is continuous — the tail estimate is used only to verify continuity of the limit, not to assume uniform convergence.

Now apply Dini's theorem: $[0,1]$ is compact, $S_N$ are continuous and increase pointwise to $S$, and $S$ is continuous as just shown. Hence $S_N \to S$ uniformly on $[0,1]$.