Dini's Theorem

Theorem
Let $K \subseteq \R$ be compact.

Let $\sequence {f_n}$ be a sequence of continuous real functions defined on $K$.

Let $\sequence {f_n}$ converge pointwise to a continuous function $f$.

Suppose that:
 * $\forall x \in K : \sequence {\map {f_n} x}$ is monotone.

Then the convergence of $\sequence {f_n}$ to $f$ is uniform.

Proof
Let:
 * $d_n := \size {f_n - f}$

We have:
 * $(1): \quad \sequence {d_n}$ converge pointwise to $0$

since $\sequence {f_n}$ converges pointwise to $f$.

We also have:
 * $(2): \quad \forall x \in K : \sequence {\map {d_n} x}$ is monotonically decreasing

since $\sequence {\map {f_n} x}$ is monotone.

Let $\epsilon > 0$.

Consider $\sequence {A_n}$ defined by:
 * $A_n := \set {x \in K : \map {d_n} x < \epsilon}$

As $d_n$ are continuous, $A_n$ are open sets.

In view of $(1)$, we have:
 * $K = \bigcup A_n$

Since $K$ is compact, there exists a finite subcover:
 * $K = A_{n_1} \cup \cdots \cup A_{n_m}$

for some $m \in \N_{>0}$.

Let $N := \max \set {n_1, \ldots, n_m}$ be the maximum element among $n_1, \ldots, n_m$.

Then, since $\sequence {A_n}$ is increasing in view of $(2)$, we have:
 * $\forall n \ge N : K = A_n$

That is:
 * $\forall n \ge N : \forall x \in K :\quad \map {d_n} x = \size{\map {f_n} x - \map f x} < \epsilon$