Equicontinuity + pointwise boundedness implies uniform boundedness on compact sets

Let $(K,d)$ be a compact metric space and let $\mathcal{F}\subseteq C(K,\mathbb{R})$ be a family of continuous functions . Assume:

• $\mathcal{F}$ is equicontinuous on $K$, and
• $\mathcal{F}$ is pointwise bounded on $K$ (for each $x\in K$, $\sup_{f\in\mathcal{F}}|f(x)|<\infty$).

Lemma: Then $\mathcal{F}$ is uniformly bounded on $K$; i.e., there exists $M<\infty$ such that $|f(x)|\le M\quad \text{for all } f\in\mathcal{F},\ x\in K.$

This lemma is a standard step in the proof of the Arzelà–Ascoli theorem .

Proof sketch: Apply equicontinuity with $\varepsilon=1$: for each $x\in K$ there exists $\delta_x>0$ such that $d(x,y)<\delta_x \implies |f(y)-f(x)|<1 \quad \text{for all } f\in\mathcal{F}.$ By pointwise boundedness at $x$, choose $M_x$ with $|f(x)|\le M_x$ for all $f\in\mathcal{F}$. Then for all $y\in $B$ (x,\delta_x)$ and all $f\in\mathcal{F}$, $|f(y)|\le |f(y)-f(x)|+|f(x)|<1+M_x.$ The balls $\{B(x,\delta_x)\}_{x\in K}$ form an open cover of $K$. Compactness gives a finite subcover $B(x_i,\delta_{x_i})$ for $i=1,\dots,N$. Let $M=\max_{1\le i\le N} (1+M_{x_i}),$ which is finite. Then $|f(y)|\le M$ for all $y\in K$ and all $f\in\mathcal{F}$.