Proof of A Continuous Function on a Compact Set Admits a Nondecreasing Modulus of Continuity

Conventions. The order $\le$ and the arithmetic of $\mathbb{R}$ are those of the ordered field of real numbers. Two inequalities may be added: if $a\le b$ and $c\le d$ then $a+c\le b+c\le b+d$ by the compatibility of $\le$ with addition and the transitivity of $\le$.

Step 1: the oscillation of $f$ is bounded. Since $f$ is continuous on $K$, for every $x\in K$ and every positive $\varepsilon\in\mathbb{R}$ there is a positive $\delta\in\mathbb{R}$ such that every $y\in K$ with $d(x,y)<\delta$ satisfies $d_{\mathbb{R}}(f(y),f(x))<\varepsilon$, that is, $|f(y)-f(x)|<\varepsilon$. This is exactly the continuity property required in Extreme Value Theorem on a Compact Subset of a Metric Space, which therefore provides $x_{\min},x_{\max}\in K$ with

$$f(x_{\min})\le f(z)\le f(x_{\max})\qquad\text{for every }z\in K .$$

Put $C=f(x_{\max})-f(x_{\min})$. Let $x,y\in K$. From $f(x_{\min})\le f(y)$ and claim 4 of Elementary Order Arithmetic in an Ordered Field we get $-f(y)\le-f(x_{\min})$, and adding this to $f(x)\le f(x_{\max})$ gives $f(x)-f(y)\le C$. Exchanging $x$ and $y$ gives $f(y)-f(x)\le C$, hence $-C\le f(x)-f(y)$ by claim 4 again, using $-\bigl(f(y)-f(x)\bigr)=f(x)-f(y)$. By claim 6 of Properties of the Absolute Value in an Ordered Field, $|f(x)-f(y)|\le C$.

Step 2: construction of $\omega$. For $t\in T$ put

$$S_{t}=\bigl\{\,|f(x)-f(y)|\ :\ x,y\in K\ \text{with}\ d(x,y)\le t\,\bigr\}\subseteq\mathbb{R}.$$

Since $K$ is nonempty, choose $x_{0}\in K$. Then $d(x_{0},x_{0})=0\le t$ by the metric axioms, and $|f(x_{0})-f(x_{0})|=|0|=0$ by claim 1 of Properties of the Absolute Value in an Ordered Field, so $0\in S_{t}$ and $S_{t}$ is nonempty. By Step 1 the number $C$ is an upper bound for $S_{t}$, so $S_{t}$ is bounded above, and Least Upper Bound Property of the Real Numbers provides a least upper bound of $S_{t}$, which is unique by Uniqueness of the Supremum and of the Infimum. Define $\omega:T\to\mathbb{R}$ by $\omega(t)=\sup S_{t}$.

Step 3: $\omega$ is a modulus of continuity. Since $0\in S_{t}$ and $\omega(t)$ is an upper bound for $S_{t}$, we have $0\le\omega(t)$ for every $t\in T$, which is condition 1 of Modulus of Continuity.

For condition 2, let $\varepsilon\in\mathbb{R}$ be positive. By Heine-Cantor Theorem: Continuity on a Compact Metric Space or a Compact Subset Implies Uniform Continuity, applied with the metric spaces $(M,d)$ and $(\mathbb{R},d_{\mathbb{R}})$ and the compact set $K$, the map $f$ is uniformly continuous on $K$; hence there is a positive $\delta_{0}\in\mathbb{R}$ such that all $x,y\in K$ with $d(x,y)<\delta_{0}$ satisfy $d_{\mathbb{R}}(f(x),f(y))=|f(x)-f(y)|<\varepsilon$. Put $\delta=\delta_{0}\cdot2^{-1}$, which by claim 8 of Elementary Order Arithmetic in an Ordered Field is positive and satisfies $\delta<\delta_{0}$. Let $t\in T$ satisfy $t\le\delta$, and let $x,y\in K$ satisfy $d(x,y)\le t$. Then $d(x,y)\le\delta$ and $\delta<\delta_{0}$, so $d(x,y)<\delta_{0}$ by claim 2 of Elementary Order Arithmetic in an Ordered Field, whence $|f(x)-f(y)|<\varepsilon$ and in particular $|f(x)-f(y)|\le\varepsilon$. Thus $\varepsilon$ is an upper bound for $S_{t}$, and since $\omega(t)$ is the least upper bound of $S_{t}$ we get $\omega(t)\le\varepsilon$. This is condition 2, so $\omega$ is a modulus of continuity.

Step 4: claim 1. Let $s,t\in T$ with $s\le t$. If $x,y\in K$ satisfy $d(x,y)\le s$, then $d(x,y)\le t$ by transitivity; hence $S_{s}\subseteq S_{t}$. Consequently $\omega(t)$, being an upper bound for $S_{t}$, is an upper bound for $S_{s}$, and since $\omega(s)$ is the least upper bound of $S_{s}$ we get $\omega(s)\le\omega(t)$.

Step 5: claim 2. Let $x,y\in K$ and $t\in T$ satisfy $d(x,y)\le t$. Then $|f(x)-f(y)|\in S_{t}$, and $\omega(t)$ is an upper bound for $S_{t}$, so $|f(x)-f(y)|\le\omega(t)$.