Proof of Linear Moduli of Continuity

Conventions. The order $\le$ and the arithmetic of $\mathbb{R}$ are those of the ordered field of real numbers. Multiplication by a nonnegative real number preserves $\le$: if $a\le b$ and $0\le\lambda$, then either $a=b$, and the two products are equal, or $a<b$, and then $\lambda a\le\lambda b$ by claim 10 of Elementary Order Arithmetic in an Ordered Field when $0<\lambda$, while $\lambda=0$ makes both products $0$.

Claim 2. Let $s,t\in T$ with $s\le t$. Multiplying by the nonnegative $c$ gives $cs\le ct$, that is, $\omega_{c}(s)\le\omega_{c}(t)$.

Claim 1. We verify the two conditions of Modulus of Continuity.

Condition 1. Let $t\in T$, so $0\le t$. Multiplying by the nonnegative $c$ gives $c\cdot0\le ct$, and $c\cdot0=0$ by Zero Products and Elementary Identities in a Field; hence $0\le\omega_{c}(t)$.

Condition 2. Let $\varepsilon\in\mathbb{R}$ be positive. Suppose first that $c=0$. Then $\omega_{c}(t)=0\cdot t=0$ for every $t\in T$ by Zero Products and Elementary Identities in a Field, so $\omega_{c}(t)\le\varepsilon$ for every $t\in T$, and $\delta=1$ is positive by claim 6 of Elementary Order Arithmetic in an Ordered Field.

Suppose instead that $c\ne0$. Then $0<c$, since $0\le c$ and $0\ne c$. By claim 7 of Elementary Order Arithmetic in an Ordered Field the multiplicative inverse $c^{-1}$ exists and is positive, so $\delta=\varepsilon c^{-1}$ is positive by claim 5 there. Let $t\in T$ satisfy $t\le\delta$. Multiplying by the nonnegative $c$ gives $ct\le c\delta$, and

$$c\delta=c\bigl(\varepsilon c^{-1}\bigr)=\varepsilon\bigl(cc^{-1}\bigr)=\varepsilon$$

by the commutativity and associativity of multiplication and the defining property of the multiplicative inverse. Hence $\omega_{c}(t)\le\varepsilon$.

In both cases a positive $\delta$ has been produced with $\omega_{c}(t)\le\varepsilon$ for every $t\in T$ with $t\le\delta$, which is condition 2. Therefore $\omega_{c}$ is a modulus of continuity.