Proof of Sums, Nonnegative Multiples, Monotonicity and Quadratic Reparametrisation of Moduli of Continuity

Each result cited is universally quantified over the data in its own statement. Write $\sigma$ for the function of claim 1 and $\tau$ for the function of claim 2, and recall that a modulus of continuity satisfies the two conditions of Modulus of Continuity: it is nonnegative on $T$, and for every positive $\varepsilon$ there is a positive $\delta$ such that every $t\in T$ with $t\le\delta$ satisfies a value at most $\varepsilon$.

Claim 1. Let $t\in T$. Since $0\le\omega(t)$ and $0\le\omega'(t)$, claim 2 of Elementary Arithmetic in an Ordered Field gives $0\le\sigma(t)$, which is condition 1.

For condition 2, let $\varepsilon\in\mathbb{R}$ be positive; then $\tfrac{\varepsilon}{2}$ is positive by claim 8 of Elementary Order Arithmetic in an Ordered Field. Choose a positive $\delta$ such that every $t\in T$ with $t\le\delta$ satisfies $\omega(t)\le\tfrac{\varepsilon}{2}$, and a positive $\delta'$ such that every $t\in T$ with $t\le\delta'$ satisfies $\omega'(t)\le\tfrac{\varepsilon}{2}$. Let $\delta''$ be the lesser of $\delta$ and $\delta'$, as provided by claim 9 of Elementary Order Arithmetic in an Ordered Field; it is one of the two, hence positive, and satisfies $\delta''\le\delta$ and $\delta''\le\delta'$. Let $t\in T$ satisfy $t\le\delta''$. By transitivity $t\le\delta$ and $t\le\delta'$, so $\omega(t)\le\tfrac{\varepsilon}{2}$ and $\omega'(t)\le\tfrac{\varepsilon}{2}$, so claim 3 of Elementary Arithmetic in an Ordered Field gives $0\le\tfrac{\varepsilon}{2}-\omega(t)$ and $0\le\tfrac{\varepsilon}{2}-\omega'(t)$; their sum is nonnegative by claim 2 of that lemma, and it equals $\bigl(\tfrac{\varepsilon}{2}+\tfrac{\varepsilon}{2}\bigr)-\bigl(\omega(t)+\omega'(t)\bigr)$, so claim 3 of that lemma applied in the other direction gives

Thus $\sigma$ is a modulus of continuity.

Claim 2. Let $t\in T$. From $0\le c$ and $0\le\omega(t)$, claim 5 of Elementary Arithmetic in an Ordered Field gives $0\le c\,\omega(t)=\tau(t)$, which is condition 1.

For condition 2, let $\varepsilon\in\mathbb{R}$ be positive. If $c=0$ then $\tau(t)=0$ for every $t\in T$, and $\tau(t)\le\varepsilon$ holds for every $t\in T$, so any positive $\delta$ serves. Suppose instead that $c$ is positive. Then $\tfrac{1}{c}$ is positive by claim 7 of Elementary Order Arithmetic in an Ordered Field, so $\tfrac{\varepsilon}{c}$ is positive by claim 5 of that lemma. Choose a positive $\delta$ such that every $t\in T$ with $t\le\delta$ satisfies $\omega(t)\le\tfrac{\varepsilon}{c}$. For such a $t$, multiplying by the nonnegative number $c$ with claim 5 of Elementary Arithmetic in an Ordered Field gives

Thus $\tau$ is a modulus of continuity.

Claim 3. Let $s,t\in T$ satisfy $s\le t$. If $\omega$ and $\omega'$ are nondecreasing then $\omega(s)\le\omega(t)$ and $\omega'(s)\le\omega'(t)$, so $\sigma(s)\le\sigma(t)$ by the same combination of claims 3 and 2 of Elementary Arithmetic in an Ordered Field used in claim 1. If $\omega$ is nondecreasing then $\omega(s)\le\omega(t)$, and multiplying by the nonnegative number $c$ with claim 5 of Elementary Arithmetic in an Ordered Field gives $\tau(s)\le\tau(t)$.

Claim 4. Let $M$ be as in the claim, and write $Q(s)$ and $\omega^{[c]}$ as in the statement; the supremum defining $\omega^{[c]}(s)$ is the least upper bound of $Q(s)$, as recorded in The Real Numbers: Standing Notation and Background §bounds.

The three inequalities. Let $s,t\in T$ satisfy $t^{2}\le c\,s$. Then $\omega(t)\in Q(s)$, and a supremum is an upper bound of its set, so $\omega(t)\le\omega^{[c]}(s)$; this is the displayed inequality of the claim. Taking $t=0$, which is admissible because $0^{2}=0\le c\,s$, gives $0\le\omega(0)\le\omega^{[c]}(s)$, since $\omega$ is nonnegative. Finally $M$ is an upper bound of $Q(s)$ by hypothesis, so $\omega^{[c]}(s)\le M$, a supremum being the least upper bound.

Monotonicity. Let $s,s'\in T$ satisfy $s\le s'$. Multiplying by the nonnegative number $c$ with claim 5 of Elementary Arithmetic in an Ordered Field gives $c\,s\le c\,s'$, so every $t\in T$ with $t^{2}\le c\,s$ satisfies $t^{2}\le c\,s'$ by transitivity; hence every member of $Q(s)$ is a member of $Q(s')$. As $\omega^{[c]}(s')$ is an upper bound of $Q(s')$, it is an upper bound of $Q(s)$, and therefore $\omega^{[c]}(s)\le\omega^{[c]}(s')$.

Condition 2. Let $\varepsilon\in\mathbb{R}$ be positive and choose a positive $\theta$ such that every $t\in T$ with $t\le\theta$ satisfies $\omega(t)\le\varepsilon$. Let $\delta$ be $1$ if $c=0$, and $\tfrac{\theta^{2}}{c}$ if $c$ is positive; in the second case $\theta^{2}$ is positive by claim 5 of Elementary Order Arithmetic in an Ordered Field, $\tfrac{1}{c}$ is positive by claim 7 of that lemma, and so $\delta$ is positive by claim 5 of it again. In either case $c\,\delta\le\theta^{2}$: the product is $0$ when $c=0$, and equals $\theta^{2}$ otherwise. Let $s\in T$ satisfy $s\le\delta$ and let $t\in T$ satisfy $t^{2}\le c\,s$. Multiplying $s\le\delta$ by the nonnegative $c$ with claim 5 of Elementary Arithmetic in an Ordered Field gives $c\,s\le c\,\delta$, so $t^{2}\le\theta^{2}$ by transitivity. Were $\theta<t$, claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field would give $\theta^{2}<t^{2}$, contradicting $t^{2}\le\theta^{2}$; so $t\le\theta$ by trichotomy, and hence $\omega(t)\le\varepsilon$. Thus $\varepsilon$ is an upper bound of $Q(s)$, and $\omega^{[c]}(s)\le\varepsilon$ because a supremum is the least upper bound. So $\omega^{[c]}$ is a modulus of continuity, and it is nondecreasing by the previous paragraph.