Let $f : I\subset\mathbb{R} → \mathbb{C}$ be a continuous function on the closed interval $I$. A modulus of continuity of $f$ is any real-extended valued function $\omega: [0, ∞] → [0, ∞]$, vanishing at $0$ and continuous at $0$, that is $\lim_{t\to0}\omega(t)=\omega(0)=0.$
We say $f$ admits $\omega$ as modulus of continuity if and only if, $$\forall x,x'\in I: \|f(x)-f(x')\|\leq\omega(|x-x'|).$$ Here is my question: Does any function $f$ as above admit a modulus of continuity?
Yes, you can define it as $$\omega (t) := \sup_{0 < |x - x'|<t} \frac{\|f(x)-f(x')\|}{|x-x'|}$$ for $t>0$ and $\omega(0)=0$. Since a continuous function in the closed interval is uniformly continuous, $\omega$ will be continuous at $0$.