Any function with a modulus of continuity proportional to any preassigned $\epsilon>0$ is Lipschitz-continuous The following is from Courant & John's Introduction to Calculus and Analysis Volume I, p. 43.

Lipschitz-continuity means that the "difference quotient"
$$ \frac{f(x_2)-f(x_1)}{x_2-x_1} $$
formed for any two distinct points of the interval never exceeds a fixed finite value $L$ in absolute value or that the mapping $y=f(x)$ magnifies distances of points on the $x$-axis at most by the factor $L$. Clearly, for a Lipschitz-continous function the expression $\delta(\epsilon)=\epsilon/L$ is a modulus of continuity since $|f(x_2)-f(x_1)|<\epsilon$ for $|x_2-x_1|<\epsilon/L$. Conversely, any function with a modulus of continuity proportional to $\epsilon$, say $\delta(\epsilon)=c\epsilon$, is Lipschitz-continuous, with $L=1/c$.

How can the last claim be established? I've tried to show that $f(x_2)-f(x_1)<(x_2-x_1)/c+\epsilon$ for arbitrary $\epsilon>0$ but without success.
Edit:

Our definition of continuity of function $f(x)$ at $x_0$ requires that for every degree of precision $\epsilon>0$ there exist quantities $\delta>0$ (so-called moduli of continuity) such that $|f(x)-f(x_0)|<\epsilon$ for all $x$ in the domain of $f$ for which $|x-x_0|<\delta$.

 A: This is mostly an exercise in syntactic manipulation; which fact seems to have been obscured by specifics. For clarity let us first give some definitions, following Courant-John.
Let $X$ and $Y$ be metric spaces, $f:X\to Y$ be a function. Then a function $\mathfrak{m}:X\times \mathbb{R}_{>0}\to\mathbb{R}_{>0}$ is called a pointwise modulus of continuity for $f$ if
$$\forall x_0\in X,\forall \epsilon\in\mathbb{R}_{>0},\forall x\in X: d_X(x,x_0)<\mathfrak{m}(x_0,\epsilon)\implies d_Y(f(x),f(x_0))<\epsilon.$$
Similarly a function $\mathfrak{m}:\mathbb{R}_{>0}\to\mathbb{R}_{>0}$ is called a uniform modulus of continuity for $f$ if
$$\forall \epsilon\in\mathbb{R}_{>0},\forall x_1,x_2\in X: d_X(x_1,x_2)<\mathfrak{m}(\epsilon)\implies d_Y(f(x_1),f(x_2))<\epsilon.$$
It's clear that $f$ is continuous iff it has a pointwise modulus of continuity and it is uniformly continuous iff it has a uniform modulus of continuity. Denote by $\operatorname{PMC}(f)$ and $\operatorname{UMC}(f)$ the sets of all pointwise and uniform moduli of continuity for $f$, respectively. It's straightforward that when $f$ is continuous and uniformly continuous, respectively,
$$\left[\overline{\mathfrak{m}}:(x,\epsilon)\mapsto \sup_{\mathfrak{m}\in\operatorname{PMC}(f)}\mathfrak{m}(x,\epsilon)\right]\in\operatorname{PMC}(f)\quad\text{ and } \quad\left[\overline{\mathfrak{m}}:\epsilon\mapsto \sup_{\mathfrak{m}\in\operatorname{UMC}(f)}\mathfrak{m}(\epsilon)\right]\in\operatorname{UMC}(f).$$

Claim: Let $X$ and $Y$ be metric spaces, $f:X\to Y$ and $\mathfrak{m}:\mathbb{R}_{>0}\to\mathbb{R}_{>0}$ be two functions. Suppose $\mathfrak{m}$ is strictly increasing (hence is a bijection) with inverse $\omega=\mathfrak{m}^{-1}:\mathbb{R}_{>0}\to\mathbb{R}_{>0}$ (thus $\mathfrak{m}(\epsilon)=\eta \iff \epsilon=\omega(\eta)$ and $\omega$ is also strictly increasing). Then the following are equivalent:

*

*$\forall x_1,x_2\in X: d_Y(f(x_1),f(x_2))\leq \omega(d_X(x_1,x_2))$.

*$\mathfrak{m}\in\operatorname{UMC}(f)$.

(Note that we haven't defined $\omega(0)$; for the inequality in 1. to make sense it cannot be negative. Let's put $\mathfrak{m}(0)=0=\omega(0)$.)
Proof: (1.$\Rightarrow$2.) Let $\epsilon\in\mathbb{R}_{>0}$ and suppose $x_1,x_2\in X$ are such that $d_X(x_1,x_2)<\mathfrak{m}(\epsilon)$, whence $\omega(d_X(x_1,x_2))<\omega\circ\mathfrak{m}(\epsilon)=\epsilon$ as $\omega$ is strictly increasing. Then $d_Y(f(x_1),f(x_2))\leq \omega(d_X(x_1,x_2))<\epsilon$.
(1. $\Leftarrow$2.) Conversely, suppose $\mathfrak{m}$ is a uniform modulus of continuity for $f$. Let $x_1,x_2\in X$ be such that $x_1\neq x_2$. Then $\eta^\ast=d_X(x_1,x_2)\in\mathbb{R}_{>0}$. Then for any $\epsilon\in]\omega(\eta^\ast),\infty[$ we have $d_X(x_1,x_2)=\eta^\ast=\mathfrak{m}\circ \omega(\eta^\ast)<\mathfrak{m}(\epsilon)$. As $\mathfrak{m}$ is a uniform modulus of continuity for $f$, this implies $d_Y(f(x_1),f(x_2))<\epsilon$, whence
$$d_Y(f(x_1),f(x_2))\leq \inf(]\omega(\eta^\ast),\infty[)=\omega(\eta^\ast)=\omega(d_X(x_1,x_2)).$$

Let us now apply this to Hölder functions. For any $\theta\in\mathbb{R}_{>0}$, a function $f:X\to Y$ is called (globally) $\theta$-Hölder if for some $C\in\mathbb{R}_{>0}$:
$$\forall x_1,x_2\in X: d_Y(f(x_1),f(x_2))\leq C d_X(x_1,x_2)^\theta. \quad\quad\quad(\ast)$$
Note that for any $(C,\theta)\in\mathbb{R}_{>0}^2,$
$$\omega^{(C,\theta)}:\mathbb{R}_{\geq0}\to \mathbb{R}_{\geq0}, \eta\mapsto C\eta^\theta$$
sends $0$ to $0$, and is strictly increasing with inverse
$$\mathfrak{m}^{(C,\theta)}:\epsilon\mapsto\left(\omega^{(C,\theta)}\right)^{-1}(\epsilon)=\left(\dfrac{\epsilon}{C}\right)^{1/\theta},$$
thus by the statement in the earlier section $\mathfrak{m}^{(C,\theta)}$ is a uniform modulus of continuity for any function $f$ satisfying the (global) $\theta$-Hölder estimate $(\ast)$. (See https://www.desmos.com/calculator/cnvmwhr7ug for a humble graph of the family $\mathfrak{m}^{(\bullet,\bullet)}$.)

Adapting the above to the pointwise moduli of continuity is straightforward.
Finally note that often instead of an $\mathfrak{m}$ as above an $\omega$ as above seems to be what is referred to as a modulus of continuity. It should also be noted that often a modulus of continuity is by definition taken to be increasing, and its behavior near $0$ is further specified; e.g. Stein in Singular Integrals and Differentiability Properties of Functions (p.175) defines a modulus of continuity as a function $\omega:\mathbb{R}_{>0}\to\mathbb{R}_{>0}$ such that

*

*$\omega$ is strictly increasing,

*$\frac{\omega(\eta)}{\eta}$ is increasing as $\eta\to 0$, and

*$\omega(2\eta)\leq c\omega(\eta)$.

(There are other moduli of continuity defined in the book, and elsewhere.)
