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Any function with a modulus of continuity proportional to any preassigned $\epsilon>0$ is Lipschitz-continuous

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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 $|x-x_0|<\delta$.


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