Suppose $\ f:\mathbb{R}\to \mathbb{R}\ $ has the following property:
$$ \exists\ c>0\quad \text{ such that }\quad \forall\ x,y\in\mathbb{R},\ \forall\ \delta > 0,\quad \left\lvert x - y \right\rvert < \delta \implies \left\lvert f(x) - f(y) \right\rvert < c\delta\qquad (1). $$
I can easily prove the following:
If $f$ has property $\ (1),\ $ then $\ f\ $ is uniformly continuous ( "Take $\ \delta = \varepsilon/c.$" ), and $f$ is therefore continuous also.
If $\ f\ $ is uniformly continuous, then $\ f\ $ does not necessarily have property $\ (1).\ $ Counter-example: Consider $\ f(x) = \sqrt{x}\ $ on $\ [0,1].$
If $\ f\ $ is Lipschitz continuous, then $\ f\ $ has property $\ (1)\ $ because, assuming $f$ is Lipschitz continuous with Lipschitz constant $K\geq 0$, if $\ \left\lvert x - y \right\rvert < \delta,\ $ then $\ \left\lvert f(x) - f(y) \right\rvert \leq K \left\lvert x - y \right\rvert < K\delta,\ $ and so taking $\ c=K\ $ shows that $\ f\ $ satisfies $\ (1).$
However, I am having trouble proving the last thing on my list: that if $\ f\ $ satisfies property $\ (1)\ $ then $\ f\ $ is Lipschitz continuous (which would show that property $(1)$ is equivalent to Lipschitz continuity). Now, my mind is turning to unbounded variation and Weierstrass function stuff for coutner-examples, or more likely, I'm overcomplicating things and overlooking an obvious affirmative proof. Thoughts?
Suppose that $f$ satisfies $(1)$ and choose a $c$ according to $(1)$. For arbitrary $x,y \in \mathbb{R}$ we set $\delta_n = |x-y| + \frac1n$, where $n\in \mathbb{N}$. Now since $|x-y| < \delta_n$ we get that $$|f(x)-f(y)| < c \delta_n =c\left(|x-y|+\frac1n\right) \quad $$ and thus that $$|f(x)-f(y)| \leq \inf_{n\in \mathbb{N}}c\left(|x-y|+\frac1n\right) = c|x-y|,$$ so we conclude that $f$ is Lipschitz with Lipschitz constant $K=c$.