Lipschitz sufficiency condition Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$ is a function and $0 < M < \infty$, prove that $f$ is Lipschitz with constant $M$, if for each $x \in \mathbb{R}$ we have $Df(x) \subset [-M,M]$.
Given a function $f : \mathbb{R} \rightarrow \mathbb{R}$, an extended real number $\lambda$ is called a derived number for $f$ at $x_0$ if there exists a sequence $h_n \rightarrow 0 (h_n \neq 0)$ such that $$\lim_{n \to \infty} \frac{f(x+h_n) - f(x)}{h_n} = \lambda \text{ or } \lambda = Df(x_0).$$
$Df(x_0)$ can have infinite values - consider the example $f(x) = |x|$, where $Df(0) = [-1,1]$
I think mean value theorem cannot be applied here, because differentiability is not guaranteed. Any help with this problem is appreciated. 
 A: For any pair of points $x \leq y$, we have $|f(x)-f(y)|=|\int_x^yf'(t)dt| \leq \int_x^y |f'(t)|dt \leq \int_x^y M dt=M(y-x)=M|x-y|$
I have used the triangle inequality for integrals.
A: Remark: According to the definition of $Df$ given in the question, the function $f(x)=|x|$ should have  $Df(0)=\{-1,1\}$ (a two-element set), not $[-1,1]$ (an interval).
Remark 2. Under the given assumptions  $f$ is continuous, because at a point of discontinuity we would have an infinite derived number. 
We cannot apply the mean value theorem, but we can mimic its proof. Given $a<b$, let $\lambda=\dfrac{f(b)-f(a)}{b-a}$ and consider $g(x)=f(x)-\lambda x$. Since $g(a)=g(b)$, the function $g$ has an interior point of maximum or minimum, say $c$. I'll do the case of maximum. There is a neighborhood of $c$ in which $g(x)\le g(c)$; that is, $$f(x)\le f(c)+\lambda (x-c)\tag1$$ 
By choosing an appropriate  sequence $x_n\to c$, we conclude from (1)  that $Df(c)$ contains a derived number $w$ with $|w|\ge |\lambda|$. Thus, $|\lambda|\le M$, which says exactly 
$$|f(a)-f(b)|\le M|a-b|\tag2$$ 
