If $g$ is continous on $[a,b]$ with bounded upper and lower derivatives on $(a,b)$, will $g$ be Lipschitz? By the Mean Value Theorem from ordinary calculus one knows that, if $f$ is continous on $[a,b]$ and differentiable on $(a,b)$ with bounded derivatives, then $f$ has to be Lipschitz on $[a,b]$.
Now I ask a more general question: If $g$ is continous on $[a,b]$ with bounded upper and lower derivatives on $(a,b)$, will $g$ be Lipschitz? If not, what would be a counterexample?
I really don't know how to proceed. One idea I had was approximating $g$ with a piecewise linear function $\phi$, but I don't know if this gets us anywhere.
 A: Suppose that 
$$M = 1 + \sup_{x\in (a,b)}\{|\overline{D}f(x)|,|\underline{D}f(x)|\}$$
and $f$ is continuous.
Fact.
Pick $[c,d]\subseteq (a,b)$. Then
$$|f(d) - f(c)|\leq M(d-c)$$
Proof. Define 
$$S = \big\{x\in [c,d]\,\big|\,|f(x) - f(c)|\leq M(x-c)\big\}$$
Clearly $S$ is closed as $f$ is continuous and $c\in S$. Note that for every $x_0\in [c,d]$ there exists $\delta_{x_0}>0$ such that
$$\bigg|\frac{f(x)-f(x_0)}{x-x_0}\bigg|< M$$
for all $x\in (x_0-\delta,x_0+\delta)$. We can rewrite it to
$$\big|f(x)-f(x_0)\big| \leq M\cdot |x-x_0|$$
Pick $u = \sup S$. If $u < d$, then
$$\bigg|f\left(u+\frac{\delta_{u}}{2}\right) - f(c)\bigg| \leq \bigg|f\left(u+\frac{\delta_{u}}{2}\right) - f(u)\bigg| + \bigg|f(u) - f(c)\bigg|\leq M\cdot \frac{\delta_{u}}{2} + M\cdot (u - c) = M\left(u+\frac{\delta_{u}}{2} - c\right)$$
Then $u+\frac{\delta_{u}}{2}\in S$. This is a contradiction with $u = \sup S$. This implies $u = d$ and hence
$$|f(d) - f(c)|\leq M\cdot (d-c)$$
Now Fact implies that $f$ is Lipschitz.
Remark.
Pick a sequence $\{x_n\}_{n\in \mathbb{N}}$ of elements of $S$ such that $\lim x_n = x \in [c,d]$. Then
$$|f(x_n) - f(c)| \leq M\cdot (x_n-c)$$
By taking limits of both sides of these inequality and by continuity of $f$ we deduce that
$$|f(x) - f(c)|\leq M\cdot (x-c)$$
Hence $x\in S$ and this shows that $S$ is closed.
A: It suffices to prove the following:

Let $f:[c, d] \to \Bbb R$ be a continuous function with an upper derivative that is bounded above:
  $$
\forall x \in [c, d]: \overline Df(x) \le L  \, .
$$
  Then 
  $$
 f(d) -f(c) \le L(d-c) \, .
$$

If both the upper derivative and the lower derivative are bounded
$$
 -L \le \underline D f(x) \le \overline D f(x) \le L
$$
then the above can be applied to $f$ and $-f$, and it follows that
$$
  |f(d) - f(c) | \le L (d-c)
$$
on any subinterval $[c, d] \subset (a,b)$, so that $f$ is Lipschitz continuous.
The proof of the above claim resembles that of the mean-value theorem (and Rolle's theorem): Consider the function
$$
g(x) = f(x) - (x-c)\frac{f(d)-f(c)}{d-c} \, .
$$
Then $g(c) = g(d)$, so that $g$ attains its maximum at a point $\xi \in (c, d]$. It follows that
$$
 \overline Dg(\xi) \ge \lim_{\delta \to 0} \sup \bigl\{ \frac{g(\xi +h)-g(\xi)}{h} \bigm\vert -\delta < h < 0 \bigr\} 
 \ge 0
$$
and therefore
$$
 0 \le \overline Dg(\xi) =  \overline Df(\xi) - \frac{f(d)-f(c)}{d-c}
\le L - \frac{f(d)-f(c)}{d-c} \\
\implies f(d) - f(c) \le L(d-c) \, .
$$
