A function with a non-negative upper derivative must be increasing?

Solution 1:

The statement is not true. The Weierstrass function has upper derivative greater than zero everywhere, it is continuous, and it is not an increasing function. This question comes as an error in Royden, and should read that the lower derivative of $f$ is greater than or equal to zero, per this errata.

http://www2.math.umd.edu/~pmf/docs/Real%20Analysis.pdf

Solution 2:

Probably not the best approach, but here is an idea: show taht MVT holds in this case:

Lemma Let $[c,d]$ be a subinterval of $[a,b]$. Then there exists a point $e \in [c,d]$ so that

$$\frac{f(d)-f(c)}{d-c}=\overline{D}f(e)$$

Proof:

Let $g(x)=f(x)-\frac{f(d)-f(c)}{d-c}(x-c) \,.$

Then $g$ is continuous on $[c,d]$ and hence it attains an absolute max and an absolute minimum. Since $g(c)=g(d)$, then either $g$ is constant, or one of them is attatined at some point $e \in (c,d)$.

In the first case you can prove that $\overline{D}g=0$ on $[c,d]$, otherwise it is easy to conclude that $\overline{D}g(e)=0$.

Your claim follows immediately from here.