Let $f$ be continuous on $[a,b]$ with $\bar D f \geq 0$ (upper Dini derivative of $f$) on $(a,b)$. Show that $f$ is increasing on $[a,b]$.
Hint: Show this is true for $g$ with $\bar D g \geq \epsilon > 0$ on $[a,b]$. Apply this to the function $g(x) = f(x) + \epsilon x$.
This is question 19 from chapter 6.2 of Royden-Fitzpatrick Analysis 4th edition.
My approach is as follows
- $g$ is continuous as it is the linear combination of 2 continuous functions.
- $\bar D g = \bar D f + \epsilon \geq \epsilon > 0$ which means $g$ is strictly increasing on $(a,b)$.
- $f = g – \epsilon x$ and $\bar D f = \bar D g – \epsilon \geq 0$ implies $f$ is increasing (it is not decreasing) on $(a,b)$.
Does it make sense? Thanks for any help. The question is also related to Continuous function on $[a, b]$ with bounded upper and lower derivatives on $(a, b)$ is Lipschitz.
Best Answer
How do you know that $2$ holds? In fact, this is the gist of the proof, unless I am misreading your question, you need to do a bit of work. (Drawing a picture will help!) First suppose that $\bar D f >0$ on $(a,b)$. If there are $a<c<d<b$ such that $f(c)>f(d)$ then we may choose $f(c)>\mu>f(d)$. Let $S=\{t\in (c,d):f(t)>\mu\}$ and consider $\xi=\sup S.$ Note that $c<\xi<d$. Take an increasing sequence $(t_n)\subseteq (c,d)$ such that $t_n\to \xi.$ Then, $f(t_n)\to f(\xi)$. If $f(\xi)\neq \mu$ then there is a $\mu<\alpha<f(\xi)$. Continuity of $f$ now implies that there is an interval $I=(\xi,\xi+\delta)$ such that $t\in I\Rightarrow f(t)>\alpha>\mu$. But this contradicts the definition of $\xi.$ Thus, $f(\xi)= \mu.$
We have shown that for each $t\in (\xi,d),\ \frac{f(t)-f(\xi)}{t-\xi}\le0$, and we conclude that $ D^+ f(\xi)\le 0$, which is a contradiction. Thus, the claim is true for the strict inequality and $now$ we define $g_{\epsilon}(t)=f(t)+\epsilon t$. It follows that $\bar D g_{\epsilon} >0$ on $(a,b)$ so $g_{\epsilon}$ is non-decreasing there, and as $\epsilon$ is arbitrary, $f$ is also non-decreasing.