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

Question

I am trying to show that if $f$ is continuous on the interval $[a,b]$ and its upper derivative $\overline{D}f$ is such that $ \overline{D}f \geq 0$ on $(a,b)$, then $f$ is increasing on the entire interval. Here $\overline{D}f$ is defined by $$ \overline{D}f(x) = \lim\limits_{h \to 0} \sup\limits_{h, 0 < |t| \leq h} \frac{f(x+t) - f(x)}{t} $$

I am not sure where to begin, though. Letting $x,y \in [a,b]$ be such that $x \leq y$, suppose for contradiction that $f(x) > f(y)$, then continuity of $f$ means that there is some neighbourhood of $y$ such that $f$ takes on values strictly less than $f(x)$ on this neighbourhood. Now I think I would like to use this neighbourhood to argue that the upper derivative at $y$ is then negative, but I cannot see how to complete this argument.

Any help is appreciated! :)

Answer 1

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.

Answer 2

This is based on David's comment above.

Choose $u,v, \epsilon$ such that $ a < u < v < b$ and $ \epsilon > 0 $.

Let $$ S = \{ x \in [u,v] : f(x) + \epsilon x \geq f(u) + \epsilon u \}.$$ $S$ is not empty as $ u \in S$ and $S$ is closed as $f$ is continuous.

Let $\sup S = t$, $t$ is in $S$ as $S$ is closed. We will show $t=v$.

If $ t < v$, then define for $ \delta \in (0, v- t] $, $g(\delta) = \sup_{0 < h \leq \delta} \dfrac{f(t+h) - f(t)}{h}$.

Since $g(\delta)$ decreases to $\overline{D}f(t) \geq 0$ as $ \delta \to 0^+$, we must have $ g(\delta) > -\epsilon $ for sufficiently small positive $\delta$, and this means, the set $g(\delta)$ is a supremum of, must contain an element greater than $-\epsilon$. Hence there is a $t_1 \in (t,v]$ with $\dfrac{f(t_1) - f(t)}{t_1 - t} > -\epsilon$, we have $f(t_1) + \epsilon t_1 > f(t) + \epsilon t \geq f(u) + \epsilon u$, i.e., $t_1 \in S$. This contradiction implies $v = \sup S$ and $f(v) \geq f(u ) + \epsilon (v - u)$. Letting $\epsilon \to 0$, we have $f(v) \geq f(u)$. Hence $f$ is non-decreasing on $(a,b)$. This can be extended to all of $[a,b]$ by keeping $v$ fixed and letting $u \to a$ and keeping $u$ fixed and letting $v \to b$.

Answer 3

Let $ f:[a,b]→\Bbb{R}$ be a continuous function such that one of Dini's one-sided derivatives exists and positive. Then it can be shown $f$ is monotonically non-decreasing on $[a,b]$.

Assume, $D^{+} f(t):=lim sup_{\delta \to 0+} {\frac{f(t+\delta )−f(t)}{\delta}}$ exists and positive i.e $D^+{f(x)} >0, \space \forall x\in [a, b]$

Claim: $f$ is monotonically non-decreasing on $[a,b]$.

Suppose, $f$ fails to be non-decreasing on $[a,b]$.

Then, $\exists c, d $ with $a\le c<d\le b$ such that $f(c) >f(d) $

Let, $y\in (f(d), f(c)) $

Since, $f\in C[a, b]$ , by Intermediate Value Theorem, $\exists t \in [c, d]$ such that $ f(t) =y$

Then, the set $\{x : f(x) =y\}\cap [c, d] $ is non empty.

Suppose, $x_o=Sup\{x : f(x) =y\}\cap [c, d] \}$

Now, $f(d) <y$ and $f$ is continuous implies $x_o<d$

Thus, $f(x) <y, \forall x\in (x_o, d]$

And the set $\{x : f(x) =y \}$ is closed as $f$ is continuous, so $f(x_o) =y$

And, this implies that $D^+ { f(x_o) } \le 0$

Hence, a contradiction.

Hence, if $f$ is continuous and $D^+{f(x)} > 0,\space \forall x\in [a, b] $ then $f$ must be non -decreasing on $[a, b]$ $($ in fact, $f$ is strictly increasing on $[a, b]) $