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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! :)

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This might be a way: Show the result is true assuming $\overline D f>0$ on $[a,b]$. Then, assuming $f$ is not increasing, there is an $\epsilon>0$ such that $f_\epsilon(x)=f(x)+\epsilon x$ is not increasing. But $\overline D f_\epsilon=\overline D f+\epsilon>0$ on $[a,b]$. – David Mitra Jan 20 '13 at 18:53
up vote 1 down vote accepted

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



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.

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N.S., I just corrected a small typo error. Everything looks alright now. :) – Haskell Curry Jan 20 '13 at 18:33
@HaskellCurry Ty. – N. S. Jan 20 '13 at 18:34
Thank you for the help, but I am not seeing how to conclude that $\overline{D}g(e) = 0$ if $g(e)$ is, for example, an absolute max. This upper derivative is certainly not positive since $g(e)$ is an absolute max, but how can we conclude it is $0$? – Nicole Jan 20 '13 at 22:04

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.

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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 $v \to b$.

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