Are there any differentiable not $C^1$ function $f: [0,1] \to \mathbb{R}$, $f'(0)<f'(1)$ such that there exists $c \in (f'(0),f'(1))$ that $f'$ doesn't reach value $c$?
Classical example of differentiable but not $C^1$ function doesn't work here.
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1I do not understand: any differentiable function has the Darboux property. Can you explain better your question? – Siminore Aug 25 '15 at 19:55
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He's talking about the Darboux property on $f'(x)$, he explained it correctly – Smurf Aug 25 '15 at 19:56
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9No, every derivative has the intermediate value property/Darboux property. – Daniel Fischer Aug 25 '15 at 19:57
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2Look here: https://en.wikipedia.org/wiki/Darboux%27s_theorem_(analysis) – Siminore Aug 25 '15 at 19:57
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Every derivative has the intermediate value property/Darboux property. That follows from the mean value theorem (and the intermediate value theorem for continuous functions):
Let $I\subset \mathbb{R}$ be an interval and $g\colon I \to \mathbb{R}$ differentiable. Let $a < b$ such that $[a,b] \subset I$. If $g'(a) = g'(b)$, there is nothing to prove. Otherwise, we may assume that $g'(a) < g'(b)$ [replace $g$ with $-g$ if necessary]. Choose $c \in (g'(a),g'(b))$. Consider the quotient
$$m = \frac{g(b) - g(a)}{b-a}.$$
If $m = c$, the mean value theorem asserts the existence of a $t\in (a,b)$ with $g'(t) = c$. If $m < c$, then define
$$h(x) = \begin{cases}\quad g'(b) &, x = b \\ \dfrac{g(b) - g(x)}{b-x} &, a \leqslant x < b,\end{cases}$$
and if $m > c$, define
$$h(x) = \begin{cases}\quad g'(a) &, x = a\\ \dfrac{g(x) - g(a)}{x-a} &, a < x \leqslant b.\end{cases}$$
Then $h$ is continuous with $h(a) < c < h(b)$, and the intermediate value theorem asserts the existence of an $s \in (a,b)$ with $h(s) = c$. Then the mean value theorem asserts the existence of a $t \in (a,s)$ resp. $t \in (s,b)$ with $g'(t) = c$.
Daniel Fischer
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