Use Darboux’s Theorem to prove a function

Question

Deﬁne a function by \begin{equation*} g(x) = \begin{cases} x,&x<1\\ x-1, &x\geq 1. \end{cases} \end{equation*} Prove that there does not exist a function $f(x)$ that is diﬀerentiable on all of $\mathbb{R}$ such that $f(x) = g(x)$. And my instructor requires us to use Darboux’s Theorem to prove this question.

Could someone help me to figure it out? I'd appreciate it!

"f(x)=g(x)"? Why don't you just say "g(x) is not differentiable on all of $\Bbb R$"? — DHMO, Apr 13 '17 at 17:02

score 1 · Accepted Answer · answered Apr 13 '17 at 17:06

1

Assume, for a contradiction, there exists such a function $f(x)$ everywhere-differentiable on $\Bbb R$ satisfying $f'(x) = g(x)$.

Since $g(x)$ is the derivative of $f(x)$, Darboux's theorem states that the image of an interval under $g$ is also an interval.

However, $g([0.5,1]) = [0,0] \cup [0.5,1)$ which is not an interval.

Hence, a contradiction is reached.

answered Apr 13 '17 at 17:06

DHMO

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could you tell me what is the meaning of g([0.5,1])=[0,0]∪[0.5,1)?Thanks – YUANFENG Apr 15 '17 at 00:12
@YUANFENG The image of $[0.5,1]$ under $g$ is ${0} \cup [0.5,1)$ – DHMO Apr 15 '17 at 07:10
okay!I got it!Thanks – YUANFENG Apr 17 '17 at 17:30

Use Darboux’s Theorem to prove a function

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