How can I prove for certain that there exists no differentiable function $g(x)\colon [-1,1]\rightarrow \mathbb{R}$ with
$$g'(x) = \operatorname{sgn}(x) = \begin{cases} +1, & \text{for $x > 0$} \\
0, & \text{for $x=0$} \\ -1, & \text{for $x<0$} \end{cases} $$
I'm currently jumping to the absolute value function but that's not differentiable at $x = 0$ and that doesn't exactly disprove the existence of others.
Looking at the limits
\begin{align}
\lim_{h\to 0^{+}} g'(h) &=1\\
&= \lim_{h\to 0^{+}} \frac{g(0+h)-g(0)}{h}\\
&\neq \lim_{h\to 0^{-}}\frac{g(0+h)-g(0)}{h}\\
&= -1\\
&= \lim_{h\to 0^{-}}g'(h)\\
&\neq 0\\
&= g'(0)
\end{align}
which implies discontinuity at $h = 0$.
Continuity and differentiability are topics that I still don't quite grasp properly. Can I use the contraposition from Darboux's Theorem for this specific case?
Best Answer
"Can I use the contraposition from Darboux's Theorem for this specific case?"
Yes. Darboux's theorem says that if $f:[a,b]\to\mathbb{R}$ is a differentiable function and suppose $f'(a)<\lambda<f'(b)$, then there is a point $x\in(a,b)$ such that $f'(x)=\lambda$.
The contraposition says that if $f'$ does not satisfy the "intermediate property above", then $f$ cannot be differentiable on $[a,b]$.
The existence of $g$ in your question would contradict this theorem: you may consider for instance $\lambda=\frac12$.