I'm trying to prove Darboux's theorem via the method suggested in a question. The question is as follows:
Suppose g is differentiable on $[a,b]$ and $g'(a) < k < g'(b)$. By considering the following:
$$h(x)=\frac{g(x)-g(a)}{x-a} \text{ for } x \ne a \text{ and } h(a)=g'(a)$$
$$f(x)=\frac{g(b)-g(x)}{b-x} \text{ for } x \ne b \text{ and } f(b)=g'(b)$$
show there is a subinterval $[\alpha,\beta] \subseteq [a,b]$ with
$$\frac{g(\beta)-g(\alpha)}{\beta – \alpha} = k$$
(Obiously from here all that's required is a statement of MVT to finish the proof but i can't get the given result). I can see that $h$ and $f$ are continuous on $[a,b]$, and i've tried using the intermediate value theorem, but i don't see how to 'get rid' of the a and b from the definitions in order to reach the final result.
Best Answer
Let $r(\alpha,\beta):=\frac{g(\beta)-g(\alpha)}{\beta-\alpha}$. Notice that $r(a,\beta)<k$ and $r(\alpha,b)>k$ for $\beta$ near $a$ and $\alpha$ near $b$ (say, within $\epsilon$). Pick any parametrization $\alpha(t),\beta(t)$ $(t\in[0,1])$,that goes from $(a,a)$ to $(b,b)$ and $\alpha(t)\neq\beta(t)$ for $t\in(0,1)$ (this avoids the denominator of $r$ from being 0). Clearly $r(\alpha(t),\beta(t))$ is continuous, so by the IVT, there will be a $t_*$ such that $r(\alpha(t_*),\beta(t_*))=k$.