I wrote a proof of Darboux's theorem and would appreciate it if someone could review it.
The idea of the proof is to instead prove the following equivalent reformulation of the theorem:
If $f:(a,b)\to\mathbb{R}$ is differentiable and nonconstant, then for any $x_1,x_2\in (a,b)$, if $f'(x_1)$ and $f'(x_2)$ have opposite sign (the only way for this to make sense is if $f$ is nonconstant), then there is a $c\in(x_1,x_2)$ with $f'(c)=0$.
The equivalence of this statement to Darboux's can be seen by defining $g(x)=f(x)-\mu x$, where $\mu$ is a number between $f'(x_1)$ and $f'(x_2)$, and noting that $g'(x_1)=f'(x_1)-\mu$ and $g'(x_2)=f'(x_2)-\mu$ are of opposite sign. Thus, $g'$ will have a root if and only if there's a $c\in(x_1,x_2)$ with $f'(c)=\mu$.
The proof then proceeds by arguing that because $f'(x_1)$ and $f'(x_2)$ have opposite sign, say $f'(x_1)<0$ and $f'(x_2)>0$, $f$ must decrease near $x_1$ and increase near $x_2$, so the minimum of $f$ cannot occur at the endpoints. At this minimum, the derivative of $f$ will be zero, proving the statement.
Proof: Pick $x_1,x_2\in(a,b)$ arbitrarily and suppose that $f'(x_1)<0<f'(x_2)$ (the idea of the proof for the case $f'(x_1)>0>f'(x_2)$ is identical). Since $f$ is differentiable on $(a,b)$, it must be continuous on $[x_1,x_2]$, so the extreme value theorem guarantees that $f$ has a minimum on $[x_1,x_2]$.
We will now argue that the minimum of $f$ on $[x_1,x_2]$ cannot occur at $x_1$ or $x_2$. We assumed that $f'(x_1)<0$, or $-f'(x_1)>0$, so the definition of $\lim_{x\to x_1}\frac{f(x)-f(x_1)}{x-x_1}=f'(x_1)$ implies the following bound holds in some deleted neighborhood of $x_1$:
$$\left|\frac{f(x)-f(x_1)}{x-x_1}-f'(x_1)\right|<-f'(x_1)$$
This is equivalent to
$$2f'(x_1)<\frac{f(x)-f(x_1)}{x-x_1}<0\tag{$\star$}$$
so if $x$ is sufficiently small and $x>x_1$, then $x-x_1>0$, so multiplying $(\star)$ through by $x-x_1$ implies $f(x)-f(x_1)<0$, that is, $f(x)<f(x_1)$. This shows that thare are values of $f(x)$ smaller than $f(x_1)$, so $f(x_1)$ cannot be the minimum of $f$.
Proceeding similarly for $x_2$, we get the following bound for all $x$ in a deleted neighborhood of $x_2$:
$$0<\frac{f(x)-f(x_2)}{x-x_2}<2f'(x_2)$$
It follows that for sufficiently small $x$ with $x<x_2$, we have $x-x_2<0$, so multiplying throughout this bound by $x-x_2$ gives $f(x)-f(x_2)<0$, that is, $f(x)<f(x_2)$. We conclude that $f(x_2)$ cannot be the minimum either.
It follows from the above that the minimum of $f$ is located in $(x_1,x_2)$, say $c$. Since $f$ is differentiable at $c$, its derivative there must be zero from Fermat's theorem on local extrema. $\blacksquare$
I appreciate any and all feedback.
Best Answer
The proof is generally correct. These are my remarks: