Some confusion about the proof for Darboux’s Theorem

continuityderivativesproof-explanationreal-analysis

I'm having some confusion in the proof of Darboux's Theorem. It appears similar questions have been asked before, but I'm still confused by the replies, so I thought I would ask my own.

Here is my proof.
Let $g(x) = f(x) – \gamma x$

Assume $f'(a) < f'(b)$ w.l.o.g.

We know
$f'(a) < \gamma < f'(b)$ by hypothesis.

So,
$f'(a) – \gamma = g'(a) < 0$ and $f'(b) – \gamma = g'(b) > 0$

Since
$g'(a) < 0$ and $g'(b) > 0$, (opposite signs) we know $\exists c$ such that $g'(c) = 0$

That step right there is my confusion. I am basically using the IVT to claim there is a value in between. However, to use the IVT, the function has to be continuous. That is not an assumption in the problem, only that $f$ is continuous. I've found this question asked a couple of times, but the common reply seems to be that the derivative need not be continuous to have the intermediate value property because of Darboux's Theorem. But I am trying to prove Darboux's Theorem! So while I believe that fact, I can't use the theorem within its proof. I cannot seem to justify that step in the event that $g'$ is discontinuous.

I have been told there is another version of the proof combining the MVT and IVT. However, I've found it online in a few places, and I'm having a hard time following it. So I am trying to figure out how to do it this way since I don't understand the other way. Can someone explain to me why I can use the IVT without the derivative being continnuous?

Best Answer

Since $g'(a)<0$, $g(x)<g(a)$ when $x>a$ and $x$ is close enough to $a$. And, since $g'(b)>0$, $g(x)<g(b)$ when $x<b$ and $x$ is close enough to $b$. So, $g$ has a minimum on $[a,b]$ which is attained at some $x_0\in(a,b)$. And so $g'(x_0)=0$, since:

  • if $x>x_0$, then $\frac{g(x)-g(x_0)}{x-x_0}\geqslant0$, and therefore $g'(x_0)=\lim_{x\to x_0^{\,+}}\frac{g(x)-g(x_0)}{x-x_0}\geqslant0$;
  • if $x<x_0$, then $\frac{g(x)-g(x_0)}{x-x_0}\leqslant0$, and therefore $g'(x_0)=\lim_{x\to x_0^{\,-}}\frac{g(x)-g(x_0)}{x-x_0}\leqslant0$.

And $g'(x_0)=0\iff f'(x_0)=\gamma$.

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