I tried to prove Darboux's theorem. It is the following theorem:
Let $f: [a,b]\to \mathbb R$ be a differentiable function and let $f'(a) < \alpha < f'(b)$. Then there exists $c \in [a,b]$ with $f'(c) = \alpha$.
Please could somebody check my proof?
Define $g(x) = f(x) – \alpha x$. Then $g$ is continuous and because $[a,b]$ is compact $g$ attains its minimum on $[a,b]$. Let $x_m \in [a,b]$ be such that $g(x_m) \le g(x)$ for all $x\in [a,b]$. If $x_m \in (a,b)$ then $g'(x_m) = 0 = f'(x_m) – \alpha$ which shows the claim.
If $x_m = a$ then $g'(x_m) = g'(a) < 0$. Because $g$ is continuous and $g'(a) < 0$ there exists $\delta>0$ such that if $x \in (a,a+\delta)$ then $g(x) < g(a) = g(x_m)$. But this is a contradiction because $x_m$ is the minimum. If $x_m = b$ then again there is $\delta>0$ such that if $x \in (b-\delta, b)$ then $g(x)<g(b)=g(x_m)$ because $g'(b) > 0$. Again this is a contradiction. It follows that the minimum is attained in the interior $(a,b)$.
Best Answer
You do not need continuity of $g$ here. In fact, this step step is valid even if $g$ is differentiable in $a$ but not continuous anywhere (except at $a$ where it must be continuous because it is differentiable). It is immediate from $g'(a)=\lim_{h\to0^+}\frac{g(a+h)-g(a)}{h}<0$ that for at least some $h>0$ we have $\left|\frac{g(a+h)-g(a)}{h}-g'(a)\right|<\frac12|g'(a)|$ and hence $\frac{g(a+h)-g(a)}{h}<0$ and finally $g(a+h)<g(a)$, contradicting the minimum at $a$.
(Exactly, and in that case we can conclude $g'(x_m)=0$, hence $f'(x_m)=\alpha$ ...)
So, basically, your proof is valid. However, since you added the unnecessary (for that single step) condition that $g$ is continuous I was afraid that you were not completely sure what you really used there.