I am reading the proof of this theorem from Rudin and I have some questions.
1) Why denominator in (18) is not equal to $0$? I think that if $g(x)=g(y)$ then by Rolle's theorem $g'(\xi)=0$ for some point $\xi$. But for applying this theorem $g$ must be continuous at $[a,b]$. But our $g$ is differentiable in $(a,b)$ $\Rightarrow$ $g$ is continuous in $(a,b)$.
2) Also why denominator in (19) i.e. $g(y)\neq 0$?
Can anyone help with these questions please?
Best Answer
1) If $g(x)=g(y)$ then there exists (by the mean value theorem) a $t$ between $x$ and $y$ with $g'(t)=0$, contradicting line one of the theorem.
2) Again, if $g(y)=0$, then there exists (by the mean value theorem) a $t$ between $a$ and $y$ such that $g'(t)=0$, contradicting line one of the theorem.