Ok.. so I am trying to prove that: If $f:[0,1]\to [0,1]$ is an orientation preserving diffeomorphism that does not have fixed points at $(0,1)$ and $f'(0)\neq1, f'(1)\neq1$, then $f$ is $C^1$-structurally stable (i.e. there exists $V$ an $C^1$ neighborhood of $f$, such that for all $g$ in $V$, $g$ and $f$ are topologically conjugate).
We use this theorem: Any two orientation preserving homeomorphisms of $[0,1]$ that do not have fixed points in $(0,1)$, are topologically conjugate.
So all there is left to prove is that there is a neigborhood of $f$ where any $g$ there satisfies the hypothesis of the previous theorem.
This is what they do:
They assert that there exists $U_1$ neighborhood (from now on, neighborhood means $C^1-$neighborhood when talking about $f$), $W_0$ and $W_1$ neighborhoods of $0$ and $1$ respectively such that any $g$ in $U_1$ is also orientation preserving and has unique fixed points at $W_0$ and $W_1$, which are $1$ and $0$ respectivelly.
Since $[0,1]-W_0-W_1$ is compact, $\mid f(x)-x \mid$ has a positive minimum at $[0,1]-W_0-W_1$, hence there exists a neighborhood $U_2$ of $f$, such that any element there has no fixed points at $[0,1]-W_0-W_1$
Then they just take the intersection of $U_1$ and $U_2$.
Ok, so my question is regarding 1.
Why is this true? When do you use the derivative hypothesis?
Thanks in advanced.
Assume, for definiteness, that $f'(0) > 1$. Since $f$ is $C^1$, there are $\varepsilon > 0$ and $K \in (0, 1)$ such that $$ f'(x) \ge 1 + \varepsilon \quad \text{ for all } x \in [0, K]. $$
For any $g$ whose distance from $f$ in the $C^1$-metric is less than $\varepsilon/2$ we have $$ g'(x) \ge 1 + \frac{\varepsilon}{2} \quad \text{ for all } x \in [0, K]. $$ An application of the MVT gives that $$ g(x) > x \text{ for all } x \in (0, K]. $$ So, you can take $[0, K)$ as a neighborhood $W_0$ of $0$. The case of $f'(0) < 1$ is considered in an analogous way.