# Prove the Lipschitz constant must be less than 1.

I've been set this problem recently and I'm having a lot of trouble with it. Any help would be much appreciated!

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a function with continuous derivatives of all orders and suppose that, for some $x_{0}\in \mathbb{R}$ the derivative $f'(x_{0})$ is non-zero. Write $f(x_{0})=y_{0}$.

(a) Show there exists an open interval $D$ containing $x_{0}$ such that $f'(x)\neq 0$ for all $x\in D$.

(b) Define $F_{y_{0}}:D\rightarrow \mathbb{R}$ by $F_{y_{0}}(x)=x-\frac{f(x)-y_{0}}{f'(x)}$. Show that $F_{y_{0}}$ is a Lipschitz function with Lipschitz constant less than 1.

I've managed to prove the first part, but I'm having trouble with (b). I first used the Mean Value Theorem to get $|F_{y_{0}}(y)-F_{y_{0}}(x)| = | y-x | | F'_{y_{0}}(c)|$. I then found that $F'_{y_{0}}(c)=\frac{(f(c)-y_{0})f''(c)}{(f'(c))^2}$, but I'm unsure where to go from here.

• If $D$ is small enough, then $|f(c)-y_0|$ is very small which leads to small $|F_{y_0}'(c)|$. – Jochen Aug 21 '15 at 10:36
• Yes I understand that, but why must it be less than 1? I tried to bound it but that doesn't seem to work. – jackwo Aug 21 '15 at 10:43
• You can make the Lipschitz constant as small as you like. Strictly smaller than $1$ is needed for the Banach contraction principle. – Jochen Aug 21 '15 at 11:53
• How do you show that the Lipschitz constant can be made as small as you like? – jackwo Aug 21 '15 at 16:47

## 1 Answer

Hope this rough idea can be useful. You can always assume that $x_0=y_0=0$. Near $x=0$, $f(x) = f'(0)x+o(x)$. Hence, near $x=0$, $$F_0(x) = x - \frac{f'(0)x+o(x)}{f'(x)}.$$ But $f'(x)=f'(0)+O(x)$ near $x=0$. Hence $F_0(x) =\ldots$