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.

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

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 $F_0(x) =\ldots$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.