0
$\begingroup$

I have a fixed-point question similar to the Banach fixed-point theorem.

Let $f\colon \mathbb{R}^n \rightarrow \mathbb{R}^n$ be a given function and let $x^\star \in \mathbb{R}^n$ be a known fixed-point of $f$, i.e. $f(x^\star) = x^\star$.

It is also known that $f$ satisfies: \begin{eqnarray} \sup_{y\neq0} \frac{\|f(x^\star+y) - f(x^\star) \|_2}{\|y\|_2} \leq \alpha. \hspace{1in} (1) \end{eqnarray} for some value of $\alpha \in (0,1)$. Can we conclude that (1) $x^\star$ is the only fixed-point of $f$ and (2) that iterating $f$ will yield $x^\star$.

If (1) was true for all $x$ rather just $x^\star$ then $f$ would be Lipschitz and a contraction so $x^\star$ would be unique (and not needed to be assumed) and iterating $f$ would yield it. However, (1) assumes less about the continuity of $f$ but does bound it to a cone with slope $\alpha<1$ around $x^\star$. Is this enough or where does it fall apart?

$\endgroup$

2 Answers 2

Reset to default
1
$\begingroup$

You can't use the theorem, but you don't need to. Start at any $y_0$ and take $y_{n+1} = f(y_{n})$. What can you say about $\|y_n - x^*\|_2$?

$\endgroup$
0
1
$\begingroup$

You don't even need to consider a sequence. If $z$ is a different fixed point you can take $y=z-x^*$ to obtain the contradiction $$\|y\|=\|z+x^*\|=\|f(z)-f(x^*)\|\le \alpha \|y\|.$$

$\endgroup$
2
  • $\begingroup$ This is a nice way of showing that $x^\star$ is unique, but does this argument also imply that iterating $f$ yields it? $\endgroup$
    – curiousStudent
    Jul 23, 2015 at 13:41
  • $\begingroup$ Yes, as Robert Israel points out, it does. $\endgroup$
    – Jochen
    Jul 23, 2015 at 13:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .