Converse of the Banach fixed point theorem - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T09:21:08Z http://mathoverflow.net/feeds/question/54846 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/54846/converse-of-the-banach-fixed-point-theorem Converse of the Banach fixed point theorem Nima 2011-02-09T04:16:00Z 2011-02-23T07:22:16Z <p>Let $f:X \to X$ ($X = R^d$ for some $d$) be a mapping such that $f^n (x) \to x^\ast$ for all $x \in X$ as $n \to \infty$ ($x^\ast$ is unique). Can we say anything about the spectral structure of the gradient matrix $\nabla f (x^\ast)$ ? Do we know that the spectral or the operator norm of this matrix is less than one?</p> http://mathoverflow.net/questions/54846/converse-of-the-banach-fixed-point-theorem/54854#54854 Answer by Dick Palais for Converse of the Banach fixed point theorem Dick Palais 2011-02-09T06:33:13Z 2011-02-09T06:45:29Z <p>I think this is a counter-example to what you are asking. Choose a smooth function $a : R \to R$ with $a(0) = 0$, $a(-x) = a(x)$, $a(x) = x^2$ near $x = 0$, $a(x)$ monotonically increasing for $x > 0$, and $a(x)$ everywhere less than $1$. Then if we let $f(x) := (1 - a(x)) x$ we have $f'(0) = 1$, but $f^n(x)$ converges to $0$ for all $x$. This is because $f^n(x)$ is monotonic and bounded, and so approaches a limit, which must be a fixed point and so can only be $0$.</p>