Converse of the Banach fixed point theorem

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?

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Of course, a priori, there is no reason why $f$ should be differentiable at $x^*$ –  Dick Palais Feb 9 '11 at 5:43

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 $0$.
An example of such an $f$ is $\frac{x}{1+x^2}$. But what if we demand that $f_n(x)$ converges to $x^*$ uniformly? –  Vít Tuček Feb 9 '11 at 13:32
@Vít Tuček: for instance the iterates of your $f$ do converge uniformly to $0$; $\|f^{n}\|_\infty=f^{n}(1)=o(1)$. –  Pietro Majer Oct 27 '13 at 8:47