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Hi

I have a function $F:\mathbb{R} ^ n\rightarrow \mathbb{R}^n$ for which I know there exist a unique fixed point $x ^ *$ (say). I also know that the Jacobian of $F$ at each point $x$ in $\mathbb{R} ^ n$ has all of its eigenvalues in $[0,1)$ (but they are different for each $x$). Are these facts enough for me to say that the iterative sequence $x _ {n+1} = F(x_ n)$ converges to $x ^ *$ independently of the initial point $x_ 0$? (I know that if $x_0$ is close enough to $x ^ *$ then the sequence coverges but my question concerns any $x_0$ in $\mathbb{R} ^ n$.) Whatever the answer is, could you give me a reference to some theorem that justifies that?

Thank you

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Do you mean, all eigenvalues of the jacobian are real and in $[0,1)$, or just their modulus is in $[0,1)$? –  Pietro Majer Nov 28 '12 at 17:10
    
yes, real and in [0,1) –  Gili Nov 28 '12 at 17:56
    
I think you need to specify exact smoothness assumptions. Do you mean that $F$ is at least $C^1$ ? If not, what exactly the $J$ means? –  Alexandre Eremenko Nov 28 '12 at 17:56
    
More than that, F is infinitely differentiable. –  Gili Nov 28 '12 at 18:05

1 Answer 1

up vote 2 down vote accepted

Your question is stated as a Conjecture just before Theorem 2.1.5 in the book MR1015711 Belitskiĭ, G. R. and Lyubich, Yu. I. Matrix norms and their applications. Birkhäuser Verlag, Basel, 1988.

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Thanks for pointing out the conjecture. I don't have the book but I found a lot here and there about this type of problems. However my function has an additional property: Its range is a compact subset of $R^n$. So, there is some $M\in(0,1)$ which is larger than the eigenvalues of all Jacobians and consequently there is a matrix norm such that the norms of all Jacobians are bounded away from one, right?. Does this help towards a positive answer? –  Gili Dec 5 '12 at 19:14

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