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Why is this function onto?

  1. Sep 23, 2009 #1
    Suppose you have a continuously differentiable function f: R -> R with |f'(x)| <= c < 1 for all x in R. Define a second function [tex]F: R^2 \rightarrow R^2[/tex] by

    [tex]F(x,y) = (x + f(y), y + f(x)).[/tex]

    F is supposed to be onto. Why is that so?

    Intuitively, I would say that F is locally onto by the Inverse Function Theorem (since det(DF) = 1 - c^2 > 0). And globally the identity portion of F dominates over the little perturbation from f, so I would expect the function to be onto. But that's far away from a proof. Any suggestions?
     
  2. jcsd
  3. Sep 23, 2009 #2
    Given (a,b) we want to solve x+f(y) = a, y+f(x) = b . This suggests a "fixed point" problem in the plane, and the derivative hypothesis should show it is contractive.
     
  4. Sep 23, 2009 #3
    Got it. Thanks!
     
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