# Why is this function onto?

1. Sep 23, 2009

### sin123

Suppose you have a continuously differentiable function f: R -> R with |f'(x)| <= c < 1 for all x in R. Define a second function $$F: R^2 \rightarrow R^2$$ by

$$F(x,y) = (x + f(y), y + f(x)).$$

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. Sep 23, 2009

### g_edgar

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.

3. Sep 23, 2009

### sin123

Got it. Thanks!