Summary: Two examples illustrating the Inverse Function Theorem.
Example One. Let L(x) = x for x R, a = 0,let
p(x) =



0 if x -1,
x + 1 if -1 < x 0,
1 - x if 0 < x 1,
0 1 < x.
Note that Lip (p) = 1 on any interval containing 0 and that
= inf{|L(x)| : |x| = 1} = 1.
Let f(x) = L(x) + p(x). Since f is constant on [0, 1] we see that f is not invertible on any set whose interior
contains 0. Thus the hypothesis < in the Inverse Function Theorem cannot be weakened.
Example Two. Let
f : R2
R2
be such that f(x) = (sin(x1 + x2), exp(x1) + exp(-2x2)) for x R2
. Evidently, f(0) = 0 and
m(f(x)) =

Source: Allard, William K. - Department of Mathematics, Duke University

Collections: Mathematics