- THE IMPLICIT FUNCTION THEOREM 329
that (DJ (x 0 ))-^1 is a bounded linear operator. Without loss of generality we may
assume, by replacing f by the map
x H (DJ (xo))-^1 (f (x + xo) - f (xo)),
that X = Y, xo = 0, f (0) = 0, and DJ (xo) = idx.
Define g : U ---+ X by g (x) ~ x - f (x), which is the difference of f and its
linearization at 0. We have Dg (0) = 0 and since g is C^1 , there exists 5 > 0 such
that(K.1)1
llDg (x)ll :::; 2
for x E B (0, 5) ~ { x E X : llxl l :::; 5}. Note that this implies for x E B (0, 5) that
(K.2)1
llDJ (x) (v)ll ~ llvll - llDg (x) (v)ll ~ 2 llvllfor all v E X. By the fundamental theorem of calculus and (K.1), we have that forany x 1 , x2 EB (0, 5),
(K.3) Ilg (x1) - g (x2)ll = 111
1
Dg (tx1 + (1 - t) x2) (x1 - x2) dt"
:::; llx1 - x2ll fo1
IJDg (tx1 + (1 - t) x2)ll dt
1
:::; 2 llx1 - x2ll ·Since g (0) = 0, we obtain that g (x) E B (0, 5 /2) for x E B (0, 5).
Given y EX, we define gy: U---+ X by
(K.4) gy ( x) ~ y + x - f ( x) = y + g ( x).
In particular, f (x) = y if and only if gy (x) = x. For y E B (0, 5/2) we have
by the triangle inequality, gy (.B (0, 5)) C B (0, 5). Moreover, by (K.4) and (K.3),
for any y E B (0, 5/2) the map gYl.B(o,o) : B (0, 5) ---+ B (0, 5) is a contraction
mapping of a complete metric space. Therefore Lemma K.2 implies that there
exists a unique fixed point x EB (0, 5) of gyl.B(o,o); that is , we have an inverse map
1-^1 : .B (o, 5/2)---+ .B (0, 5).
Next, we show that 1-^1 is Lipschitz continuous. For any x 1 , x2 EB (0, 5),
II! (x1) - f (x2)ll ~ llx1 - x2ll - Ilg (x1) - g (x2)ll
1
~ 2 llx1 - x2JJ.Hence, given any y 1 , Y2 EB (0, 5/2), we have
(K.5)Finally, we show that 1-^1 is differentiable. By (K.2), we h ave
s~p ll(Df (x))-111:::; 2.
xEB(O,o)