328 K. IMPLICIT FUNCTION THEOREM
PROOF. The idea is to iterate the map f (starting at any point) and obtain
a Cauchy sequence which converges to the fixed point. Let xo E X and for each
i 2: 0 define Xi EX inductively by Xi+l ~ f (xi)· We have
d (x i, Xi+1) ~ >.d (xi-1, xi) ~ · · · ~ >.id (xo, x1)for all i E N and hence for any j > i,
d (xi, Xj) ~ d (xi, Xi+l) + ... + d (Xj-1, Xj)
~ (>.i + · · · + >.J-^1 ) d(xo,xi)
,\ i~ l-,\d(xo,x1).
Since >. E [O, 1), we have that {xi} :, 0 is a Cauchy sequence. Since X is complete,
the limit x 00 ~ limi--+oo Xi exists. Taking the limit as i ---t oo of Xi+1 = f (xi) and
using the continuity off, we have x 00 = f (x 00 ).
The fixed point x 00 is unique since if x~ is also a fixed point of f , then
d (x 00 , x~) = d (f (x 00 ), f (x:XJ) ~ >.d (xoo, x'.x,),
which implies that d (x 00 , x~) = 0. D
1.3. Inverse function theorem.
We first discuss the differentiability of maps between Banach spaces. Let
(X, II · llx) and (Y, II · llY) be Banach spaces. Let Lk (X, Y) denote the Banach
space of k-multilinear continuous maps from Xk ~ X x · · · x X to Y with the
standard norm:llAll ~sup{llA(x1, ... ,xk)llY: llx1llx ~ l,. .. ,llxkllx ~ 1};
Awe write L (X, Y) = L^1 (X, Y). There are natural isomorphisms between L^2 (X, Y)
and L(X, L(X, Y)), between L^3 (X, Y) and L(X, L(X, L(X, Y))), etc.
Let U c X be an open set. We say that a map f : U ---t Y is (Frechet)
differentiable at x EU if there exists D f (x) EL (X, Y) such that
lim llf(x+h)-f(x)-Df(x)(h)llY =O.
h-+O llhllxWe call D f ( x) the (Frech et) derivative of f at x.
We say that f is C^1 if D f : U ---t L (X, Y) is continuous. If D f is differentiable
at every point in U , then we have D^2 f : U ---t L^2 (X, Y) and we say that f is
C^2 if D^2 f is continuous. Continuing this way we may define the k-th derivative
Dk f : U ---t Lk (X, Y) of f and what it means for f to be Ck.
Next we state and prove the Banach space inverse function t heorem.THEOREM K.3. Let (X, 11 · llx) and (Y, 11 · llY) be Banach spaces and let UC X
be an open set. If f : U ---t Y is a Ck map where k 2: 1 and xo E U is such that
D f (xo) : X ---t Y is a bijection, then there exists an open neighborhood W of x 0
such that flw : W ---t f (W) is a bijection onto an open neighborhood off (xo) and
the inverse of flw' i.e., Ulw )-^1 : f (W) ---t w , is a ck map.PROOF. We prove the case where k = 1; the cases where k 2: 2 is left as