Statement

Let XX be a , r>0r>0, and B={x:xx0r}B=\{x:\|x-x_0\|\le r\}. Suppose T:BXT:B\to X satisfies

T(x)T(y)qxy,q<1,T(x0)x0(1q)r.\|T(x)-T(y)\|\le q\|x-y\|,\qquad q<1, \quad \|T(x_0)-x_0\|\le(1-q)r.

Then T(B)BT(B)\subseteq B, since T(x)x0qr+(1q)r\|T(x)-x_0\|\le qr+(1-q)r. The ball is complete, so the gives a unique fixed point in it.

Small-parameter form

For T(x)=x0+εN(x)T(x)=x_0+\varepsilon N(x), it suffices that εsupBNr|\varepsilon|\sup_B\|N\|\le r and εLipBN<1|\varepsilon|\operatorname{Lip}_B N<1. A contraction estimate without the self-map condition is insufficient to apply the theorem on the ball.