Statement

Let XX be a , r>0r>0, and B={x:∥x−x0∥≤r}B=\{x:\|x-x_0\|\le r\}. Suppose T:B→XT:B\to X satisfies

∥T(x)−T(y)∥≤q∥x−y∥,q<1,∥T(x0)−x0∥≤(1−q)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)−x0∥≤qr+(1−q)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 ∣ε∣sup⁡B∥N∥≤r|\varepsilon|\sup_B\|N\|\le r and ∣ε∣Lip⁡BN<1|\varepsilon|\operatorname{Lip}_B N<1. A contraction estimate without the self-map condition is insufficient to apply the theorem on the ball.