An inexact Newton residual gives a local error recurrence

If $n=0$, the space ${\mathbb{R}}^0$ is a singleton by [F6]. Its unique map $A$ is linear and satisfies $A(A(0))=0$, so taking $B=A$ verifies both inverse identities in the Statement's convention. Thus $J(x)=J_{\ast }=A$ is invertible, and $x=x_{\ast }=p=r=F(x)=0$ in this space. Take $r_0=R$, $C_F=1$, and $C=0$. Every claimed estimate is $0\le 0$, so the assertion holds. For the remainder assume $n\ge 1$, so the imported positive-dimensional Newton lemmas and operator-norm comparisons apply.

Put $K_{\ast }:=\mathrm{\parallel }{J}_{\ast }^{-1}{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$. The inverse identity gives $\mathrm{\parallel }e{\mathrm{\parallel }}_2=\mathrm{\parallel }{J}_{\ast }^{-1}{J}_{\ast }e{\mathrm{\parallel }}_2\le K_{\ast }\mathrm{\parallel }e{\mathrm{\parallel }}_{\ast }$, and the operator-norm definition gives $\mathrm{\parallel }e{\mathrm{\parallel }}_{\ast }\le \mathrm{\parallel }{J}_{\ast }{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\mathrm{\parallel }e{\mathrm{\parallel }}_2$. By [F3], choose a smaller radius $r_1\le R$ and $M>0$ such that every $J(x)$ on ${\bar{B}}_2(x_{\ast },r_1)$ is invertible and $\mathrm{\parallel }J(x{)}^{-1}{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le M$. Decrease $r_1$ if necessary so $r_1\le 1$, and set $r_0:=r_1$.

Let $e:=x-x_{\ast }$. By [F4] and $\mathrm{\parallel }J(x)-J_{\ast }{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le L\mathrm{\parallel }e{\mathrm{\parallel }}_2$, $\mathrm{\parallel }F(x)-J_{\ast }e{\mathrm{\parallel }}_2\le \mathrm{\parallel }F(x)-J(x)e{\mathrm{\parallel }}_2+\mathrm{\parallel }(J(x)-J_{\ast })e{\mathrm{\parallel }}_2\le \frac{3L}{2}\mathrm{\parallel }e{\mathrm{\parallel }}_2^2$. Since $\mathrm{\parallel }e{\mathrm{\parallel }}_2\le K_{\ast }\mathrm{\parallel }e{\mathrm{\parallel }}_{\ast }$ and $\mathrm{\parallel }e{\mathrm{\parallel }}_{\ast }\le \mathrm{\parallel }{J}_{\ast }{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}{r}_0$, this gives $\mathrm{\parallel }F(x){\mathrm{\parallel }}_2\le (1+\frac{3L}{2}{K}_{\ast }^2\mathrm{\parallel }{J}_{\ast }{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}{r}_0)\mathrm{\parallel }e{\mathrm{\parallel }}_{\ast }=:C_F\mathrm{\parallel }e{\mathrm{\parallel }}_{\ast }$.

The linear residual equation and steps 1.2–2.1 imply $\mathrm{\parallel }p{\mathrm{\parallel }}_2\le M\mathrm{\parallel }J(x)p{\mathrm{\parallel }}_2\le M(1+\eta )\mathrm{\parallel }F(x){\mathrm{\parallel }}_2\le M(1+\bar{\eta })C_F\mathrm{\parallel }e{\mathrm{\parallel }}_{\ast }$.

Put $e_{+}:=x+p-x_{\ast }$. From $J(x)p=-F(x)+r$, $J_{\ast }{e}_{+}=J_{\ast }e-F(x)+r+(J_{\ast }-J(x))p$. Step 1.2 and [F4] give $\mathrm{\parallel }F(x)-J_{\ast }e{\mathrm{\parallel }}_2\le (3L\mathrm{/}2)\mathrm{\parallel }e{\mathrm{\parallel }}_2^2\le (3L\mathrm{/}2)K_{\ast }^2\mathrm{\parallel }e{\mathrm{\parallel }}_{\ast }^2$, so $\mathrm{\parallel }r{\mathrm{\parallel }}_2\le \eta \mathrm{\parallel }F(x){\mathrm{\parallel }}_2\le \eta \mathrm{\parallel }e{\mathrm{\parallel }}_{\ast }+(3\eta L{K}_{\ast }^2\mathrm{/}2)\mathrm{\parallel }e{\mathrm{\parallel }}_{\ast }^2$. Also $\mathrm{\parallel }{J}_{\ast }-J(x){\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le L{K}_{\ast }\mathrm{\parallel }e{\mathrm{\parallel }}_{\ast }$ and step 3.1 bounds $\mathrm{\parallel }p{\mathrm{\parallel }}_2\le M(1+\bar{\eta })C_F\mathrm{\parallel }e{\mathrm{\parallel }}_{\ast }$. The triangle inequality and $\eta \le \bar{\eta }$ yield $\mathrm{\parallel }{e}_{+}{\mathrm{\parallel }}_{\ast }\le \eta \mathrm{\parallel }e{\mathrm{\parallel }}_{\ast }+(\frac{3(1+\bar{\eta })L{K}_{\ast }^2}{2}+L{K}_{\ast }M(1+\bar{\eta })C_F)\mathrm{\parallel }e{\mathrm{\parallel }}_{\ast }^2$. Set $C$ equal to this nonnegative coefficient; this is the claimed recurrence, and step 1.2 supplies invertibility.