Anchored IRGNM error identity and fixed-reference residual inequality

For $h=x-u$, let $\gamma (s)=F(u+sh)$. The segment lies in $B$. If $h\mathrm{\ne }0$ and $0<s<1$, then for real $r\to 0$ with $s+r\in [0,1]$, same-field Fréchet differentiability gives $\parallel \frac{F(u+(s+r)h)-F(u+sh)}{r}-F^{\mathrm{\prime }}(u+sh)h\parallel =\frac{\mathrm{\parallel }F(u+(s+r)h)-F(u+sh)-F^{\mathrm{\prime }}(u+sh)(rh)\mathrm{\parallel }}{\mathrm{\mid }r\mathrm{\mid }}⟶0.$ Thus ${\gamma }^{\mathrm{\prime }}(s)=F^{\mathrm{\prime }}(u+sh)h$. For $h=0$ the curve is constant. In either case differentiability makes $\gamma$ continuous by the Fréchet remainder definition, and this derivative is continuous on the segment with endpoint extensions because the assumed continuous Fréchet differentiability makes $s\mapsto F^{\mathrm{\prime }}(u+sh)$ continuous in operator norm, including at $s=0,1$, and evaluation at the fixed vector $h$ is continuous. Applying the Banach-valued fundamental theorem [F3] and subtracting the constant integral of $Ah$ gives $F(x)-F(u)-Ah={\int }_0^1(F^{\mathrm{\prime }}(u+sh)-A)hds\mathrm{.}$ Linearity of $R_{u+sh}$ and its norm bound make the integrand norm at most $\kappa \mathrm{\parallel }Ah\mathrm{\parallel }$, so the integral has that bound. The triangle and reverse triangle inequalities give the two-sided estimate. Subtracting $(F^{\mathrm{\prime }}(u)-A)h$ bounds the linearization remainder by $2\kappa \mathrm{\parallel }Ah\mathrm{\parallel }$, proving the cone constant. This integrates the continuous vector $F^{\mathrm{\prime }}(u+sh)h$, not a possibly unmeasurable choice of the operators $R_{u+sh}$. Taking $u=x^{†}$ also gives $\mathrm{\parallel }t\mathrm{\parallel }\le 2\kappa v$.

Let $A_x=F^{\mathrm{\prime }}(x)$. Since $A_x$ is bounded and $\alpha >0$, [F2] applied to $A_x$ supplies the inverse defining the unique correction in [F1] for every anchor $x_0\in X$; no evaluation at $x_0$ is required. Adding $(A_x^{\ast }{A}_x+\alpha I)e$ to that correction equation gives $(A_x^{\ast }{A}_x+\alpha I)e_{+}=\alpha e_0+A_x^{\ast }(y_{\delta }-y+t)$. By factorization and [F4], $A_x^{\ast }{A}_x=S+A^{\ast }HA$ and $A_x^{\ast }=A^{\ast }(I+R{)}^{\ast }$. Moving $A^{\ast }HA{e}_{+}$ to the right gives the fixed-reference identity with $z$ in the Statement. The inverse used here exists independently of where $x_{+}$ lies.

By [F4], $\mathrm{\parallel }H\mathrm{\parallel }\le 2\kappa +{\kappa }^2=\theta$ and $\mathrm{\parallel }(I+R{)}^{\ast }\mathrm{\parallel }\le 1+\kappa$. Combining these bounds with $\mathrm{\parallel }t\mathrm{\parallel }\le 2\kappa v$ proves $\mathrm{\parallel }z\mathrm{\parallel }\le c_0\delta +d_{0}v+\theta v_{+}$.

Apply $A(S+\alpha I{)}^{-1}$ to step 1.2 and use [F2] with $e_0=S^{\nu }w$. Then $v_{+}\le W{\alpha }^{\nu +1\mathrm{/}2}+\mathrm{\parallel }z\mathrm{\parallel }$; moving $\theta v_{+}$ to the left gives the residual inequality. Applying $(S+\alpha I{)}^{-1}$ without $A$ instead gives $\mathrm{\parallel }{e}_{+}\mathrm{\parallel }\le W{\alpha }^{\nu }+\mathrm{\parallel }z\mathrm{\parallel }\mathrm{/}(2\sqrt{\alpha })$, proving the error bound. This controls the fixed source term through its spectral filter, not through a crude ${\alpha }^{-1}\mathrm{\parallel }{e}_0\mathrm{\parallel }$ estimate.