The regularized LM step and residual-forcing parameter exist

Fix $\alpha >0$ and write $B_{\alpha }:=T^{\ast }T+\alpha I_X$. By [F2], $B_{\alpha }$ is bijective, so $h_{\alpha }:=B_{\alpha }^{-1}{T}^{\ast }r$ is defined and satisfies $T^{\ast }(T{h}_{\alpha }-r)+\alpha h_{\alpha }=0$.

For any $h\in X$, set $d:=h-h_{\alpha }$. Expanding with [F4] gives ${\mathrm{\Phi }}_{\alpha }(h)-{\mathrm{\Phi }}_{\alpha }(h_{\alpha })=⟨T{h}_{\alpha }-r,Td{⟩}_Y+\alpha ⟨h_{\alpha },d{⟩}_X+\frac{1}{2}\mathrm{\parallel }Td{\mathrm{\parallel }}_Y^2+\frac{\alpha }{2}\mathrm{\parallel }d{\mathrm{\parallel }}_X^2\mathrm{.}$ The first two terms sum to $⟨T^{\ast }(T{h}_{\alpha }-r)+\alpha h_{\alpha },d{⟩}_X=0$ by [F1] and step 1.1. The remaining terms are nonnegative, and because $\alpha >0$ their sum vanishes only when $d=0$. Thus $h_{\alpha }$ is the unique global minimizer and ${\mathrm{\Phi }}_{\alpha }(h_{\alpha })\le {\mathrm{\Phi }}_{\alpha }(0)$, so $\mathrm{\parallel }{h}_{\alpha }{\mathrm{\parallel }}_X\le \mathrm{\parallel }r{\mathrm{\parallel }}_Y\mathrm{/}\sqrt{\alpha }$.

Define $f(\alpha ):=\mathrm{\parallel }T{h}_{\alpha }-r{\mathrm{\parallel }}_Y$. For $\alpha ,\beta >0$, the inverse identity $B_{\alpha }^{-1}-B_{\beta }^{-1}=(\beta -\alpha )B_{\alpha }^{-1}{B}_{\beta }^{-1}$ and [F2] give $h_{\alpha }-h_{\beta }=(\beta -\alpha )B_{\alpha }^{-1}{h}_{\beta }$. Therefore, when $\alpha ,\beta$ lie near a fixed ${\alpha }_0>0$, $\mathrm{\parallel }{h}_{\alpha }-h_{\beta }{\mathrm{\parallel }}_X\le \mathrm{\mid }\beta -\alpha \mathrm{\mid }\frac{\mathrm{\parallel }r{\mathrm{\parallel }}_Y}{\alpha \sqrt{\beta }},$ which tends to zero with $\mathrm{\mid }\beta -\alpha \mathrm{\mid }$. Thus $h_{\alpha }$ and $f(\alpha )$ are continuous on $(0,\mathrm{\infty })$, using [F6] for continuity of the norm.

By step 2.1 and [F3], $\mathrm{\parallel }T{h}_{\alpha }{\mathrm{\parallel }}_Y\le \mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }r{\mathrm{\parallel }}_Y\mathrm{/}\sqrt{\alpha }\to 0$ as $\alpha \to \mathrm{\infty }$. The reverse triangle inequality from [F6] gives $\mathrm{\mid }f(\alpha )-\mathrm{\parallel }r{\mathrm{\parallel }}_Y\mathrm{\mid }\le \mathrm{\parallel }T{h}_{\alpha }{\mathrm{\parallel }}_Y$, so $f(\alpha )\to \mathrm{\parallel }r{\mathrm{\parallel }}_Y>\sigma \mathrm{\parallel }r{\mathrm{\parallel }}_Y$.

Comparing the minimizer with the fixed vector $v$ gives $f(\alpha {)}^2+\alpha \mathrm{\parallel }{h}_{\alpha }{\mathrm{\parallel }}_X^2\le \mathrm{\parallel }Tv-r{\mathrm{\parallel }}_Y^2+\alpha \mathrm{\parallel }v{\mathrm{\parallel }}_X^2\mathrm{.}$ Because $\mathrm{\parallel }Tv-r{\mathrm{\parallel }}_Y\le (\sigma \mathrm{/}\gamma )\mathrm{\parallel }r{\mathrm{\parallel }}_Y<\sigma \mathrm{\parallel }r{\mathrm{\parallel }}_Y$, there is a strict squared-norm gap below ${\sigma }^2\mathrm{\parallel }r{\mathrm{\parallel }}_Y^2$. Also $v\mathrm{\ne }0$, since $v=0$ would contradict the range bound and $r\mathrm{\ne }0$. Choose $a>0$ small enough that $\mathrm{\parallel }Tv-r{\mathrm{\parallel }}_Y^2+a\mathrm{\parallel }v{\mathrm{\parallel }}_X^2<{\sigma }^2\mathrm{\parallel }r{\mathrm{\parallel }}_Y^2$. The minimizer comparison in this step gives $f(a{)}^2\le \mathrm{\parallel }Tv-r{\mathrm{\parallel }}_Y^2+a\mathrm{\parallel }v{\mathrm{\parallel }}_X^2<{\sigma }^2\mathrm{\parallel }r{\mathrm{\parallel }}_Y^2$, so $f(a)<\sigma \mathrm{\parallel }r{\mathrm{\parallel }}_Y$.

By step 3.2, choose $b>a$ with $f(b)>\sigma \mathrm{\parallel }r{\mathrm{\parallel }}_Y$. Continuity from step 3.1 and [F5] give $\alpha \in [a,b]$ such that $f(\alpha )=\sigma \mathrm{\parallel }r{\mathrm{\parallel }}_Y$. Since $a>0$, this is an admissible positive parameter and proves the forcing claim.