A strong Wolfe step exists on a bounded-below descent ray

The chain rule in [F1] makes $\varphi$ continuously differentiable on a neighborhood of each finite point of $[0,\mathrm{\infty })$. Define $\psi (t):=\varphi (t)-\varphi (0)-c_{1}t{\varphi }^{\mathrm{\prime }}(0)$. Then $\psi (0)=0$ and $d:={\psi }^{\mathrm{\prime }}(0)=(1-c_1){\varphi }^{\mathrm{\prime }}(0)<0$. By [F3], $\psi (t)\mathrm{/}t\to d$ as $t\to 0$. Choose $\delta >0$ so that $\psi (t)\mathrm{/}t<d\mathrm{/}2<0$ for $0<t<\delta$; hence $\psi (t)<0$ there.

Since $\varphi$ is bounded below and $-c_1{\varphi }^{\mathrm{\prime }}(0)>0$, $\psi (t)\to +\mathrm{\infty }$ as $t\to +\mathrm{\infty }$. Thus $S:=\{t>0:\psi (t)\ge 0\}$ is nonempty and bounded below by $0$, so [F4] gives $t_{\ast }:=\inf S$. Step 1.1 shows that $S$ has no points near $0$, hence $t_{\ast }>0$. Every $s\in (0,t_{\ast })$ has $\psi (s)<0$, while the defining approximation property of the infimum gives points of $S$ arbitrarily close above $t_{\ast }$. Continuity of $\psi$ therefore forces $\psi (t_{\ast })=0$.

Every $s\in (0,t_{\ast })$ has $\psi (s)<0$, so Armijo holds strictly throughout that interval. Apply [F2] to $\varphi$ on $[0,t_{\ast }]$ to get $\xi \in (0,t_{\ast })$ with ${\varphi }^{\mathrm{\prime }}(\xi )=(\varphi (t_{\ast })-\varphi (0))\mathrm{/}{t}_{\ast }=c_1{\varphi }^{\mathrm{\prime }}(0)$. Hence $\mathrm{\mid }{\varphi }^{\mathrm{\prime }}(\xi )\mathrm{\mid }=c_1\mathrm{\mid }{\varphi }^{\mathrm{\prime }}(0)\mathrm{\mid }<c_2\mathrm{\mid }{\varphi }^{\mathrm{\prime }}(0)\mathrm{\mid }$, so the strong curvature inequality holds strictly at $\xi$.

The strict Armijo inequality holds at every point of $(0,t_{\ast })$, including $\xi$, and both $\varphi$ and ${\varphi }^{\mathrm{\prime }}$ are continuous there. Therefore both strict inequalities persist on some open interval about $\xi$ contained in $(0,t_{\ast })$. Every point of that interval is a strong Wolfe step, proving the asserted nonemptiness.