= Solution
For $p=-g\ne0$, part (ii) simplifies to $\alpha_{\max}=2(1-\beta)/\gamma$. If the initial step is accepted, $\alpha_k=\alpha_{\mathrm{init}}$. Otherwise the immediately preceding trial $\alpha_k/\tau$ failed and hence exceeded $\alpha_{\max}$. Therefore
$$
\boxed{\alpha_k\geq\min\left\{\alpha_{\mathrm{init}},\frac{2\tau(1-\beta)}\gamma\right\}=\alpha_{\min}>0.}
$$
The fixed global Lipschitz constant and search parameters make this lower bound uniform across iterations. At $g=0$, the zero update passes the first trial and the sequence can be kept stationary.
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