= Solution
From $B_kp_k=-g_k$ and the <matrix 2-norm> bound, $\|g_k\|_2\leq\kappa_b\|p_k\|_2$, so $\|p_k\|_2\geq\|g_k\|_2/\kappa_b$. Let $C=2\tau\varepsilon(1-\beta)/\gamma>0$. The summand in the supplied decrease bound satisfies
$$
\min\left(\alpha_{\mathrm{init}},C\frac{\|g_k\|_2}{\|p_k\|_2}\right)\|g_k\|_2\|p_k\|_2
=\min\left(\alpha_{\mathrm{init}}\|g_k\|_2\|p_k\|_2,C\|g_k\|_2^2\right)
\geq\min(\alpha_{\mathrm{init}}/\kappa_b,C)\|g_k\|_2^2.
$$
Thus, with the fixed positive constant $c=\beta\varepsilon\min(\alpha_{\mathrm{init}}/\kappa_b,C)$,
$$
c\sum_{k=0}^j\|g_k\|_2^2\leq f(x_0)-f(x_{j+1})\leq f(x_0)-\inf f<\infty.
$$
The nonnegative series therefore converges, forcing its terms to zero:
$$
\boxed{\lim_{k\to\infty}g(x_k)=0.}
$$
The argument uses the given angle condition and upper bound on $\|B_k\|_2$; it does not need an additional uniform lower eigenvalue bound on $B_k$.
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