= Solution
The <Cauchy point> provides both an inexpensive fallback step and a standard against which a better step can be judged. Requiring a fixed <Fraction of Cauchy decrease> gives
$$
m_k(0)-m_k(s_k)\geq\frac\kappa2\|\nabla f(x^k)\|\min\{\Delta_k,\|\nabla f(x^k)\|/\|H_k\|\},\qquad\kappa>0.
$$
Consequently an exact <global minimum> of each <trust-region subproblem> is unnecessary. A step from any inner solver is adequate if it meets this bound.
The convergence argument combines this decrease with model accuracy. For example, suppose the <gradient> is uniformly <Lipschitz continuous> on a neighborhood of the visited <level set>, the model <Hessian matrices> are uniformly bounded, and the usual acceptance and radius safeguards are used. If $\|\nabla f(x^k)\|\geq\epsilon>0$, the <Taylor expansion> gives a model error bounded by $K\|s_k\|^2$. For sufficiently small $\Delta_k$, division by the <Cauchy decrease bound> makes $|1-\rho_k|$ small. Such trials are successful and cannot keep shrinking the radius. With the <gradient> bounded away from zero, the radius therefore has a positive lower bound, and each accepted step decreases $f$ by a fixed positive amount. There must be infinitely many accepted steps, since repeated rejection would force the radius below the small-radius acceptance threshold. This contradicts boundedness below of $f$ and proves $\liminf_k\|\nabla f(x^k)\|=0$.
Stronger standard conclusions can be obtained with further regularity and safeguards. Merely being twice <continuously differentiable> and bounded below does not supply uniform <Hessian> bounds on an unbounded <level set>; those uniform assumptions must be stated when using this argument.
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