 Solution

Write $D=f(x)-m(s)$ for the predicted decrease, assuming a nonstationary point and a step at least as good as the Cauchy point. When $\Delta \le \Vert g{\Vert }_2$, part (ii) gives $D\ge \frac{1}{2}\Vert g{\Vert }_2\Delta$. Part (iv), using its weaker requested bound, then givesConsequently $\Delta \le \frac{1}{2}(1-\eta )\Vert g{\Vert }_2$ implies $\boxed{\rho \ge \eta }$.

For the radius conclusion, use the standard update rules from part (i). Suppose no subsequence of the generated gradient norms tends to zero. Then some $\epsilon >0$ bounds those norms below eventually. If ${\Delta }_k\to 0$, eventually ${\Delta }_k\le \frac{1}{2}(1-{\eta }_2)\epsilon$, so every subsequent step is very successful by the preceding estimate. The radius is then maintained or increased, contradicting convergence to zero. Hence a radius tending to zero forces a subsequence of gradient norms to tend to zero. The conclusion uses the trust-region update rules as well as the model estimates.