= Solution
The <Cauchy point> minimizes the model along the negative-gradient ray within the ball. For $g\ne0$, write $s=-tg$, $0\leq t\leq\Delta/\|g\|_2$. Then
$$
m(-tg)=f(x)-t\|g\|_2^2+\frac12t^2\|g\|_2^2,
$$
whose scalar minimum occurs at $t=1$ unless the ball cuts the ray short. Therefore
$$
\boxed{s_c=-\min\{1,\Delta/\|g\|_2\}\,g.}
$$
If $\|g\|_2\leq\Delta$, its predicted decrease is $\tfrac12\|g\|_2^2$. Otherwise it is $\Delta\|g\|_2-\tfrac12\Delta^2\geq\tfrac12\Delta\|g\|_2$. Thus every feasible step with $m(s)\leq m(s_c)$ satisfies the <Cauchy decrease bound>
$$
\boxed{f(x)-m(s)\geq\frac12\|g\|_2\min\{\|g\|_2,\Delta\}.}
$$
For $g=0$, take $s_c=0$; the bound reduces to zero.
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