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Use the induced spectral norm for the symmetric matrix $H$ and interpret $\Vert g\Vert /\Vert H\Vert =+\infty$ when $H=0$. If $q=g^{T}Hg\le 0$, the Cauchy point gives a decrease at least $\Delta \Vert g\Vert$. If $q>0$, its step size satisfies $\tau q\le \Vert g{\Vert }^2$ and $q\le \Vert H\Vert \Vert g{\Vert }^2$, soThus the numerical constant $1/2$ is independent of the model data. The bound links a nonzero gradient to a definite predicted decrease, including for an indefinite matrix $H$.

For $g\ne 0$, the Cauchy point minimizes the quadratic model along the feasible part of the negative gradient ray. Writing $q=g^{T}Hg$, its step size isIndeed $m(-tg)=a-t\Vert g{\Vert }^2+\frac{1}{2}{t}^{2}q$ on $0\le t\le \Delta /\Vert g\Vert$. For $q\le 0$ it decreases throughout that interval; otherwise its unconstrained minimum occurs at $t=\Vert g{\Vert }^2/q$. The constant $a$ does not affect the minimizing step.

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 givesConsequently 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 $\Vert \nabla f(x^k)\Vert \ge ϵ>0$, the Taylor expansion gives a model error bounded by $K\Vert s_k{\Vert }^2$. 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 ${\mathrm{liminf}}_k\Vert \nabla f(x^k)\Vert =0$.

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.

The Cauchy point minimizes the model along the negative gradient ray inside the trust region. With $q=g^{T}Hg$, writeIf $q\le 0$, this last quadratic function decreases throughout the interval. If $q>0$, its unconstrained minimum occurs at $t=\Vert g{\Vert }^2/q$. ThusThe additive constant $a$ has no effect on the Cauchy point. To obtain the Cauchy decrease bound, first suppose $q>0$. Since $\tau q\le \Vert g{\Vert }^2$,The spectral norm inequality $q\le \Vert H\Vert \Vert g{\Vert }^2$ gives $\tau \ge \min \{\Delta /\Vert g\Vert ,1/\Vert H\Vert \}$. HenceIf $q\le 0$, the decrease is at least $\Delta \Vert g\Vert$, which also implies the displayed bound. When $H=0$, interpret $\Vert g\Vert /\Vert H\Vert =+\infty$. Thus the universal choice is $c=1/2$.