Fraction-to-boundary steps preserve positivity and exact primal-dual feasibility certifies a gap

Because $s$ is interior, a small positive segment from $s$ in direction $\mathrm{\Delta }s$ remains in $K$. The set of feasible fractions in $[0,1]$ is closed by [F2], so its maximum ${\bar{\alpha }}_s>0$ is attained and $s+{\bar{\alpha }}_s\mathrm{\Delta }s\in K$. Any $0<\alpha \le \eta {\bar{\alpha }}_s$ has $\theta =\alpha \mathrm{/}{\bar{\alpha }}_s<1$, and the update is $(1-\theta )s+\theta (s+{\bar{\alpha }}_s\mathrm{\Delta }s)$. An open ball about $s$ lies in $K$; scaling that ball by $1-\theta >0$ and adding $\theta (s+{\bar{\alpha }}_s\mathrm{\Delta }s)\in K$ shows the updated point is interior. The same argument applies to $z$ in $K^{\ast }$. In the orthant, the coordinate-ratio formula is the maximal nonnegative boundary fraction. For a negative coordinate direction, $\alpha \le \eta (-x_i\mathrm{/}\mathrm{\Delta }{x}_i)$ gives $x_i+\alpha \mathrm{\Delta }{x}_i\ge (1-\eta )x_i>0$; nonnegative directions preserve positivity, and the same two cases apply to $s$.

For a Lorentz factor, write $s=(t,u)$ and $\mathrm{\Delta }s=(\dot{t},v)$, with $t>\mathrm{\parallel }u{\mathrm{\parallel }}_2$. Set $q=t^2-\mathrm{\parallel }u{\mathrm{\parallel }}_2^2>0$ and $c=\mathrm{\mid }\dot{t}\mathrm{\mid }+{\sum }_i\mathrm{\mid }{v}_i\mathrm{\mid }$. If $c=0$ the step is unchanged. Otherwise $t-\mathrm{\parallel }u{\mathrm{\parallel }}_2=q\mathrm{/}(t+\mathrm{\parallel }u{\mathrm{\parallel }}_2)\ge q\mathrm{/}(2t)$, while $\mathrm{\mid }\dot{t}\mathrm{\mid }+\mathrm{\parallel }v{\mathrm{\parallel }}_2\le c$. The algorithm's safe fraction $\beta =\min (1,q\mathrm{/}(2tc))$ therefore gives $\alpha =\eta \beta$ with $\alpha (\mathrm{\mid }\dot{t}\mathrm{\mid }+\mathrm{\parallel }v{\mathrm{\parallel }}_2)<t-\mathrm{\parallel }u{\mathrm{\parallel }}_2$. The triangle inequality then yields $t+\alpha \dot{t}>\mathrm{\parallel }u+\alpha v{\mathrm{\parallel }}_2$, so the updated Lorentz vector is interior. For a PSD factor $Y\succ 0$ and symmetric direction $D$, let $C=L(Y^{-1})L(D)$ as in the algorithm, where $L(M)={\sum }_{i,j}\mathrm{\mid }{M}_{ij}\mathrm{\mid }$. If $C=0$, then $D=0$. Otherwise set $B=Y^{-1\mathrm{/}2}D{Y}^{-1\mathrm{/}2}$. Submultiplicativity of the operator norm and [F7] give $\mathrm{\parallel }B{\mathrm{\parallel }}_2\le \mathrm{\parallel }{Y}^{-1}{\mathrm{\parallel }}_2\mathrm{\parallel }D{\mathrm{\parallel }}_2\le \mathrm{\parallel }{Y}^{-1}{\mathrm{\parallel }}_F\mathrm{\parallel }D{\mathrm{\parallel }}_F\le C$. Since $\alpha =\eta \min (1,1\mathrm{/}C)$, $\alpha C<1$, and for every nonzero $w$, $w^T(I+\alpha B)w\ge (1-\alpha \mathrm{\parallel }B{\mathrm{\parallel }}_2)\mathrm{\parallel }w{\mathrm{\parallel }}_2^2>0$. Thus $Y+\alpha D=Y^{1\mathrm{/}2}(I+\alpha B)Y^{1\mathrm{/}2}\succ 0$. The same bounds apply separately to primal and dual factors, so each computed safe fraction is no larger than its maximal boundary-preserving fraction.

For any feasible conic pair, [F1] and [F2] give $b^{T}y-⟨h,z⟩\le c^{T}x$. Taking the infimum over primal feasible $x$ and the supremum over dual feasible pairs shows that a feasible dual value is a lower bound on $p^{\ast }$. Hence a computed primal-dual feasible pair with exact gap $⟨z,s⟩\le \epsilon$ is an $\epsilon$-suboptimality certificate.

In LP/QP mode, exact $Ax=b$, $x\ge 0$, $s\ge 0$, and $Qx+c-A^{T}y-s=0$ make ${\mathrm{\nabla }}_{w}L(x,y,s)=0$. To apply [F4], use the smooth inequality convention $g_i(w)=-w_i\le 0$, multiplier $\lambda =s$, and equality multiplier $\nu =-y$. Since $Q⪰0$, $L$ is convex, so [F5] implies $L(w,y,s)\ge L(x,y,s)$ for every $w$; therefore $q(y,s)=L(x,y,s)=f_0(x)-x^{T}s$. By [F4], $q(y,s)\le p^{\ast }$, and primal feasibility gives $p^{\ast }\le f_0(x)$. Thus $0\le f_0(x)-p^{\ast }\le x^{T}s$.

In LP/QP mode, [F6] supplies a unique exact Newton direction under the stated rank and strict-interiority hypotheses. Symmetric-cone mode may have a singular derivative, which the algorithm tests and reports.

Every dimension-bounded loop terminates after finitely many operations, and the outer loop increments its update counter until it reaches the finite cap $N$ or returns an earlier status. Therefore every invocation terminates with one of its declared statuses. A tolerance status based on nonzero residuals is explicitly residual-candidate, so no step above treats it as a certificate.