The conic phase-I margin separates strict feasibility but may leave weak infeasibility undecided

If ${\rho }_{\ast }<0$, then by the definition of an infimum some feasible pair $(x,\tau )$ has $\tau <0$. Write $h-Gx+\tau e=k\in K$. Then $h-Gx=k+(-\tau )e$. An open ball around $(-\tau )e\in \mathrm{int}K$ lies in $K$; translating that ball by $k\in K$ keeps it in $K+K\subseteq K$. Hence $k+(-\tau )e\in \mathrm{int}K$, so $Ax=b$ and $h-Gx\in \mathrm{int}K$.

If the original system has a feasible $x$, then $\tau =0$ is feasible for the Phase-I set, hence ${\rho }_{\ast }\le 0$. This proves infeasibility when ${\rho }_{\ast }>0$ or the Phase-I set is empty. If ${\rho }_{\ast }=0$ is attained, the attaining pair cannot have $\tau <0$, since that would force ${\rho }_{\ast }<0$; hence it has $\tau =0$ and its $x$ is originally feasible. Strict feasibility at some $x$ would give $s=h-Gx\in \mathrm{int}K$. Because $\mathrm{int}K$ is open and $e\in \mathrm{int}K$, there is $\delta >0$ small enough that $s-\delta e\in K$, making $\tau =-\delta$ feasible and contradicting ${\rho }_{\ast }=0$. If ${\rho }_{\ast }=0$ is not attained, original feasibility would itself provide an attaining pair at $\tau =0$, a contradiction. Thus zero is inconclusive until attainment is checked.