A pure witness suffices for a QMA completeness threshold

By [F1]–[F2], choose an orthonormal eigenbasis $u_1,\dots ,u_d$ for $\rho$, with eigenvalues ${\lambda }_j\ge 0$. Basis independence [F6] and its diagonal-sum definition give ${\sum }_j{\lambda }_j=\mathrm{Tr}(\rho )=1$. Thus $\rho ={\sum }_j{\lambda }_j\mathrm{\mid }{u}_j⟩⟨u_j\mathrm{\mid }$, with zero eigenvalue terms allowed.

Linearity [F5] gives $\mathrm{Tr}(E_x\rho )={\sum }_j{\lambda }_j\mathrm{Tr}(E_x\mathrm{\mid }{u}_j⟩⟨u_j\mathrm{\mid })$. By [F3], these terms are precisely the verifier's pure-state acceptance probabilities. By [F4], in an orthonormal basis beginning with $u_j$, the rank-one factor has only its first diagonal entry nonzero after multiplication by $E_x$; [F6] therefore gives $\mathrm{Tr}(E_x\mathrm{\mid }{u}_j⟩⟨u_j\mathrm{\mid })=⟨E_x{u}_j,u_j⟩$. This is a convex average; if every positive-weight term were below $c$, its finite weighted average would be below $c$. Hence some $j$ with ${\lambda }_j>0$ has acceptance at least $c$, so [F7] gives a pure witness meeting the QMA completeness threshold.