 Solution

Write $x_i=⟨{\psi }_i|X|{\psi }_i⟩$, which is real because $X$ is a Hermitian operator. Setwith $\mathrm{sgn}(0)=0$. This Hermitian operator satisfies $-I\le T\le I$. The trace-norm variational principle for Hermitian operators gives the diagonal absolute-sum bound for the trace norm:

For $X=\rho -P$, with $P=|\psi ⟩⟨\psi |$, extend $|\psi ⟩$ to an orthonormal basis $|{\psi }_1⟩=|\psi ⟩,|{\psi }_2⟩,\dots$. Put $r=⟨\psi |\rho |\psi ⟩$. The first diagonal entry is $r-1\le 0$, and all the others are nonnegative. Their sum is $1-r$, because $\mathrm{Tr}\rho =1$. The bound therefore yields $\Vert \rho -P{\Vert }_1\ge 2(1-r)$. Using the definitions of trace distance and quantum fidelity,This pure-target lower bound on trace distance is attained whenever $\rho$ has no coherence between $|\psi ⟩$ and its orthogonal complement. The proof used the trace-norm variational principle for Hermitian operators, together with positivity and normalization of a density operator.