If $\varepsilon=0$, then $\rho_{AB}=\sigma_{AB}$ and the desired continuity bound is immediate, so assume $\varepsilon>0$. Apply the <positive-negative decomposition of a Hermitian operator> to
$$
X=\rho_{AB}-\sigma_{AB}=X_+-X_-.
$$
Because $\operatorname{Tr}X=0$ and $\lVert X\rVert_1=2\varepsilon$, one has
$$
\operatorname{Tr}X_+=\operatorname{Tr}X_-=\varepsilon.
$$
Thus $\Delta_{AB}=X_+/\varepsilon$ is positive with trace one, hence is a <density operator>. Define
$$
\omega_{AB}=\frac{\sigma_{AB}+\varepsilon\Delta_{AB}}{1+\varepsilon}
=\frac{\sigma_{AB}+X_+}{1+\varepsilon}
=\frac{\rho_{AB}+X_-}{1+\varepsilon}.
$$
It is a <convex combination> of states. The equation $\varepsilon\Delta'_{AB}=(1+\varepsilon)\omega_{AB}-\rho_{AB}$ gives
$$
\Delta'_{AB}=X_-/\varepsilon,
$$
which is likewise positive and has trace one.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 323 / 4 / c / i / Solution

The <spectral theorem for normal operators> gives a unique decomposition
$$
A=A_+-A_-,
$$
where $A_+,A_-\geq0$ and their supports are orthogonal. They are called the positive and negative parts of $A$. If $\operatorname{Tr}A=0$, then $\operatorname{Tr}A_+=\operatorname{Tr}A_-=\lVert A\rVert_1/2$.

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 Incoming links: Positive-negative decomposition of a Hermitian operator