Combining parts (iii) and (iv), then multiplying by $1+\varepsilon$, gives
$$
H(A|B)_\sigma-H(A|B)_\rho
\leq\varepsilon\bigl(H(A|B)_{\Delta'}-H(A|B)_\Delta\bigr)
+(1+\varepsilon)H\!\left(\frac{\varepsilon}{1+\varepsilon}\right).
$$
The <dimension bound for quantum conditional entropy> is
$$
-\log d_A\leq H(A|B)_\tau\leq\log d_A.
$$
Indeed, <Subadditivity of Von Neumann entropy> gives $H(A|B)_\tau\leq S(\tau_A)\leq\log d_A$, while the <Araki–Lieb inequality> gives $H(A|B)_\tau\geq-S(\tau_A)\geq-\log d_A$. Hence
$$
H(A|B)_\sigma-H(A|B)_\rho
\leq2\varepsilon\log d_A
+(1+\varepsilon)H\!\left(\frac{\varepsilon}{1+\varepsilon}\right).
$$
Interchanging $\rho$ and $\sigma$ proves the <continuity bound for quantum conditional entropy>:
$$
\boxed{|H(A|B)_\rho-H(A|B)_\sigma|
\leq2\varepsilon\log d_A
+(1+\varepsilon)H\!\left(\frac{\varepsilon}{1+\varepsilon}\right)}.
$$

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 / v / Solution

If $\varepsilon=\lVert\rho_{AB}-\sigma_{AB}\rVert_1/2$ and $d_A=\dim\mathcal H_A$, then
$$
|H(A|B)_\rho-H(A|B)_\sigma|
\leq2\varepsilon\log d_A+(1+\varepsilon)H\!\left(\frac{\varepsilon}{1+\varepsilon}\right).
$$
The proof couples the two states through their positive and negative differences, then combines <concavity of quantum conditional entropy> with the <entropy bound for a binary mixture>.

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 Incoming links: Continuity bound for quantum conditional entropy