Partial trace contracts trace distance

Put $X=\rho -\sigma$ and $Y={\mathrm{Tr}}_{H_B}X$. Both are Hermitian. For $Y$, this follows directly from [F1]: conjugating its $(i,i^{\mathrm{\prime }})$ entry and using $X_{(i,j),(i^{\mathrm{\prime }},j)}=\stackrel{‾}{X_{(i^{\mathrm{\prime }},j),(i,j)}}$ gives $\stackrel{‾}{Y_{i{i}^{\mathrm{\prime }}}}=Y_{i^{\mathrm{\prime }}i}$. By [F2], the two partial traces of $\rho ,\sigma$ are density operators, so their trace distance is defined.

For any finite-dimensional Hermitian $Z$, diagonalize it in an orthonormal basis with eigenvalues ${\lambda }_i\in \mathbb{R}$. If $A=A^{\ast }$ and $\mathrm{\parallel }A\mathrm{\parallel }\le 1$, then each diagonal entry $⟨A{e}_i,e_i⟩$ is real and lies in $[-1,1]$: [F7] gives $\mathrm{\mid }⟨A{e}_i,e_i⟩\mathrm{\mid }\le \mathrm{\parallel }A{e}_i\mathrm{\parallel }\mathrm{\parallel }{e}_i\mathrm{\parallel }\le \mathrm{\parallel }A\mathrm{\parallel }\le 1$, where the middle inequality follows from [F6] by scaling a unit vector. Hence $\mathrm{\mid }\mathrm{tr}(ZA)\mathrm{\mid }=\mathrm{\mid }{\sum }_i{\lambda }_i⟨A{e}_i,e_i⟩\mathrm{\mid }\le {\sum }_i\mathrm{\mid }{\lambda }_i\mathrm{\mid }=\mathrm{\parallel }Z{\mathrm{\parallel }}_1$ by [F3]–[F4]. Equality is attained by the self-adjoint contraction $A={\sum }_{{\lambda }_i>0}{P}_i-{\sum }_{{\lambda }_i<0}{P}_i$, so $\mathrm{\parallel }Z{\mathrm{\parallel }}_1={\max }_{A=A^{\ast },\mathrm{\parallel }A\mathrm{\parallel }\le 1}\mathrm{tr}(ZA)$.

For every self-adjoint contraction $A$ on $H_A$, [F4] gives an orthonormal eigenbasis of $A$. If $Ae=\lambda e$ for a unit eigenvector, then [F7] and [F6] give $\mathrm{\mid }\lambda \mathrm{\mid }=\mathrm{\mid }⟨Ae,e⟩\mathrm{\mid }\le \mathrm{\parallel }Ae\mathrm{\parallel }\mathrm{\parallel }e\mathrm{\parallel }\le \mathrm{\parallel }A\mathrm{\parallel }\le 1$. By [F5], the product basis diagonalizes $A\otimes I_{H_B}$ with each eigenvalue repeated, so this tensor operator is a self-adjoint contraction as well. The pairing identity [F1] gives $\mathrm{tr}(YA)=\mathrm{tr}(X(A\otimes I_{H_B}))$, whose absolute value is at most $\mathrm{\parallel }X{\mathrm{\parallel }}_1$ by step 1.2 applied on $H_A\otimes H_B$. Taking the maximum over $A$ in step 1.2 on $H_A$ yields $\mathrm{\parallel }Y{\mathrm{\parallel }}_1\le \mathrm{\parallel }X{\mathrm{\parallel }}_1$.

Apply the trace-distance definition [F3] to $X=\rho -\sigma$ and $Y={\mathrm{Tr}}_{H_B}\rho -{\mathrm{Tr}}_{H_B}\sigma ={\mathrm{Tr}}_{H_B}X$. Step 2.1 gives $2D({\mathrm{Tr}}_{H_B}\rho ,{\mathrm{Tr}}_{H_B}\sigma )=\mathrm{\parallel }Y{\mathrm{\parallel }}_1\le \mathrm{\parallel }X{\mathrm{\parallel }}_1=2D(\rho ,\sigma )$; division by $2$ proves the assertion.