Partial trace contracts the trace norm on Hermitian operators

If either tensor factor is zero-dimensional, then $X=0$ and the claimed inequality follows from the empty coordinate sum in [F4]; assume both factors are nonzero for the rest of the proof. For any Hermitian $Z$ on a nonzero finite-dimensional complex-inner-product space, [F3] gives an orthonormal eigenbasis $(e_i)$ with eigenvalues ${\lambda }_i$. The adjoint identity [F11] and conjugate symmetry make ${\lambda }_i=⟨Z{e}_i,e_i⟩=⟨e_i,Z{e}_i⟩=\overline{{\lambda }_i}$, so each ${\lambda }_i$ is real. For every Hermitian $Y$ with $\mathrm{\parallel }Y\mathrm{\parallel }\le 1$, [F12] gives $\mathrm{tr}(ZY)={\sum }_i{\lambda }_i⟨Y{e}_i,e_i{⟩}_{\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}$. Again [F11] makes each diagonal entry real, and [F8]–[F9] put it in $[-1,1]$. Thus $\mathrm{\mid }\mathrm{tr}(ZY)\mathrm{\mid }\le {\sum }_i\mathrm{\mid }{\lambda }_i\mathrm{\mid }=\mathrm{\parallel }Z{\mathrm{\parallel }}_1$: in the eigenbasis, $Z^{†}Z$ has eigenvalues ${\lambda }_i^2$, so [F2] its nonnegative square root has eigenvalues $\mathrm{\mid }{\lambda }_i\mathrm{\mid }$, and [F1], [F2], [F12] give the trace norm sum. Equality is attained by the Hermitian contraction acting by $\mathrm{sgn}({\lambda }_i)$ on each $e_i$ (and by $0$ when ${\lambda }_i=0$). Hence $\mathrm{\parallel }Z{\mathrm{\parallel }}_1={\sup }_{Y=Y^{†},\mathrm{\parallel }Y\mathrm{\parallel }\le 1}\mathrm{\mid }\mathrm{tr}(ZY)\mathrm{\mid }$.

Formula [F4] shows ${\mathrm{Tr}}_{E}X$ is Hermitian: transposing and conjugating its $(i,i^{\mathrm{\prime }})$ entry gives ${\sum }_j\stackrel{‾}{X_{(i^{\mathrm{\prime }},j),(i,j)}}={\sum }_j{X}_{(i,j),(i^{\mathrm{\prime }},j)}$. If $Y=Y^{†}$ and $\mathrm{\parallel }Y\mathrm{\parallel }\le 1$, choose an orthonormal basis $(f_j)$ of $H_E$ by [F10] and write arbitrary $\xi ={\sum }_j{x}_j\otimes f_j$ and $\zeta ={\sum }_j{y}_j\otimes f_j$. The product inner product and tensor-map action [F6]–[F7] give $⟨(Y\otimes I_E)\xi ,\zeta ⟩={\sum }_j⟨Y{x}_j,y_j⟩={\sum }_j⟨x_j,Y{y}_j⟩=⟨\xi ,(Y\otimes I_E)\zeta ⟩$ by self-adjointness and [F11], so $Y\otimes I_E$ is Hermitian. Also [F6]–[F8] give $\mathrm{\parallel }(Y\otimes I_E)\xi {\mathrm{\parallel }}^2={\sum }_j\mathrm{\parallel }Y{x}_j{\mathrm{\parallel }}^2\le {\sum }_j\mathrm{\parallel }{x}_j{\mathrm{\parallel }}^2=\mathrm{\parallel }\xi {\mathrm{\parallel }}^2$, so $\mathrm{\parallel }Y\otimes I_E\mathrm{\parallel }\le 1$.

For the remaining nonzero-factor case, apply the variational identity in step 1.1 to ${\mathrm{Tr}}_{E}X$. For each allowed $Y$, [F5] and step 1.2 give $\mathrm{\mid }\mathrm{tr}(({\mathrm{Tr}}_{E}X)Y)\mathrm{\mid }=\mathrm{\mid }\mathrm{tr}(X(Y\otimes I_E))\mathrm{\mid }\le \mathrm{\parallel }X{\mathrm{\parallel }}_1$, using step 1.1 for the Hermitian contraction $Y\otimes I_E$. Taking the supremum proves the claimed inequality.