Tensor-product state errors add under trace distance

By [F5], write each ${\rho }_j$ in an orthonormal eigenbasis with nonnegative eigenvalues. By [F9] their sum is $\mathrm{tr}{\rho }_j=1$. The tensor-product basis from [F6] diagonalizes ${⨂}_j{\rho }_j$ with eigenvalues all products of those nonnegative eigenvalues. Their sum is ${\prod }_j\mathrm{tr}{\rho }_j=1$, so the product is a density operator; the same argument applies to ${⨂}_j{\sigma }_j$.

Add and subtract the products with their first $r$ factors changed, for $r=0,\dots ,L$, to obtain ${⨂}_{j=1}^L{\rho }_j-{⨂}_{j=1}^L{\sigma }_j={\sum }_{r=1}^L\left({⨂}_{j<r}{\rho }_j\right)\otimes ({\rho }_r-{\sigma }_r)\otimes \left({⨂}_{j>r}{\sigma }_j\right)\mathrm{.}$ This is a finite telescoping identity.

Apply [F7] to the trace-norm triangle inequality for the finite sum in step 1.2. By [F1], [F5], [F8], and [F9], each unchanged state factor has trace norm one; by [F3], the $r$th summand has trace norm exactly $\mathrm{\parallel }{\rho }_r-{\sigma }_r{\mathrm{\parallel }}_1$. Divide by $2$ and use [F2] to obtain the asserted sum.

Apply [F4] to the product output states verified in step 1.1 and the declared finite POVM. The resulting outcome total variation is at most their trace distance, hence at most the sum from step 2.1.