Fixed marginals do not determine global trace-norm convergence

For $N\ge 2$, index each binary factor's ordered basis by $\{0,1\}$; a string $x=(x_1,\dots ,x_N)$ has even parity when ${\sum }_j{x}_j$ is even. Let ${\pi }_N$ be the diagonal matrix with entry $2^{-(N-1)}$ on even strings and zero on odd strings, and let ${\gamma }_N=I\mathrm{/}{2}^N$. There are $2^{N-1}$ even strings: the first $N-1$ bits can be arbitrary and the last bit is uniquely determined. Thus both matrices are nonnegative of trace one. Fix $0\le k<N$ coordinates and their values. Among the $N-k\ge 1$ unassigned coordinates, exactly half of the $2^{N-k}$ completions have the required parity, namely $2^{N-k-1}$. Summing by [F1], the marginal diagonal entry of each fixed $k$-bit string is $2^{N-k-1}\mathrm{/}{2}^{N-1}=2^{-k}$; all off-diagonal entries vanish. Thus $({\pi }_N{)}_k=I\mathrm{/}{2}^k=({\gamma }_N{)}_k$ for every proper block.

On an even string the diagonal difference ${\pi }_N-{\gamma }_N$ is $2^{-(N-1)}-2^{-N}=2^{-N}$; on an odd string it is $-2^{-N}$. There are $2^N$ strings in total, so the absolute eigenvalues sum to $2^N{2}^{-N}=1$. Therefore $\mathrm{\parallel }{\pi }_N-{\gamma }_N{\mathrm{\parallel }}_1=1$ for every $N$.

Let $P_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}$ be the diagonal matrix with entry one on even strings and zero on odd strings. Both $P_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}$ and $I-P_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}$ are nonnegative and sum to $I$, so they form a two-outcome POVM by [F2]. Their first-effect trace pairings are $\mathrm{Tr}({\pi }_N{P}_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}})=1$ and $\mathrm{Tr}({\gamma }_N{P}_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}})=1\mathrm{/}2$, a gap of $1\mathrm{/}2$. Hence equality of all proper local marginals does not entail global trace-norm convergence or equality of global POVM trace pairings. This is a mathematical separation; no physical probability interpretation is adopted.