Entropy density agreement does not force local state agreement

On $V={\mathbb{C}}^2$, the supplier [F3] identifies the stated canonical one-site density as $\gamma =I\mathrm{/}2$ and its $N$-site density as ${\gamma }_N=(I\mathrm{/}2{)}^{\otimes N}$. For each $N\ge 1$, set ${\gamma }_N=(I\mathrm{/}2{)}^{\otimes N}$ and ${\tau }_N=\mathrm{\mid }1⟩⟨1\mathrm{\mid }\otimes (I\mathrm{/}2{)}^{\otimes (N-1)}$. Both are positive of trace one, so are finite tensor densities. Their positive eigenvalues are respectively $2^{-N}$ with multiplicity $2^N$ and $2^{-(N-1)}$ with multiplicity $2^{N-1}$. By [F1], $s_N({\gamma }_N)=N\log 2$ and $s_N({\tau }_N)=(N-1)\log 2$; therefore $\mathrm{\mid }{s}_N({\tau }_N)-s_N({\gamma }_N)\mathrm{\mid }\mathrm{/}N=(\log 2)\mathrm{/}N\to 0$.

By [F2], the one-site reductions are $\gamma$ and $\mathrm{\mid }1⟩⟨1\mathrm{\mid }$. Their difference in the ordered basis is $\mathrm{diag}(1\mathrm{/}2,-1\mathrm{/}2)$, whose eigenvalues have absolute sum one. Thus $\mathrm{\parallel }{\tau }_{N,1}-\gamma {\mathrm{\parallel }}_1=1$ for every $N$. Entropy-density agreement alone therefore does not imply local trace-norm convergence.