If $\text{Tr}_B \rho^{AB}$ is almost pure, then $\rho^{AB}$ is almost a product state? Let $\rho^{AB}$ be a bipartite state, and let $\rho^A$ denote the partial trace. Suppose 
$$ \lVert \rho^A - |\sigma\rangle\langle\sigma|^A \rVert_1 \leq \varepsilon $$ 
for some pure state $|\sigma\rangle$. Here $\lVert\cdot\rVert_1$ denotes the trace norm. Is it true that
$$ \lVert \rho^{AB} - |\sigma\rangle\langle\sigma|^A\otimes \rho^B \rVert_1 = O(\varepsilon) $$
This holds in the exact case, that is if the partial trace of a bipartite state is pure, then it is a product state (see reference). I am wondering if this approximate version is true as well. Any thoughts are appreciated!
 A: Let us purify $\rho_{AB}$ to $\psi_{ABR} \in H_A\otimes H_B\otimes H_R$. You are given that
$$ \lVert \rho_A - |\sigma\rangle\langle\sigma|_A \rVert_1 \leq \epsilon $$
By a tight version of Fannes inequality, we have
$$S(\rho_A) \leq \epsilon\log d + H(\epsilon, 1-\epsilon), $$
where $S(A)$ is the von Neumann entropy, $H(X)$ is the binary entropy and $d$ is the dimension of $H_A$. This gives us a lower bound on the largest eigenvalue of $\rho_A$ i.e. $\lambda_1 > 1 - \delta$. I have not worked out $\delta = \delta(\epsilon)$ here but I think this should be possible.
Meanwhile, the Schmidt decomposition of $\psi_{ABR}$ is
$$\psi_{ABR} = \sum_i \sqrt{\lambda_i}\vert i\rangle_A\otimes \vert \tilde{i}\rangle_{BR}$$
Taking the trace over $R$ of $\vert\psi\rangle\langle\psi\vert_{ABR}$ and denoting $\text{Tr}_R \vert\tilde{i}\rangle\langle\tilde{i}\vert = \omega^i_B$, we have
\begin{align}
\rho_{AB} &= \lambda_1\vert 1\rangle\langle 1\vert\otimes\omega^1_B + \sum_{(i,j)\neq (1,1)} \sqrt{\lambda_i\lambda_j} \vert i\rangle\langle j\vert\otimes\text{Tr}_R(\vert\tilde{i}\rangle\langle\tilde{j}\vert) \\
&= \lambda_1\vert 1\rangle\langle 1\vert\otimes\omega^1_B + O(\delta)
\end{align}
Thus, we have
$$\lVert \rho_{AB} -  \vert 1\rangle\langle 1\vert\otimes\omega^1_B \rVert_1 \leq O(\delta)$$
