About an inequality derived from Uhlmann's theorem

Question

Consider two density operators $\rho, \xi \in \mathrm{D}(\mathcal{H} \otimes \mathcal{L})$ where $\mathrm{dim}(\mathcal{H}) \le \mathrm{dim}(\mathcal{L})$. Define

\begin{align*} \epsilon := \mathrm{min}\{\|\,|\phi\rangle - |\psi\rangle\,\| : &|\phi\rangle,|\psi\rangle\,\in \mathcal{H} \otimes\mathcal{L}, \\&\mathrm{Tr}_{\mathcal{L}}(|\phi\rangle\langle\phi|)=\mathrm{Tr}_{\mathcal{L}}(\rho), \mathrm{Tr}_{\mathcal{L}}(|\psi\rangle\langle\psi|)=\mathrm{Tr}_{\mathcal{L}}(\xi) \}. \end{align*} Then it follows from Uhlmann's theorem that $\mathrm{F}(\mathrm{Tr}_{\mathcal{L}}(\rho), \mathrm{Tr}_{\mathcal{L}}(\xi)) = (1 - \frac{\epsilon^2}{2})^2 $.

Now, in this paper, the authors define $\eta \in \mathbb{R}_{\ge 0}$ by requiring $\mathrm{F}(\mathrm{Tr}_{\mathcal{L}}(\rho), \mathrm{Tr}_{\mathcal{L}}(\xi)) = (1 - \frac{\eta^2}{2})^2$. They proceed to claim, using the above result, that there must exist purifications $|\phi\rangle, |\psi\rangle \in \mathcal{H}\otimes\mathcal{L}$ of $\mathrm{Tr}_{\mathcal{L}}(\rho)$ and $\mathrm{Tr}_{\mathcal{L}}(\xi)$ respectively such that $\|\,|\phi\rangle - |\psi\rangle\,\| \le \eta$.

However, shouldn't the inequality be the other way around, since non-ideal purifications would result in $\|\,|\phi\rangle - |\psi\rangle\,\|$ being greater than the minimum, which in this case must be $\eta$? And why can't we go for the stronger equality, since we can always pick the $|\phi\rangle, |\psi\rangle$ that give $\mathrm{min}(\|\,|\phi\rangle - |\psi\rangle\,\|) = \eta$?

I am sure I am having some trivial misunderstanding of elementary algebra here, but I genuinely can't seem to figure out what.

Answer 1

This is just a case of $x=y$ implies $x \leq y$.

By what you argue in the question, there exist purifications such that $\| |\phi\rangle - |\psi\rangle \| = \eta$ and hence there also exist purifications (the same ones) such that $\| |\phi\rangle - |\psi\rangle \| \leq \eta$. The authors don't need to use the fact that it is an equality simply because it does not give them any advantage in the later proof, assuming an equality would lead to the same result.