Let $C
\_p$ be the Schatten-p-classes on a separable Hilbert spaces, $p\ge 1$.
Let ${\rm Tr}$ be the standard trace.
Let $y\in C\_p$ be a self-adjoint operator (or even a positive operator) and let $x\_n \in C\_{p'}$, $\frac1p+\frac1{p'}=1$, be a sequence of self-adjoint operators with $\|x\_n\|\_{p'}=1$ such that ${\rm Tr}(x\_n^\*y)\to \|y\|\_p$.
Let $u\_n |x\_n|=x\_n $ be the polar decomposition.
Do we have  

$$\|u\_n^\* y - |y|\| \_p \to 0, ~{
\rm as~}n\to \infty?$$



norm estimates for Schatten class
I understand that you are happy with the case $y\geq 0$, so let's assume that for simplicity (I expect that small modifications should deal with arbitrary self-adjoint $y$). In that case, **and if** $1<p<\infty$, it is true that $\| u\_n^\* y - y\|\_p \to 0$. I am not sure about the extreme cases $p=1,\infty$, where the uniform convexity argument breaks down.


Here is the argument. It should work more generally in the non-commutative $L\_p$ space of an arbitrary von Neumann algebra. Without loss of generality, we can assume that $\|y\|\_p=1$. In that case, the unique $x \in C\_{p'}$ of norm one such that $Tr(x^\* y) = 1$ is $y^{\frac{p}{p'}}$. By the uniform convexity of $C\_{p'}$, the assumption that $\|x\_n\|\_{p'} \leq 1$ and $Tr(x\_n^\* y) \to 1$ implies that $\| x\_n - y^{\frac{p}{p'}}\|\_{p'} \to 0$.


Now consider the Mazur map $z=u|z| \in C\_{p'} \mapsto M\_{p',p}(z) := u |z|^{\frac{p'}{p}} \in C\_p$. It is known that it is continuous (and even uniformly continuous when restricted to the unit ball, see for example the papers by [Yves Raynaud](https://jot.theta.ro/jot/archive/2002-048-001/2002-048-001-003.pdf) or [Eric Ricard](https://arxiv.org/pdf/1407.8334.pdf) for the optimal estimates). Therefore, we obtain
$$\| u\_n |x\_n|^{\frac{p}{p'}} - y \|\_p = \| M\_{p',p}(x\_n) - M\_{p',p}(y^{\frac{p}{p'}})\|\_p \to 0.$$


$x\_n$ being self-adjoint, we have $u\_n = P\_n^+-P\_n^-$ for orthogonal projections $P\_n^+$ and $P\_n^-$ commuting with $x\_n$ and $|x\_n|$. Multiplying on both sides by $P\_n^-$ the preceding inequality, we get
$$ \| P\_n^- |x\_n|^{\frac{p}{p'}} + P\_n^- y P\_n^-\|\_p \to 0.$$
But since $0 \leq P\_n^- |x\_n|^{\frac{p}{p'}} \leq P\_n^- |x\_n|^{\frac{p}{p'}} + P\_n^- y P\_n^-$ ($y$ is positive), we get
$$\||x\_n|^{\frac{p}{p'}}- u\_n |x\_n|^{\frac{p}{p'}}\|\_p =2\|P\_n^- |x\_n|^{\frac{p}{p'}} \|\_p \leq 2\| P\_n^- |x\_n| + P\_n^- y P\_n^-\|\_p \to 0.$$


This allows to conclude:
$$ \|u\_n^\* y - y\|\_p \leq \|u\_n^\*(y - u\_n |x\_n|^{\frac{p}{p'}})\|\_p + \| |x\_n|^{\frac{p}{p'}} - u\_n |x\_n|^{\frac{p}{p'}}\|\_p + \| u\_n |x\_n|^{\frac{p}{p'}}-y\|\_p$$
goes to zero because each term does.


