norm estimates for Schatten class 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
$$\|u_n^* y - |y|\| _p \to 0, ~{
\rm as~}n\to \infty?$$
 A: 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 or Eric Ricard 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.
