On closest unitary matrix In this question $\|A\|_p$ is the normalized $p$-th Schatten norm which is defined to be $\left(\mathbb E_{i} \lambda_i^p\right)^{1/p}$, where $\lambda_i$ are singular values of matrix $A$.
Suppose that $A, B\in M_n(\mathbb C)$ are matrices with operator norm at most 1. Suppose that $\|AB-I\|_p < \varepsilon$. Can you prove that there is a unitary matrix $U$ such that $\|A-U\|_p < \varepsilon$?
I already know proof for this when $p=1$ or $2$ or $\infty$, but not for any other $p$.
 A: It turns out the the previous answer had the right ingredients, but in the wrong combination. Here is a cleaner proof.
Notation: Let $s_j(X)$ denote the $j$-th singular value of a matrix $X$ (we assume that singular values are arranged in decreasing order). Similarly, let $\lambda_j(X)$ denote the $j$-th eigenvalue of a Hermitian matrix $X$. Let $S(X)$ denote the diagonal matrix of singular values of $X$.

Lemma 1: If $B \in M_n(\mathbb{C})$ is a contraction, then for any $A \in M_n(\mathbb{C})$ we have
  \begin{equation*}
  s_j(A) \ge s_j(AB),\qquad 1\le j \le n.
\end{equation*}

Proof: Since $B$ is a contraction, we have
\begin{eqnarray*}
  I &\ge& BB^*\\
  AA^* &\ge& ABB^*A^*\\
  \lambda_j(AA^*) &\ge& \lambda_j(ABB^*A^*)\\
  \lambda_j^{1/2}(AA^*) &\ge& \lambda_j^{1/2}(ABB^*A^*)\\
  s_j(A) &\ge& s_j(AB).
\end{eqnarray*}

Theorem 2: Let $A, B \in M_n(\mathbb{C})$ be contractions. Then,
  \begin{equation*}
  \|I - S(A)\| \le \|I-S(AB)\| \le \|I-AB\|,
\end{equation*}
  for any unitarily invariant norm $\|\cdot\|$.

Proof: Using Lemma 1 and that $A$ is a contraction, we have $0 \le 1 - s_j(A) \le 1 - s_j(AB)$ for all $j$. Consequently, it follows that
\begin{equation*}
  \|I-S(A)\| \le \|I-S(AB)\|,
\end{equation*}
for any unitarily invariant norm. Now using a corollary of Lidkskii's majorization (see e.g., [Theorem IV.3.4 in Bha97]), it follows that the following inequality
\begin{equation*}
  \|I-S(AB)\| = \|S(I)-S(AB)\| \le \|I-AB\|,
\end{equation*}
holds for all unitarily invariant norms.

Corollary 3: If for contractions $A$ and $B$, we have $\|AB-I\| < \epsilon$, then there exists a unitary matrix $U$ such that $\|A-U\| < \epsilon$.

Proof: Let $A=UP$ be the polar decomposition of $A$. Then for any unitarily invariant norm $\|\cdot\|$, 
\begin{equation*}
  \|A-U\|=\|P-I\|=\|S(A)-I\|.
\end{equation*}
Combining this equality with Theorem 2, the result is immediate.
A: EDIT The claim below is false (stupid error, as noted in the comments). I will leave the answer below as a potential approach; am checking if the idea can be fixed!

Here is a proof that holds under a slight additional hypothesis, namely, suppose that we also have $\|A^*B^*-I\| < \epsilon$. The conclusion is also stronger than the one requested, as under this additional hypothesis, the claim holds for all unitarily invariant norms, not just Schatten-$p$ norms.

Theorem ([Thm. IX.7.2, Bha97]). Let $A=UP$ be the polar decomposition of $A$. Then, for every unitarily invariant norm $\|\cdot\|$ and every unitary matrix $V$
  \begin{equation*}
  \|A-U\| \le \|A-V\|.
\end{equation*}

Applying this theorem to the direct sum $A\oplus (-A^*)$ we have
\begin{equation*}
 \left\|\begin{pmatrix}
   A & 0\\
   0 & -A^*
  \end{pmatrix} - 
  \begin{pmatrix}
   U & 0\\
   0 & -U^*
  \end{pmatrix}\right\| 
  \le
\left\|\begin{pmatrix}
   A & 0\\
   0 & -A^*
  \end{pmatrix} - 
  V^*\right\|
\end{equation*}
for any unitary matrix $V$. Since $B$ is a contraction, we can dilate it to the unitary matrix
\begin{equation*}
 V := \begin{pmatrix}
  B & (I-BB^*)^{1/2}\\
  (I-B^*B)^{1/2} & -B^*
  \end{pmatrix}.
\end{equation*}
Since the norms involved are unitarily invariant, we therefore obtain
\begin{equation*}
\|(A-U)\oplus(U-A)^*\| \le 
\left\|\begin{pmatrix}
   A & 0\\
   0 & -A^*
  \end{pmatrix}V - 
  I\right\| = \|(AB-I)\oplus (A^*B^*-I)\|.
\end{equation*}
But notice that $s_j(A-U)=s_j(U^*-A^*)$, for each singular value $s_j$ for $1\le j \le n$. 
Therefore, we can conclude the following weak-majorization between the singular values:
\begin{equation*}
 s(A-U) \prec_w \max\{s_j(AB-I),s_j(A^*B^*-I)\}_{j=1}^n.
\end{equation*}
From this under our hypotheses $\|AB-I\|<\epsilon$ and $\|A^*B^*-I\| < \epsilon$, it follows that $\|A-U\| < \epsilon$.
Note: It should be possible to get rid of the extra assumption, but am not sure as of now.
