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This is a question on how to relate two different distances in the matrix setting. Everywhere below, $M_n$ denotes the square matrices $n\times n$ whose entries are in $\mathbb C$. We consider the operator norm $\|\cdot\|$ on $M_n$.

By a subalgebra we mean a $C^*$-subalgebra of $M_n$. If $A\subseteq M_n$ is a subalgebra and $x\in M_n$, define the distance from $x$ to $A$ as the real number $d(x,A)=\inf_{a \in A}\|x-a\|$.

The question is the following: suppose that $A$ is a unital $C^*$-subalgebra of $M_n$ and that $x\in M_n$ is a contraction with $d(x,A)\geq\frac{1}{4}$. Does there exist $u\in A'$ (the commutant of $A$) such that $\|[x,uxu^*]\|\geq\frac{1}{16}$ (or, for what matter, $\frac{1}{64}$, the only important thing is that this number doesn't depend on the choice of $n$, $A$, or $x$)?

Beware, I am not asking whether I can find a $u$ with $\|[u,x]\|\geq\frac{1}{16}$. This is clearly possible since $y=\int_{\mathcal U(A')}uxu^*d\mu(u)$, when I am integrating over the Haar measure on the unitary group of $A'$, is the conditional expectation onto $A''=A$. Since $y\in A$, we have $\|x-y\|\geq\frac{1}{4}$, and thefore there must be a unitary $u\in \mathcal U(A')$ such that $\|x-uxu^*\|\geq\frac{1}{16}$, or in other words, such that $\|[x,u]\|\geq\frac{1}{16}$.

Thanks and best,

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Take $A=\mathbb C\oplus\mathbb C\subset M_2(\mathbb C)$, and $x=\begin{bmatrix} 0&1/4\\0&0\end{bmatrix}.$

Then $A'=A$. If $u\in A'=A$, then $u=\begin{bmatrix} \lambda&0\\0&\mu\end{bmatrix}$, with $\lambda,\mu\in\mathbb T$. Then $$ uxu^*=\begin{bmatrix} 0&\lambda\overline\mu/4\\ 0&0\end{bmatrix}, $$ and $xuxu^*=uxu^*x=0$. In particular, $[x,uxu^*]=0$.

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  • $\begingroup$ Right, make sense, thanks! To follow up though, what if $x$ is positive, or even just self-adjoint? $\endgroup$ – Alessandro Vignati Oct 16 '18 at 3:40
  • $\begingroup$ If you take the same $A$ as above and $x=\begin{bmatrix} 1&1/4\\1/4&1\end{bmatrix}$, then $x\geq0$ and $xuxu^*=uxu^*x$ for all unitary $u\in A$. $\endgroup$ – Martin Argerami Oct 16 '18 at 4:31
  • $\begingroup$ No..In particular in the top left corner of $uxu^*x$ you'll get $1+\frac{\lambda\overline\mu}{4}$ but in the top left corner you'll get it's adjoint, and unless $\lambda=\mu$ they're different. $\endgroup$ – Alessandro Vignati Oct 16 '18 at 12:19
  • $\begingroup$ Yes, you are right. $\endgroup$ – Martin Argerami Oct 16 '18 at 14:08

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