For a C*-algebra $A$ and arbitrary $x\in A$, is $\overline{x^*Ax}$ and $\overline{x^*xA x^*x}$ the same algebra?

Question

It is obvious that $\overline{x^*xx^*Axx^*x}=\overline{x^*xAx^*x}$ but I have no idea whether $\overline{x^*Ax}=\overline{x^*xAx^*x}$ or not.

Any hint would be appreciated.

Answer 1

My proof. Maybe not a right one. I will write $x^*Ax$ instead of $\overline{x^*Ax}$ in my proof.

It is obviouse that $x^*xAx^*x\subseteq x^*Ax$ , so it remains to show that $x^*Ax\subseteq x^*xAx^*x$.

Note that $x^*xAx^*x=x^*xx^*xAx^*xx^*x\subseteq x^*xx^*Ax x^*x\subseteq x^*xAx^*x$ and $x^*xx^*Axx^*x=|x|^2x^*Ax|x|^2=|x|x^*Ax|x|$.

Let $p$ be an arbitrary polymial, then $p(|x|)|x|x^*Ax|x|p(|x|)\subseteq |x|x^*Ax|x|$ since $|x|\in |x|x^*Ax|x|$.

Define $f_\delta$ such that $f_\delta(t)=\left\{\begin{array}{}0&,t\in [0,\delta/2]\\2/\delta-1/t&,t\in [\delta/2,\delta]\\1/t&,t\in [\delta,\|x\|]\end{array}\right.$

By weierstrass approximation theorem there is a polymial $p_\delta$ such that $\|p_\delta(t)-f_\delta(t)\|\leq \delta$.

Take arbitrary $x^*ax\in x^*Ax$ . We see $\begin{align}&\left\|p_\delta(|x|)|x|x^*ax|x|p_\delta(|x|)-x^*ax\right\|\\\leq&\left\|p_\delta(|x|)|x|x^*ax|x|p_\delta(|x|)-p_\delta(|x|)|x|x^*ax\right\|+\left\|p_\delta(|x|)|x|x^*ax-x^*ax\right\| \\\leq &\left\|p_\delta(|x|)|x|x^*a\right\|\cdot\left\|x|x|p_\delta(|x|)-x\right\|+\left\|p_\delta(|x|)|x|x^*-x^*\right\|\cdot\left\|ax\right\|\\\leq&\left(\delta(1+\|x\|)+1\right)\|a\|\cdot \|x\|\cdot \left\|x|x|p_\delta(|x|)-x\right\|\end{align}$

To calculate $\left\|x|x|p_\delta(|x|)-x\right\|$ , use the polar decomposition of $x$ in $A''$. Write $x=u|x|$, then $\left\|x|x|p_\delta(|x|)-x\right\|\leq \left\||x|^2p_\delta(|x|)-|x|\right\|\leq \delta\|x\|(\|x\|+1)$ .

Thus, $x^*Ax\subseteq |x|x^*Ax|x|=x^*xx^*Axx^*x\subseteq x^*xAx^*x\subseteq x^*Ax$.