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I'm working on this paper. What I'm intersted in is this theorem:

theorem 3.2

where $M_{A,B}(X)=AXB$

I don't know why we can find those two sequences $X_n$ and $x_n$, either I'm finding difficulties to show the lines highlited can someone help.

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  1. It's the definition of the norm. You have $$ \|M_{A,B}+M_{C,D}\|=\sup\{\|AXB+CXD\|:\ \|X\|=1\}, $$ which gives you the $X_n$. Then using the definition of $\|AXB+CXD\|$ gives you the $x_n$.

  2. The inequalities provided, together with the hypothesis, give you $$ \|A\|\,\|B\|+\|C\|\,\|D\|\leq\lim\|A\|\,\|Bx_n\|+\|C\|\,\|Dx_n\|\leq\|A\|\,\|B\|+\|C\|\,\|D\|. $$ The equality then forces $\|B\|=\lim\|Bx_n\|$, $\|D\|=\lim \|Dx_n\|$.

  3. From $\|Bx_n\|^2\to\|B\|^2$ you have $$ \langle B^*Bx_n,x_n\rangle\to \|B\|^2=\langle \|B\|^2x_n,x_n\rangle. $$ Because $B^*B\leq\|B\|^2$, you get $$ \langle (\|B\|^2-B^*B)x_n,x_n\rangle\to0. $$ This you can rewrite $$ \|(\|B\|^2-B^*B)^{1/2}x_n\|^2=\langle (\|B\|^2-B^*B)x_n,x_n\rangle\to0. $$ Then $$0\leq \|(\|B\|^2-B^*B)x_n\|^2\leq\|\,\|B\|^2-B^*B\|^{1/2}\,\|^2\,(\|B\|^2-B^*B)^{1/2}x_n\|^2\to0. $$ That is, $\|(\|B\|^2-B^*B)x_n\|\to0$.

  4. You have $$ z_n=x_n-\frac{B^*y_n}{\|B\|}=x_n-\frac{B^*Bx_n}{\|B\|^2}=\frac{(\|B\|^2-B^*B)x_n}{\|B\|^2}\to0. $$ Then \begin{align} \langle AX_nBx_n,CX_nDx_n\rangle &=\langle C^*AX_nBx_n,X_nDx_n\rangle\\[0.3cm] &=\langle C^*AX_n\|B\|y_n,X_nD\Big(\frac{B^*y_n}{\|B\|}+z_n\Big)\rangle\\[0.3cm] &=\langle C^*AX_ny_n,X_nD\Big(B^*y_n+z_n\|B\|\Big)\rangle\\[0.3cm] &=\langle C^*AX_ny_n,X_nDB^*y_n\rangle+\|B\|\,\langle C^*AX_ny_n,X_nD z_n\rangle\\[0.3cm] \end{align} where $$ \epsilon_n=\|B\|\,\langle C^*AX_ny_n,X_nD z_n\rangle $$ satisfies $$ \|\epsilon_n\|\leq\|B\|\,\|C\|\,\|A\|\,\|D\|\,\|z_n\|\to0. $$

  5. Since $\epsilon_n\to0$, $$ \lim \langle C^*AX_ny_n,X_nDB^*y_n\rangle =\lim\langle AX_nBx_n,CX_nDx_n\rangle=\|A\|\,\|B\|\,\|C\|\,\|D\|. $$ This last equality is due to the following. From $(3.2)$ we get \begin{align} \big(\|A\|\|B\|+\|C\|\|D\|\big)^2 &=\lim_n\!\|AX_nBx_n\|^2\!+\!\|CX_nDx_n\|^2\!+\!2\!\operatorname{Re}\!\langle AX_nBx_n,CX_nDx_n\rangle\\[0.3cm] &\leq \big(\|A\|\,\|B\|\big)^2+\big(\|C\|\,\|D\|\big)^2+2\|A\|\,\|B\|\,\|C\|\,\|D\| \end{align} This is then an equality, and since each of the three terms in the middle is bounded by its corresponding term on the right, they have to be equal. In particular, $$ \lim_n\operatorname{Re}\langle AX_nBx_n,CX_nDx_n\rangle=\|A\|\,\|B\|\,\|C\|\,\|D\|. $$ If the real part of a complex number is equal to (a bound of) its absolute value, then the number is positive and equal to such bound. Thus $$ \lim_n\langle AX_nBx_n,CX_nDx_n\rangle=\|A\|\,\|B\|\,\|C\|\,\|D\|. $$

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  • $\begingroup$ in the line where you write : This you can rewrite $$ \|(\|B\|^2-B^*B)^{1/2}x_n\|=\langle (\|B\|^2-B^*B)x_n,x_n\rangle\to0. $$ I think you might write $$ \|(\|B\|^2-B^*B)^{1/2}x_n\|^2=\langle (\|B\|^2-B^*B)x_n,x_n\rangle\to0. $$ and the lines below changes two $\endgroup$
    – Mary Maths
    Commented May 18, 2022 at 8:53
  • $\begingroup$ For the last limit, I don't see why it is equal to $\|A\|\|B\|+\|C\|\|D\|$, because we have the equality whith the norm, i.e: $lim\|AX_nB_x+CXnDx_n\|=\|A\|\|B\|+\|C\|\|D\|$ $\endgroup$
    – Mary Maths
    Commented May 18, 2022 at 9:28
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    $\begingroup$ Actually, it's not that but rather that there is no sum. I have edited more details in. $\endgroup$ Commented May 18, 2022 at 14:17

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