A matrix norm inequality II Let $\|\cdot\|$ be the spectral norm, i.e., largest singular value.  The condition number of an invertible complex matrix $X$ is defined as $\kappa(X):=\|X\|\|X^{-1}\|$.  
I am able to prove 
Proposition Let $A, B$ be $n\times n$ positive definite matrices.  If $X$ is an $n\times n$  invertible matrix such that $AXB$ is Hermitian, then \begin{eqnarray*}
  \|X^{-1}AXB\|\le \kappa(X) \|AB\|. 
 \end{eqnarray*} 
In particular, if moreover $X$ in the above proposition is unitary, then  $\|X^{-1}AXB\|\le \|AB\|$. 
I wonder if  $\kappa(X)$ in the above proposition can always be replaced with $1$.
 A: No the  norm of  the left  side  can  be very  large.
For  example $\left\| X^{-1}AXB \right\| $  is  an  unbounded  function in $(x,y)$  where  $A,X,B$ are the  following  matrices:
Put $A=B= \begin{pmatrix} 1&0\\0&2 \end{pmatrix}$  and  $X^{-1}=\begin{pmatrix} -x^3&y\\y&x \end{pmatrix}$
where $(x,y) \in \mathbb{R}^{2} \setminus \{0\}$.
In this  example  $A,B$  are  positive matrices and   $AXB$ is equal to its transpose so is  a  Hermitian matrix.
This  shows  that the  inequality $\left\| X^{-1}AXB  \right\| \leq \left\|   A  B \right\|     $  is  not  necessarily  true.  Because  the  left  side is  unbounded function in $(x,y)$ but  the  right side  is  a  constant  function.
The  left  side  is unbounded  because  of  the  following  argument:
Put  $C=X^{-1}AXB$  then $C_{21}=xy/x^{4}+y^2$. Then $|C_{21}|$  tends  to  $+\infty$  when $y=x^2$  and  $x \to 0$. So $\parallel  C \parallel_{\infty}  = Max |C_{ij}|$  goes  to  infinity.  But  the  later  norm is  equivalent  to the spectral  norm.
