inequality for operators

Question

Is it true that if $x$ is selfadjoint compact operator acting in Hilbert space $H$, $y$ is positive compact operator acting in Hilbert space $H$ and $-y\leq x\leq y$ then necessarily $|x|\leq |y|$? Here, $|x|=\sqrt{x^*x}$. Thanks, for any Hint.

Answer 1

This is not always true. Since everything is finite-dimensional below, all the operators are compact. Suppose that $$ y= \left( \begin{array}{cc} 1 & -1.5\\ -1.5 & 4 \end{array} \right) \text{ and } x= \left( \begin{array}{cc} 0.4 & 0\\ 0 & -0.4 \end{array} \right), $$ In this case, $y>0$ (being self-adjoint and having positive eigenvalues). We also have $$ |x|=\left( \begin{array}{cc} 0.4 & 0\\ 0 & 0.4 \end{array} \right) $$

You can check for yourself (by hand or with the help of a computer, calculator or website) that $$ y-x = \left( \begin{array}{cc} 0.6 & -1.5\\ -1.5 & 4.4 \end{array} \right)>0, $$ (showing $y>x$) and $$ y+x= \left( \begin{array}{cc} 1.4 & -1.5\\ -1.5 & 3.6 \end{array} \right)>0, $$ (showing $x>-y$) but $$ y-|x| =\left( \begin{array}{cc} 0.6 & -1.5\\ -1.5 & 3.6 \end{array} \right) $$ is not positive, since one of its eigenvalues is approximately $-0.02132$.