TITLE: inequality for operators
QUESTION [6 upvotes]: 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.

REPLY [4 votes]: 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 $-0.02132$.