# Is $|\langle Ax,x\rangle| \leq \langle |A|x,x\rangle$ in a complex Hilbert space?

Let $H$ be a complex Hilbert space, $A$ a bounded operator and $|A|$ the unique positive square root of $A^*A$. In a comment in Bounded self adjoint operator can be written as difference of positive operators it is suggested that $|\langle Ax,x\rangle| \leq \langle |A|x,x\rangle$, and that this is a consequence of the Polarization identity. How?

I'm trying to show that $A$ has an absolutely convergent trace $\sum\langle Ae_i,e_i\rangle$ if $|A|$ has.

• How do you define a trace class operator? Commented Apr 23, 2018 at 5:09
• @Aweygan See edit. I meant "has a trace", not is trace class. See also this question: math.stackexchange.com/questions/2749799 Commented Apr 23, 2018 at 6:44
• This is true only if A is a selfadjoint bounded linear operator. Commented Jul 16, 2021 at 11:47

$|\langle Ax, x\rangle|\leq \langle |A|x,x\rangle$ is NOT true for a general bounded operator $A$.
Example. $A=\left(\begin{array} & 0 & 1\\ 0 & 0 \end{array}\right)$,$x=\left(\begin{array} & 2\\ 1 \end{array}\right)$
We call $A$ is trace class if $|A|$ is, which is precisely the definition.
• If I am not missing something, doesn't your example give $2 \leq 4$?