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.

Related: Does $|\langle x,Ax\rangle |=\langle x,|A|x\rangle$ for bounded operators on Hilbert space?


$|\langle Ax, x\rangle|\leq \langle |A|x,x\rangle$ 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.

  • $\begingroup$ If I am not missing something, doesn't your example give $2 \leq 4$? $\endgroup$ – Ian Apr 23 '18 at 4:55
  • $\begingroup$ @lan sorry,I have corrected my answer, and there is also an example in the answer of the related problem you give. $\endgroup$ – C. Ding Apr 23 '18 at 5:02
  • $\begingroup$ +1 I'm aware of the definition, which made me ask this question: math.stackexchange.com/questions/2749799 $\endgroup$ – Bart Michels Apr 23 '18 at 6:36

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