About an inequality between the real part and absolute value of an operator

Question

When dealing with operators on a Hilbert Space, one can define for $A\in\mathcal{B} (\mathcal{H})$ two natural notions imitating the real part and absolute value of a complex number

$$ Re(A) = \frac{A+A^*}{2}, \; \vert A \vert = \sqrt{A^* A}$$

We also have an order relation on self adjoint operators. Some natural inequalities of these objects (say $\vert A \vert \leq \Vert A \Vert I$) hold, but one that I find particularly tempting is:

$$ Re(A) \leq \vert A \vert $$

or even better

$$ \vert Re(A) \vert \leq \vert A \vert $$

My question is whether the last two inequlities are true in general

For example, this would easily give the fact that any trace class operator decomposes into 4 positive trace class operators. The proof of this fact I saw is somewhat tricky and relies on different methods.

Answer 1

Neither of these are true in general. Consider $$A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}. $$ The real part is $$B = \begin{pmatrix} 0 & \frac{1}{2} \\ \frac{1}{2} & 0\end{pmatrix}, $$ And $$ |A| = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}. $$ Then $|A| - B$ has a negative eigenvalue (one can see this from plugging this into wolframalpha; $\frac{1}{2}(1 - \sqrt{2})$ is an eigenvalue), so it is not positive. Thus Re$A$ is not bounded by $|A|$ in general. Also notice that $B = |B|$, so that neither of your proposed inequalities hold.

However, one might ask if this is true in the case that $A$ is self-adjoint or normal. When $A$ is self-adjoint, we have that $|A| - A$ is positive (indeed, one can think of $A$ as the identity function $f(x) = x$ on its spectrum; and it isn't hard to convince onself that $|f| - f \geq 0$ for any real-valued function). Also clearly $|A| - |\text{Re}A| = |A| - |A| = 0 \geq 0$, so both the inequalities hold.

So what about the normal case? I honestly don't know, but I hope the above sufficiently answers your question. I might give an edit later once I give the normal case some thought.

Answer 2

As stated in PStheman's answer, the inequalities from the OP do not hold in general. However, something slightly weaker is true: For every $x\in B(H)$ there exists an isometry $v\in B(H)$ such that $\operatorname{Re}(x)_+\leq v|x|v^\ast$. This is Proposition 2.1 in Akeman, Anderson, Pedersen. Triangle inequalities in operator algebras.

For many applications such as the one stated in the OP, this is sufficient. It is also a key step in the proof of the triangle inequality for operators, which takes the form $$ |x+y|\leq v|x|v^\ast+w|y|w^\ast $$ with isometries $v, w$.