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Let $A$ be a $\mathrm{C}^\ast$-algebra and $\rho$ a positive linear functional.

Is it true that for all $a\in A$:

$$|\rho(a)|\leq \rho(|a|),$$ where $|a|=\sqrt{a^\ast a}$?

This is not difficult to show when we restrict to $A_{\operatorname{sa}}$. Indeed if $a$ is self-adjoint, $a=a_+-a_-$ and $|a|=a_++a_-$ and so:

$$|\rho(a)|=|\rho(a_+)-\rho(a_i)|\leq |\rho(a_+)|+|\rho(a_-)|=\rho(a_++a_-)=\rho(|a|).$$

I also understand that I can write $a=u|a|$ for a unitary $u$ but I can't quite seem to be able to put the two sticks together. I know also that

$$|\rho(a)|^2\leq \rho(|a|^2).$$

I want to rewrite Hanno's argument here to a slightly different situation and if the above is indeed true then I have the result. There are probably different ways of doing it but this seems the obvious way.

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This inequality need not hold if the element $a$ is not self-adjoint. Here's a counterexample.

Suppose that $A=M_2(\mathbb C)$ and that $$a=\left( \begin{array}{cc} 0 & 1 \\ 2 & 0 \end{array} \right). $$ Then $a$ has polar decomposition $$a=u|a|= \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right) \left( \begin{array}{cc} 2 & 0 \\ 0 & 1 \end{array} \right). $$ If $\rho$ is the positive linear functional on $M_2(\mathbb C)$ given by $$ \rho(x)= \left( \begin{array}{cc} \frac{3}{5} & \frac{4}{5} \end{array} \right) \left( \begin{array}{cc} x_{11} & x_{12} \\ x_{21} & x_{22} \end{array} \right) \left( \begin{array}{c} \frac{3}{5} \\ \frac{4}{5} \end{array} \right), $$ then $ \rho(|a|)=\frac{34}{25}, $ but $\rho(a)=\frac{36}{25}$.

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  • $\begingroup$ Thanks very much for this Tom. $\endgroup$ – JP McCarthy May 22 '17 at 17:13

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