Kadison's Inequality

Question

Let $\mathcal{A}$ be a C*-algebra and $\omega$ a positive linear functional.

Is there a simple proof for Kadison's inequality: $$|\omega(A)|^2\leq\|\omega\|\cdot\omega(A^*A)$$

Answer 1

Note that your inequality only is true if $\omega$ is contractive, otherwise it is missing a $\|\omega\|$ on the right-hand-side.

The canonical (and nice) proof uses $2$-positivity. But one can prove it exactly like Cauchy-Schwarz: for $A,B\in \mathcal A$ and $t\in\mathbb C$, $$ 0\leq\omega((B+tA)^*(B+tA))=|t|^2\omega(A^*A)+2\,\text{Re}\,t\omega(B^*A)+\omega(B^*B). $$ This in particular occurs for all $t$ of the form $t=re^{i\theta}$ where $e^{i\theta}\omega(B^*A)=|\omega(B^*A)|$. So we have$$ 0\leq r^2\omega(A^*A)+2\,r\,|\omega(B^*A)|+\omega(B^*B), $$ for all $r\in\mathbb R$. This can only happen if $$ |\omega(B^*A)|^2\leq \omega(A^*A)\omega(B^*B). $$ If $\mathcal A$ is unital, we can take $B=I$ and we get Kadison's inequality: $$ |\omega(A)|^2\leq \omega(A^*A)\,\omega(I)=\|\omega\|\,\omega(A^*A). $$ If $\mathcal A$ is not unital, we can use an approximation of the unit to get the result.

Answer 2

Let $\mathcal{A}$ be unital and $\omega$ be a state. Then $$ 0 \leq \omega(|a - \omega(a){\bf 1}_\mathcal{A}|^2) = \omega(a^*a - a\omega(a^*) - a^* \omega(a) + |\omega(a)|^2{\bf 1}_\mathcal{A}) = \omega(a^*a) - |\omega(a)|^2. $$