Ordering: Normalized

Question

Given a Hilbert space $1\in\mathcal{A}$.

Denote selfadjoints: $$\mathcal{S}:=\{A\in\mathcal{A}:A=A^*\}$$

Introduce an order: $$A\leq A':\ \ \sigma(A'-A)\geq0$$

Regard a projection: $$P\neq0:\quad P^2=P=P^*$$

Then one has: $$P\leq A\leq1\implies\|A\|=1$$

How can I check this?

(Operator-algebraic proof?)

Answer 1

Since you have to relate the norm with the spectrum, I don't think this has an elementary algebraic proof.

Facts needed (from the theory of Banach algebras):

1. For any $A\in\mathcal S$, $\sigma(A)=\{\phi(A):\ \phi\in S(\mathcal A)\}$, where $S(\mathcal A)$ is the state space.

2. For any $A\in\mathcal S$, $\|A\|=\text{spr}(A)=\max\{|\lambda|:\ \lambda\in\sigma(A)\}$.

From those two facts you easily deduce that $\|P\|=1$. Whenever $0\leq B\leq A$, we have $\phi(B)\leq\phi(A)$ for any state $\phi$, so $\|B\|=\text{spr}(B)\leq\text{spr}(A)=\|A\|$.

As $\|1\|=1$, you are done.