Show that $\|a\| \leq \|b\|.$

Question

Let $a$ and $b$ be self-adjoint elements in a $C^{\ast}$-algebra $A$ with $-b \leq a \leq b.$ Then show that $\|a\| \leq \|b\|.$

Since $a + b \geq 0$ and $b - a \geq 0$ there exist $c,d \in A$ such that $a + b = c^{\ast} c$ and $b - a = d^{\ast} d.$ So $a = \frac {c^{\ast} c - d^{\ast} d} {2}$ and $b = \frac {c^{\ast} c + d^{\ast} d} {2}.$ So in order to show that $\|a\| \leq \|b\|$ we need to show that $$\|c^{\ast} c - d^{\ast} d\| \leq \|c^{\ast} c + d^{\ast} d\|.$$

This boils down to the following problem $:$

Let $a$ and $b$ be two positive elements in a $C^{\ast}$-algebra $A.$ Then show that $$\|a - b\| \leq \| a + b\|.$$

But I have no idea how to show that. Could anyone please give me some suggestion?

Thanks for your time.

Answer 1

Let $e$, $r$ and $\sigma$ denote the unit, the spectral radius, and the spectrum, respectively. For each selfadjoint elements $c \in A$ we have $r(c)=||c||$. Moreover $r(b)e \pm b \ge 0$ (all spectral values are real and $\ge 0$). Thus $-r(b)e \le a \le r(b)e$ which implies $-r(b) \le \lambda \le r(b)$ $(\lambda \in \sigma(a))$. Thus $||a||=r(a) \le r(b)=||b||$.

Answer 2

Here are two more ways of proving the argument.

First, if you consider $A$ represented as $A\subset B(H)$, then the inequalities $-b\leq a\leq b$ give you $$ |\langle a\xi,\xi\rangle|\leq|\langle b\xi,\xi\rangle|,\qquad\xi\in H. $$ It follows that $$|\langle a\xi,\xi\rangle|\leq\|b\|,\qquad \xi\in H$$ Because $a$ is selfadjoint we have $$\|a\|=\sup\{|\langle a\xi,\xi\rangle:\ \|\xi\|=1\},$$ and so $\|a\|\leq\|b\|$.

A second argument (which is an abstract version of the previous one) is that, given any state $\phi$ on $a$, $$ -\phi(b)\leq\phi(a)\leq\phi(b). $$ Then $$|\phi(a)|\leq|\phi(b)|\leq\|b\|. $$ And now one uses that $$ \|a\|=\sup\{|\phi(a)|:\ \phi\ \text{ state}\}, $$ which is a direct consequence of Gelfand-Naimark.