Does $\|\cdot\|$ in a general Banach $\ast$ Algebra preserve order? Given a general Banach $\ast$-Algebra $\mathcal{B}$ and $a, b \in \mathcal{B}$ such that $a > b > 0$ (i.e. we have $a^* = a, b^* = b, (a - b)^* = a - b$ and $\sigma(a), \sigma(b), \sigma(a - b) \subseteq [0, \infty)$), is it always true that $\|a\| > \|b\|$? Does $a > b$ necessarily implies $\sigma(b) \subsetneq \sigma(a)$
In $C^*$ Algebra since the Gelfand mapping $\Gamma$ is isometry, given $a > b > 0$ we can tell $\|a\| > \|b\|$. However in a general Banach $\ast$-Algebra I do not know if this is true. 

Consider $C[0,1]$. Here $\sigma (a-b) \subseteq [0,\infty)$ iff $f(x) \geq g(x)$ for all $x$. But $\|f\| \geq \|g\|$ neither implies nor is  implied by this.
