Does $\|\cdot\|$ in a general Banach $\ast$ Algebra preserve order?

Question

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.

Answer 1

Let $\mathcal B$ be the Banach algebra $C^1_b([0,1])$ of real-valued continuously differentiable functions from $[0,1]\to \mathbb R$ with pointwise multiplication and the standard $C^1$-norm. With identity as trivial involution it is a Banach $*$-algebra.

Define $a(x):=4$, $b_n(x):=\sin(n\pi x)+2$. Then $a>b_n>0$ for all $n$. But $\|b_n\|\to\infty$ for $n\to\infty$.

Answer 2

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.