equivalent condition of an element being positive in a $C^*$-algebra Let $\mathcal{A}$ be a $C^*$-algebra. Suppose $a,b\in\mathcal{A}$ with $a,b\geq 0$ and $\Vert a\Vert\leq\Vert b\Vert$. Does it imply $a\leq b$?
Comments: I could not be able to either prove it or find a counter-example. Any comment is highly appreciated. Thanks in advance.
 A: The only C$^*$-algebras where $\|a\|\leq\|b\|\implies a\leq b$ for all positive $a,b$ are $A=\{0\}$ and $A=\mathbb C$.
In any other C$^*$-algebra, there exists a positive element $b$ with at least two points in its spectrum (a C$^*$-algebra where this is false is either $\{0\}$ or $\mathbb C$). So $\sigma(b)\subset[\lambda_1,\lambda_2]$, with $\lambda_1,\lambda_2\in\sigma(b)$. Let $f\in C(\sigma(b))$ be monotone, with $f(\lambda_1)>\lambda_1$ and $f(\lambda_2)=\lambda_2$. Put $a=f(b)$.
We have
$$
\|a\|=\max\{f(t):\ t\in\sigma(b)\}=f(\lambda_2)=\lambda_2=\max\sigma(b)=\|b\|.
$$
And if $\varphi$ is the state on $A$ obtained from extending the evaluation $g\longmapsto g(\lambda_1)$ from $C(\sigma(b))=C^*(b)$ to all of $A$, we have
$$
\varphi(b-a)=\varphi(b)-\varphi(a)=\lambda_1-f(\lambda_1)<0,
$$
showing that $b-a$ is not positive, that is $a\not\leq b$.

The above phenomenon can be seen fairly explicitly in examples. What the above is doing is to start with $b$ and then form $a$ by rising the minimum of $b$'s spectrum, without changing the maximum so that the norm is kept. For instance if $A=\mathbb C^2$, take $b=(1,0)$ and $a=(1,1)$. Then $\|b\|=\|a\|=1$, but $b-a=(0,-1)$ is not positive.
