Partial ordering of the positivity in a C-star algebra Let $A$ be a C-$\ast$ algebra. Assume that given $(a,\,b)\in A^2$, we have $$ a\geq b$$ and we also know that $a\geq0$.
Does it hold that $$\|a\|\geq\|b\|\text{ ?}$$
What I have tried:
Let $A$ be represented on a Hilbert space $\mathcal{H}$ with inner product $\langle\cdot\,,\cdot\rangle$. Then we know that $a\geq b$ implies that for all $v\in\mathcal{H}$ we have $$ \langle v, av\rangle  \geq \langle v, bv\rangle.$$
We now represent $\|a\|$ as $$\|a\|=\sup_{\|u\|=1,\|v\|=1}|\langle u,av\rangle|$$
Now I tried to reach something like for all $u, v\in\mathcal{H}$ we have $$ |\langle u, av\rangle | \geq |\langle u, bv\rangle | $$ using the polarization identity, however, I got stuck. 
I also thought about using the spectral radius formula.
Another idea was to try to prove that $$a^2 \geq b^2$$
Any ideas would be greatly appreciated. Perhaps this is false? 
 A: Yes it is true, you do not need to consider a (faithful) representation of your algebra on a Hilbert space.
Let $a$ be positive, so also hermitian. Then you have $\|a\|=\sup_{x\in\sigma(a)}|x|$. From this you find that the spectrum of $\|a\|-a$ (which is $-\sigma(a)$ translated by $\|a\|$) contains only positive elements and then $\|a\|-a$ is positive. This means
$$\|a\|≥a≥b$$
Now since $b$ is also self-adjoint you have again $\|b\|=\sup_{x\in\sigma(b)}|x|$. But since $\|a\|-b$ is positive (from $\|a\|≥b$) you have that the spectrum of $\|a\|-b$ must contain only numbers $≥0$. But the spectrum of it is precisely $\|a\|-\sigma(b)$. If $\sigma(b)$ contains only positive numbers (ie $b$ is positive) so $\|a\|≥\sup_{x\in\sigma(b)}|x|=\|b\|$ follows.
If $\sigma(b)$ has negative parts then the statement does not necessarily follow, as can be seen from considering $b=-2\|a\|$. But you can write any hermitian element as the difference of two positive elements: $b=b^+-b^-$. You can follow $\|a\|≥\|b\|$ if you can choose such a decomposition of $b$ so that $a≥b^+$ and $a≥b^-$.
