Let $\mathcal{A}$ be a $C^\star$ algebra (assume it to be unital, if necessary), and $a,b\in \mathcal{A}_{+}$, that is, $a,b$ are positive elements. It is a well known fact that $\left\Vert{a-b}\right\Vert\leq\max\{\left\Vert a\right\Vert, \left\Vert b\right\Vert\}$. This can be proved very simply by representing $\mathcal{A}$ faithfully on some Hilbert space $\mathcal{H}$: since $a,b,a-b$ are all self-adjoint, their numerical radii are equal to the norms, so $\left\Vert a-b \right\Vert = \sup_{x\in\mathcal{H}, \left\Vert x\right\Vert = 1}\{ | \langle (a-b)x,x \rangle | \} = \sup_{x\in\mathcal{H}, \left\Vert x\right\Vert = 1}\{ | \langle ax,x \rangle – \langle bx,x \rangle| \}$. But, since both $\langle ax,x \rangle$ and $\langle bx,x \rangle$ are positive real numbers, we have $| \langle ax,x \rangle – \langle bx,x \rangle| \leq \max \{ \langle ax,x \rangle , \langle bx,x \rangle \}$. So, we have $\left\Vert a-b \right\Vert \leq \sup_{x\in\mathcal{H}, \left\Vert x\right\Vert = 1}\max\{ \langle ax,x \rangle, \langle bx,x \rangle \} \leq \max \{ \sup_{x\in\mathcal{H}, \left\Vert x\right\Vert = 1} \langle ax,x \rangle, \sup_{x\in\mathcal{H}, \left\Vert x\right\Vert = 1} \langle bx,x \rangle \} = \max\{\left\Vert{a}\right\Vert, \left\Vert{b}\right\Vert \}$.
I am trying to prove this without representing $\mathcal{A}$ on some Hilbert space, if I may say, by using only techniques from spectral theory rather than operator-theoretic techniques. There is a very nice such proof when $a,b$ are projections, in another answer, the link being this : all that are used there, are some basic techniques from the spectral theory of $C^\star$ algebras, like the functional calculus. I was wondering if one could provide such a proof for any two positive elements.
Best Answer
First, since $a$ and $b$ are positive, we have $$ -\|b\|1\leq -b\leq a-b\leq a\leq \|a\|1. $$ It is then an easy application of functional calculus to show $\|a-b\|\leq \max\{\|a\|,\|b\|\}$.