Let $x,y$ be two positive elements in a $C^\ast$ algebra $\mathcal{A}$. Then $\lVert x-y\rVert\le\text{max}\{\lVert x\rVert,\lVert y\rVert\}$

Question

If $x,y$ are two positive elements in a $C^\ast$ algebra $\mathcal{A}$, then $0\le x\le \lVert x\rVert 1$ and $0\le y\le \lVert y\rVert1$. This implies $-\lVert y\rVert 1\le x-y\le\lVert x\rVert 1\implies $

$$-\text{max}\{\lVert x\rVert,\lVert y\rVert\}1\le -\lVert y\rVert 1\le (x-y)\le \lVert x\rVert1\le \text{max}\{\lVert x\rVert,\lVert y\rVert\}1$$

But from here, can I say anything about $\lVert x-y\rVert\le\text{max}\{\lVert x\rVert,\lVert y\rVert\}$?

Can anyone help me finish the proof? Thanks for your help in advance.

Answer 1

To finish the argument what you need to notice is that $x-y$ is selfadjoint, and use the following:

If $a$ is selfadjoint and $-\lambda\,1\le a\leq \lambda\,1$, then $\|a\|≤\lambda$.

There are a few ways to prove this:

• use that $\|a\|=\max\sigma(a)$, and that $\sigma(a)=\{\gamma(a):\ \gamma\in\widehat{C^*(a)}\}$.

• use the Gelfand transform, that allows you to see $a$ as the identity function over $\sigma(a)$, which gives you $\sigma(a)\subset[-\lambda,\lambda]$.

• use that there is a state $\varphi$ such that $|\varphi(a)|=\|a\|$. As $-\lambda\leq\varphi(a)\leq\lambda$, we have $|\varphi(a)|≤\lambda$.