Let $A$ be a $C^{*}$-algebra. If $0 \le a \le b$, then $\|a\| \le \|b\|$.
We may suppose that $A$ is unital. Then, let $B=C^*({1,b})$ be the $C^*$ algebra generated by $1$ and $b$. Then the Gelfand representation $$\phi : B \to C_0(\Omega(B)), \phi(b)=\hat{b}$$ is an isometric $*$-isomorphism.
Then $\hat{b}({\tau})=\tau(b) \ge 0$ since $b \ge 0$. Then $\|b\|=\ |\hat{b}\|=\sup_{\|\tau\|=1}\tau(b) \ge \tau(b)$. Thus $\tau(\|b\|-b) \ge 0$ This happens for all $\tau \in \Omega(B)$. But $$\sigma_{B}(\|b\|-b)=\sigma_A(\|b\|-b)=\{\tau(\|b\|-b): \tau \in \Omega(B)\}$$ Hence $\sigma_{B}(\|b\|-b) \subset R^{+}$. So $b \le \|b\|$ which gives that $a \le \|b\| $.
Now Let $C$ be the $C^*$-algebra generated by ${1,a}$. Then the Gelfand representation $$\psi : C \to C_0(\Omega(C)), \phi(a)=\hat{a}$$ is an isometric $*$-isomorphism. Then $||a||=||\hat{a}||=\sup_{||\tau||=1}(\tau(a)) \le \sup_{||\tau||=1}\tau(||b||) \le ||b||$
I am kind of confused here. What is the need of turning to Gelfand Representation for this ??
Is this right enough??
Thanks for the help!!
That's one way to prove it. Hard to avoid some machinery, I guess.
A shorter proof can be achieved by representing $A$ faithfully in some $B(H)$. Then $\langle a\xi,\xi\rangle\leq\langle b\xi,\xi\rangle$ for any $\xi\in H$, and so $$ \|a\|=\sup\{\langle a\xi,\xi\rangle:\ \xi\in H,\ \|\xi\|=1\} \leq \sup\{\langle b\xi,\xi\rangle:\ \xi\in H,\ \|\xi\|=1\}=\|b\|. $$