Self-adjoint elements $A_1$ and $A_2$ in a $C^\ast$-algebra

Question

Let's say that I have $0\leq A_1\leq A_2$ for self-adjoint elements $A_1,A_2$ in $C^\ast$-algebra.

Claim: $A_2\leq \|A_2\|I$ and $\|A_1\|\leq \|A_2\|$. Edit: I now see how $\|A_1\|\leq \|A_2\|$ works, however how would it be for $A_2\leq \|A_2\|I$?

I cannot see how I can show this claim?

Idea: However what I think we can do in order to show this claim is to consider (for the first claim): $C^\ast (A_2)\cong C(\sigma(A_2))$ where $A_2$ corresponds to the identity function on spectrum $A_2$ and we can also do the same for the second part in the claim, that is we can use $C^\ast (A_1)\cong C(\sigma(A_1))$ with $f(t)=\|A_2\|-t$. Correct me if the idea won't work.

Can we use the following lemma?

Lemma: If $x, y$ are two positive elements of $\mathcal{A}$, then $x+y$ is positive.

I hope that my question makes sense. Thanks in advance.

Answer 1

A self-adjoint element $A$ is positive if and only if $\sigma(A)\subset [0,\infty).$ Let $A$ be a positive element. Then $\sigma(A)\subset [0,\|A\|].$ Hence $\sigma(\|A\|I-A)=[0,\|A\|].$ Therefore $\|A\|I-A\ge 0.$

By the first part, for a positive element $A$ we have $\lambda\ge \|A\|$ if and only if $\lambda I- A\ge 0.$ If $\lambda I-A_2\ge 0 $ then by Lemma we get $\lambda I-A_1\ge 0, $ as $$\lambda I-A_1=(\lambda I-A_2)+(A_2-A_1)$$ Therefore $\lambda \ge \|A_2\|$ implies $\lambda\ge \|A_1\|.$ Hence $\|A_1\|\le \|A_2\|.$