$a,b\in A$ are selfadjoint elements of $C^*$-algebras, such that $a\le b$, why is $\|a\|\le \|b\|$

Question

Let $A$ be a unital $C^\ast$-algebra with unit $1_A$.

a) Why is $a\le \|a\|1_A$ for all selfadjoint $a\in A$ and

b) If $a,b\in A$ are selfadjoint such that $a\le b$, why is $\|a\|\le \|b\|$?

I would try it with the continuous functional calculus $$\phi:C(\sigma(a))\to A,\; f\mapsto f(a),$$ for a) with $f(x)=|x|-x$, which is a positive function on $\sigma(a)\subseteq \mathbb{R}$, is it correct?

for b) It is $a\le \|a\|$ and $b\le \|b\|$, therefore $a\le \|b\|$. How can I continue?

Regards

Answer 1

a) If $A=C(X)$, where $X$ is a compact Hausdorff space, for any $a,b\in A_{sa}$, $$ a\leq b \Leftrightarrow a(x) \leq b(x) \quad\forall x\in X $$ So if $f\in A_{sa}$ then , $f\leq \|f\|_{\infty}1_A$.

Now, if $a$ is self-adjoint, then $C^{\ast}(a,1_A)$ is abelian and unital, so by Gelfand's theorem, it is isomorphic to $C(X)$ for some compact Hausdorff $X$.

b) Now if $a \leq b$, then by (a), $a\leq \|b\|1_A$. Again look at the abelian $C^{\ast}$-algebra $C^{\ast}(a,1_A)$, and conclude that $\|a\| \leq \|b\|$

Edit: As @TrialandError points out below, part (b) only holds if $0 \leq a$ [it fails trivially even in a commutative $C^{\ast}$-algebra otherwise]

Answer 2

For a) you don't need functional calculus. Note that $\|a\|\,1_A-a$ is a selfadjoint with spectrum contained in $[0,\infty)$, so positive ($\sigma(a+k\,1_A)=\{\lambda+k:\ \lambda\in\sigma(a)$).

As stated, b) is not true: $-3<-2$, but $\|-2\|=2<\|-3\|$. It is true for positives, though. For $0\leq a \leq b$, we represent $A\subset B(H)$, then $$ \langle a\xi,\xi\rangle\leq\langle b\xi,\xi\rangle $$ for all $\xi\in H$. Then $$ \|a\|=\sup\{|\langle a\xi,\xi\rangle:\ \|\xi\|=1\} \leq \sup\{|\langle b\xi,\xi\rangle:\ \|\xi\|=1\} =\|b\|. $$ If you want to do this without representing, the same idea can be achieved using states.

Answer 3

Prologue

The spectrum is not linear in general for several reasons.
So the trick is to reduce the inequality involving only the operator and the identity.

Solution

Spectral relations: $$\sigma(A+1)=\sigma(1)+\sigma(A)\quad\sigma(\lambda A)=\lambda\sigma(A)$$

By Neumann series: $$N^*N=NN^*:\quad\|\sigma(N)\|=\|N\|$$

By spectral relations: $$A=A^*:\quad-\|A\|1\leq A\leq\|A\|1$$

Both together give: $$0\leq A\leq B\implies 0\leq A\leq\|B\|1\implies0\leq\sigma(A)\leq\|B\|\implies\|A\|\leq\|B\|$$

Concluding the proof.

Epilogue

No need of continuous calculus - it is completely elementary!
The proof is based on the very definition of the ordering merely.

Answer 4

let A be a C*-algebra and a,x in A with 0≤xx*≤a. questions: 1- Put bn=(((1/n)+a)^(-1/2))(a^(1/4))x. show that (bn) is a cauchy sequence in A. 2- Show that the limit b in A satisfies x=(a^(1/4))b and ∥b∥≤∥a∥^(1/4)

for question 1- it is easy but for question 2- how i show that ∥b∥≤∥a∥^(1/4)?