C*-Algebras: Continuity of the absolute value?

Question

In a $C^*$-algebra $\mathcal{A}$, it is possible to define the absolute value $|X| \in \mathcal{A}$ of an element $X \in \mathcal{A}$ via

$|X| := \sqrt{X^*X}$,

where the square root of a positive element $Y=X^*X$ is defined to be the unique positive $Z$ such that $Y=Z^*Z$. When restricting the mapping to the real subspace of self-adjoint elements, this formula simplifies to

$|X| = \sqrt{X^2}$.

Is $X \mapsto |X|$ (or at least its restriction to the subspace of self-adjoints) continuous with respect to the norm topology?

Answer 1

The following result should do the job, I think. It is Lemma 1.2.5 in Rordam's Introduction to K-theory for $C^{\ast}$-algebras

Let $K$ be a non-empty compact subset of $\mathbb{R}$, and $f:K\to \mathbb{C}$ be a continuous function. Let $A$ be a unital $C^{\ast}$-algebra, and let $\Omega_K$ be the set of all self-adjoint elements in $A$ with spectrum contained in $K$. The induced function $$ f:\Omega_K\to A \text{ given by } a\mapsto f(a) $$ is continuous.

Answer 2

You can write your map $|\cdot|$ as a composition:

$$|\cdot|=\sqrt{\phantom{a}}\circ\mu \circ(\mathrm{id}^*,\mathrm{id})\circ\Delta$$ where $\Delta:\mathcal A\to\mathcal A\times \mathcal A$, $x\mapsto(x,x)$, is the diagonal map, $\mathrm{id^*}$ is the involution, $\mu(x,y)=x\cdot y$ the multiplication and $\sqrt{\phantom{a}}:P\to P$ is the square root, defined on the cone of positive elements.

All maps are continuous, so $|\cdot|$ is continuous. To see why the root is continuous look at eg this answer.