Suppose $A$ is a C*-algebra, $a$ is a hermitian element of $A$. For each continous function $f:\mathbb{R}_+\to \mathbb{C}$, we say $f$ is continuous on $A$ if for every sequence $\{a_\lambda\}$ of positive elements in $A$ converging to $a$, $f(a_\lambda)\to f(a)$.
The assertion "every continuous function $f:\mathbb{R}_+\to \mathbb{C}$ is continuous on $A$" seems wrong, but I want to know the case for the square root function $x\to \sqrt{x}$. In fact, I want to know if $A\to A, a\mapsto |a|$ is continuous.
Anyone knows?
The assertion holds in general. The key fact is that if $a_\lambda$ is close enough to $a$, so is its norm. Thus we may assume that all spectra are contained in a fixed interval $[0,r]$.
Then we may approximate $f$ uniformly: given $\varepsilon>0$, there exists a polynomial $p$ with $\|f-p\|<\varepsilon$. Then \begin{align} \|f(a_\lambda)-f(a)\|&\leq \|f(a_\lambda)-p(a_\lambda)\|+\|p(a_\lambda)-p(a)\|+\|p(a)-f(a)\|\\ \ \\ &\leq \varepsilon+ \|p(a_\lambda)-p(a)\|\\ \ \\ &\leq 2 \varepsilon +\|p(a_\lambda)-p(a)\|. \end{align} With $p$ fixed, we obtain $$ \limsup\|f(a_\lambda)-f(a)\|\leq 2\varepsilon. $$ As $\varepsilon$ was arbitrary and $r$ was fixed, the limsup is zero, and so the limit exists and is zero: $$ \lim_\lambda \|f(a_\lambda)-f(a)\|=0. $$