General question about functional calculus For any normal element $a$ in a unital C*-algebra $A$ one can define the functional calculus $$\varphi\colon C(\sigma(a))\to A,$$ which is defined to be the unique $*$-homomorphism such that $\varphi(z)=a$, where $z$ is the inclusion map of $\sigma(a)$ in $\mathbb{C}$. Murphy says the following in his book on C*-algebras:

"If $p$ is a polynomial, then $\varphi(p)=p(a)$, so for $f\in C(\sigma(a))$ we may write $f(a)$ for $\varphi(f)$."

But still don't really see why we can do that. For example, if $a$ is positive (that is $a^{*}=a$ and $\sigma(a)\subset[0,\infty)$), then in particular the map $f\colon\sigma(a)\to\mathbb{C}$ defined by $f(t):=(t+1)^{-1}t^{2}$ is well-defined and continuous. Also, $f$ is not a polynomial. So according to Murphy we can write $f(a)$ for $\varphi(f)$, but I don't see why it is in fact true that $f(a)=(a+1_{A})^{-1}a^{2}$.
More generally, what justifies the notation $f(a)$ for a general $f\in C(\sigma(a))$?
 A: The point that Murphy is trying to make is that, even though $f(a)$ cannot be defined algebraically when $f$ is not a polynomial, it can still be defined analytically.  Let me try to explain
When $p$ is a polynomial, one can compute $p(a)$ algebraically.  When $f\in C(\sigma(a))$ is arbitrary, there may be no natural (i.e. algebraic) way to evaluate $f$ at $a$.  However, what we can observe is that for each polynomial $p$, we have 
$$\|p(a)\|=\sup_{\lambda\in\sigma(a)}|p(\lambda)|.$$
Thus the $*$-homomorphism $\varphi:\mathbb C[z]\to A$, $\varphi(p)=p(a)$, satisfies
$$\|\varphi(p)\|=\|p\|_{C(\sigma(a))}.$$
Since polynomials are dense in $C(\sigma(a))$, we can extend $\varphi$ to a $*$-homomorphism $C(\sigma(a))\to A$.  
The intuition for the notation "$f(a)$" is that, although $f(a)$ may not make sense at first glance, $f$ can be well-approximated by polynomials, for which $p(a)$ does makes sense.  The result in Murphy then justifies this by showing that $f(a)$ is well-defined, and has some desirable properties.
