Continuous Functional Calculus $:$ Let $A$ be a unital $C^{\ast}$-algebra and $a$ be a normal element in $A.$ Then there is a unique isometric $\ast$-homomorphism $\phi : C(\sigma(a)) \longrightarrow A$ such that $\phi (\mathbf 1) = 1$ and $\phi(f) = a$ where $\mathbf 1$ and $f$ are the functions on $\sigma (a)$ given by $z \mapsto 1$ and $z \mapsto z$ respectively.
After stating the above theorem a remark has been mentioned in the lecture note I am following $:$
Remark $:$ If a sequence of functions $\{f_n\}_{n \geq 1}$ in $C(\sigma(a))$ converges to some $f,$ then $\phi(f_n) \to \phi (f).$ $\color{red} {\text {So if}}\ \color {red} {f \in \text {Hol} (a)}$ $\color {red} {\text {then}}$ $\color {red} {\phi(f)}$ $\color {red} {\text {coincides with the image of}}$ $\color {red} f$ $\color {red} {\text {obtained from the holomorphic functional calculus of}}$ $\color {red} a$ $\color {red} {\text {in}}$ $\color {red} {A.}$
In the above remark I couldn't understand the sentence highlighted in red. Could anyone give some suggestion on it?
Thanks for your time.
Best Answer
I think the following could be meant: For $f \in Hol(a)$ and $a$ normal there are two definitions of $f(a)$ (the Continuous Functional Calculus and the Holomorphic Functional Calculus). The second is defined by $$ f(a)= \frac{1}{2\pi i} \int_\Gamma f(z)(ze-a)^{-1} dz $$ for a cycle $\Gamma$ sourrounding $\sigma(a)$ with index $1$.
First one can check that they go together for rational functions in $Hol(a)$. By Runge's Theorem there is a sequence $(f_n)$ of rational functions which is compact convergent to $f$. In particualar it is uniformly convergent on $\sigma(a)$. From the compact convergence one gets $f_n(a) \to f(a)$ in the Holomorphic Functional Calculus and uniform convergence on $\sigma(a)$ leads to $f_n(a) \to f(a)$ in the Continuous Functional Calculus. Thus both defintions of $f(a)$ lead to the same element.
Edit (concerning your comment): For a monomial $p(z)=z^k$, choose $R > r(a)$ ($r(a)$ the spectral radius of $a$) and $\Gamma = \gamma$ with $\gamma(t)=R \exp(it)$ $(t \in [0,2\pi])$. As $$ p(z)(ze-a)^{-1} = \sum_{n=0}^\infty z^{k-n-1}a^n \quad (|z| > r(a)) $$ we have $$ p(a)= \frac{1}{2\pi i} \int_\Gamma z^k(ze-a)^{-1} dz = \sum_{n=0}^\infty \frac{1}{2\pi i}\left(\int_\gamma z^{k-n-1} dz\right) a^n = a^k. $$