# Banach algebra on $C^*$-algebra

Let $$A$$ be a $$C^*$$-algebra, and let $$a$$ in $$A$$ be normal, and let $$B$$ be the $$C^*$$-algebra generated by $$a$$. Suppose that $$f:\sigma(a)\to\mathbb{C}$$ is continuous.

Show that there exists an element $$x$$ in $$B$$ such that $$\Phi(x)(s)=f(\Phi(a)(s))$$ for all $$s\in\sigma(a)$$, where $$\Phi$$ is the Gelfand homomorphism.

• What have you tried and what do you know of the Gelfand isomorphism? – Chrystomath Apr 4 at 8:19

Here is a proof. Recall the Commutative Gelfand Naimark Theorem, it states that a commutative unital $$C^*$$ algebra $$\mathfrak{A}$$ is isometrically isomorphic under the Gelfand Transfrom $$\Phi$$ to $$C(\hat{\mathfrak{A}})$$, where $$\hat{\mathfrak{A}}$$ is the multiplicative functionals on $$\mathfrak{A}$$. (I first consider the Gelfand transform as acting on multiplicative functionals, then identify it's domain with the spectrum)
we set $$\mathfrak{A}$$ to be $$B$$ as you defined, consider a multiplicative functional $$h$$ on it. since B is a $$C^*$$ algebra, we have that $$h(a^*) = \overline{h(a)}$$ (this is a known fact, one might say "every $$C^*$$ algebra is symmetric"). Now convince yourself that by continuity of $$h$$, $$h$$ is completely determined (on $$B$$) by $$h(a)$$. This gives us a bijective map $$\psi: \hat{B} \to \sigma(a)$$ (Since the image of $$\Phi(a)$$ is the spectrum of $$a$$), given by $$h \mapsto h(a) = \Phi(a)(h)$$. This is continuous in the weak * topology. And since both spaces are compact, it's a homeomorphism. This identifies the spectrum with what the Gelfand Transform acts on. Your original consideration of the Gelfand transform of $$x$$ under my definition is actually $$\Phi(x) \circ \psi^{-1}$$.
Great, look at $$f \circ \psi \in C(\hat{B})$$. Since $$a$$ is normal we can use the Commutative Gelfand Naimark Theorem and conclude that there exists $$x \in B$$ such that $$\Phi(x) = f \circ \psi$$. So:
$$\forall s \in \sigma(a) \; \Phi(x)(\psi^{-1}(s))=f(s) = f(\Phi(a)(\psi^{-1}(s)))$$
This is since the Gelfand transform of $$a$$: $$\Phi(a)(\psi^{-1}(s)) = s$$ is the identity function. The above equalities is exactly what you wanted.