In a $C^*$ algebra why $\|a\|\not =\rho(a)$ for any $a$? Where $\rho(a)$ is the spectral radius.
It can be shown that the equality holds for self-adjoint elements. Then that can be used to show that the Gelfand transform is an isometry. Thus $\|a\|=\|\hat{a}\|=max \{|\phi(a)| \hspace{0.2cm}| \phi\in Spec(A) \}$. First equality follows from the Gelfand transformation being an isometry. The second follows form the definition of the norm on $C(Spec(A))$. But we know that $spec(a)=\{\phi(a) \hspace{0.2cm}| \phi\in Spec(A) \}$. thus $\rho(a)=\rho(\hat{a})$ and the result follows. My professor said this is incorrect. What I am doing wrong here?
Edit:
I am assuming the algebra is commutative
I think your proof is correct and your professor is wrong.
A faster proof is using the full power of Gelfand transformation. Indeed, without loss of generality we may assume that $A= C(X)$ where $X$ is a compact Hausdorff space. Then if $f \in C(X)$, then $\operatorname{spec}(f) = f(X)$ and thus $$\rho(f) =\max_x |f(x)| = \|f\|_\infty.$$
For non-commutative $C^*$-algebras, the statement is false. Consider $$A:=\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}\in M_2(\mathbb{C}).$$ Then $A^2=0$ and thus $\operatorname{spec}(A) = \{0\}$ but $\|A \| \neq 0$.