This is related to Lang Real and Functional Analysis Chpt XVI, Sec 1.
Let $A$ be a banach algebra. Let $v\in A$. Spectrum of $v$ is defined as $\{z\in C\vert$ $v-ze$ is not invertible$\}$.
Thm 1.2 Let $A$ be an unital commutative normed algebra over real. Assume there is $j\in A$ s.t. $j^2=-e$. Let complex number $C=R+Rj$. Given $v\in A,v\neq 0$, there exists $c\in C$ s.t. $v-ce$ is not invertible in $A$.
Cor 1.4 The spectrum of an element in any complex Banach algebra(commutative or not) with unit element is not empty.
I am having trouble to follow the proof.
"If $A$ is a banach algebra with unit and $v\in A$, then closure of algebra generated by $v$ and unit is a commutative banach algebra. Hence, it follows from Thm 1.2 above."
$\textbf{Q:}$ Why it follows from Thm 1.2 above? Consider $v\in A$. Denote spectrum of $v$ in $A$ as $Sp_A(v)$. Denote the closure of algebra generated by $v$ and unit by $B$. It follows from Thm 1.2 that $Sp_B(v)\neq\emptyset$. Why is $Sp_A(v)\neq\emptyset$? This is more or less like the following statement. Given a ring inclusion map, $B\subset A$, non-units of $B$ are non-units of $A$. Take $B=k[x]$ and $A=k(x)$. Certainly non-units of $B$ other than $0$ are invertible in $A$.
Best Answer
The proof is indeed wrong for the reason that you say. To fix it, you have to let $B$ be the closure of the algebra generated by $v$ and the elements $(v-ce)^{-1}$ for all $c\in\mathbb{C}$. Then clearly $Sp_B(v)=\emptyset$, and $B$ is still commutative (here we use the fact that if $a$ and $b$ commute and $b$ is invertible then $a$ and $b^{-1}$ commute, since $ab^{-1}=b^{-1}bab^{-1}=b^{-1}abb^{-1}=b^{-1}a$).