I am reading the following proposition by :
FF. Bonsall and J. Duncan, Complete Normed Algebras, pg. 79
Definition: A Jordan function on $A$ is a nonzero linear functional $\phi$ on $A$ such that $\phi(a^2)=\phi(a)^2$ for every $a \in A$.
Proposition: Let $A$ be a Banach algebra. Every Jordan function $\phi$ on $A$ is multiplicative.
Proof:
Consideration the identity $$\phi((a+b)^2)=\phi(a+b)^2$$ gives $$\phi(ab + ba)=2\phi(a)\phi(b)~~~ (a,b \in A)$$ and so the result is trivial is $A$ is commutative.
I understand this part completely. It is the following part that I can not seem to see why:
If $\phi$ is not multiplicative then there exists $a,b \in A$ such that $\phi(a)=0, \phi(ab)=1$. Then $\phi(ba)=-1$. Let $c=bab$. Then $$0=2\phi(a)\phi(c)=\phi(ac+ca)=\phi((ab)^2)+\phi((ba)^2)=2$$
Why does there exist an $a,b \in A$ such that those conditions hold? Can anyone please help give me some clarity as to why that is true?
It is very easy to check that if $A$ is non-unital, a Jordan function $\phi$ extends to a Jordan function of the unitization $\tilde A$.
So, we may assume without loss of generality that $A$ is unital.
If $\phi$ is not multiplicative, there exist $x,y\in A$ with $\phi(xy)\ne\phi(x)\phi(y)$. We may write this as $\phi(xy-\phi(x)y)\ne0$. Thus, multiplying either $x$ or $y$ by an appropriate constant, we may assume $$ \phi(xy-\phi(x)y)=1. $$ Now we may write this as $\phi((x-\phi(x)1)y)=1$. So, if $a=x-\phi(x)1$, $b=y$, we get $$ \phi(a)=0,\ \ \phi(ab)=1. $$