Norm of a character in a non-unital Banach algebra without approximate identity As is shown here, the norm of a character in a non-unital Banach algebra with an approximate identity is $1$.
I wonder if this result still holds for general non-unital Banach algebras. 
Let $A$ be a non-unital Banach algebra and $\phi \in \Omega_A$($\Omega_A$ is the set of all nonzero homomorphisms from $A$ to $\mathbb C$) and $A^+$ be the unitization of $A$. There exists a unique extension $\phi^+$ of $\phi$ on $A^+$, then $\|\phi\| \le \| \phi^+ \|=1$. But do we have  $\|\phi\|=1$? 
 A: It does not always hold, here is an example with norm $<1$.
Take $A=\mathscr l^1(\mathbb N)$ with $e_n$ as the standard basis. Define multiplication via:
$$\sum_{n=1}^\infty a_n e_n \sum_{m=1}^\infty b_m e_m = \sum_{k=1}^\infty \left(\sum_{n=1}^{k-1} a_n b_{k-n}\right) e_k$$
The well definedness as a map on $\mathscr l^1 \times \mathscr l^1$ follows from the Cauchy product theorem. Associativity etc also hold.
Note that here
$$\| A \cdot B \|=\sum_{k=1}^\infty \left|\sum_{n=1}^{k-1} a_n b_{k-n}\right|≤\sum_{k=1}^\infty \sum_{n=1}^{k-1} \left|a_n b_{k-n}\right|=\|A\|\cdot \|B\|$$
for all $A,B$ so we have a Banach algebra.
By setting $\Phi \left(\sum_{n=1}^\infty a_n e_n \right)=\sum_{n=1}^\infty a_n x^{n}$ for an $x \in \mathbb C$, $|x|≤1$ you get a bounded linear map from $\mathscr l^1 \to \mathbb C$. Furthermore
$$\Phi \left(\sum_{n=1}^\infty a_n e_n \right)\Phi \left(\sum_{m=1}^\infty b_m e_m\right) = \sum_{k=1}^\infty \sum_{n=1}^{k-1} a_n b_{k-n} x^{n+(k-n)}=\Phi\left(\sum_{k=1}^\infty \sum_{n=1}^{k-1} a_n b_{k-n} e_k\right)$$
So it is a homomorphism. Also $\|\Phi\|=|x|$, which can be smaller than $1$.
