A character on a not necessarily unital Banach algebra is a non-zero homomorphism $\phi\colon A\to\mathbb{C}$. If $r(a)$ denotes the spectral radius of some element $a\in A$, then one has $|\phi(a)|\leq r(a)\leq\|a\|$. This shows that $\phi$ is bounded with $\|\phi\|\leq1$. If $A$ is unital then one can easily verify that $\phi(1)=1$, thus $\|\phi\|=1$. But what if $A$ is not unital? Do we still have $\|\phi\|=1$? Or do there exist characters with $0<\|\phi\|<1$ in that case?
EDIT: I just noticed that I also implicitly assumed the existence of a unit for proving that $\|\phi\|\leq1$. But I think one can uniquely extend $\phi$ to a (unital) character $\hat{\phi}$ on the unitization $\widetilde{A}$ here if necessary, and use the prove above to conclude that $\|\phi\|\leq\|\hat{\phi}\|\leq1$.
It can happen that the norm of a character is $< 1$ in a Banach algebra. Here is an explicit example:
Norm of a character in a non-unital Banach algebra without approximate identity
It turns out that the absence of an approximate identity is problematic here. Indeed, if $A$ admits an approximate unit $(u_i)_i$ (this is for example the case when $A$ is a $C^*$-algebra) and $\chi: A \to \mathbb{C}$ is a character on $A$, then choose $a \in A$ with $\chi(a) \neq 0$. Then $$\lim_i\chi(u_i)= \lim_i \frac{\chi(u_ia)}{\chi(a)} = \frac{\chi(a)}{\chi(a)}=1$$ which shows that $\Vert \chi \Vert \ge 1$. The other inequality is always true, as you already observed, so when the Banach algebra admits an approximate unit everything works out nicely.