[proofplan] The key observation is that a character must satisfy $|\varphi(x)| \le \|x\|$ pointwise — otherwise an element with $\|x/\varphi(x)\| < 1$ would yield, via the **Invertibility Near the Identity** theorem, an invertible element $1 - x/\varphi(x)$ whose image under $\varphi$ would be $0$, contradicting $\varphi(1) = 1$ (which a character preserves). The strategy: (1) reduce to the unital case by extending $\varphi$ to the unitisation $A^+$; (2) show $\varphi(1) = 1$, which is automatic for a non-zero character; (3) prove the pointwise bound by contradiction using **Invertibility Near the Identity**; (4) deduce $\|\varphi\|_{A^*} \le 1$, with equality in the unital case from $\varphi(1) = 1$. [/proofplan]

[step:Reduce to the unital case via the unitisation] Suppose first that $A$ is non-unital. Form the unitisation $A^+ := A \oplus \mathbb{C}$, with multiplication \begin{align*} (x, \mu)(y, \nu) := (xy + \mu y + \nu x, \mu \nu) \end{align*} and norm $\|(x, \mu)\|_{A^+} := \|x\|_A + |\mu|$, making $A^+$ a unital Banach algebra with unit $1_{A^+} = (0, 1)$, and embedding $\iota : A \hookrightarrow A^+$, $x \mapsto (x, 0)$, isometric. Define \begin{align*} \varphi^+ : A^+ &\to \mathbb{C} \\ (x, \mu) &\mapsto \varphi(x) + \mu. \end{align*} We verify $\varphi^+$ is an algebra homomorphism: $\varphi^+$ is linear (clear), $\varphi^+(1_{A^+}) = \varphi^+((0, 1)) = 0 + 1 = 1$, and for products \begin{align*} \varphi^+((x, \mu)(y, \nu)) &= \varphi(xy + \mu y + \nu x) + \mu \nu \\ &= \varphi(x)\varphi(y) + \mu \varphi(y) + \nu \varphi(x) + \mu \nu \\ &= (\varphi(x) + \mu)(\varphi(y) + \nu) = \varphi^+((x, \mu)) \varphi^+((y, \nu)), \end{align*} using $\varphi(xy) = \varphi(x)\varphi(y)$ (since $\varphi$ is a homomorphism). Since $\varphi^+(1_{A^+}) = 1 \ne 0$, $\varphi^+$ is non-zero, hence $\varphi^+ \in \Phi_{A^+}$. If we prove $\|\varphi^+\|_{(A^+)^*} \le 1$ in the unital algebra $A^+$, then for $x \in A$, \begin{align*} |\varphi(x)| = |\varphi^+((x, 0))| \le \|(x, 0)\|_{A^+} = \|x\|_A, \end{align*} which gives $\|\varphi\|_{A^*} \le 1$. Therefore for the bound $\|\varphi\|_{A^*} \le 1$ we may assume $A$ is unital. The equality $\|\varphi\|_{A^*} = 1$ in the unital case will be addressed in Step 4. From now on **assume $A$ is unital**. [/step]

[step:Show $\varphi(1) = 1$] Since $\varphi$ is non-zero, there exists $x_0 \in A$ with $\varphi(x_0) \ne 0$. Using the homomorphism property, \begin{align*} \varphi(x_0) = \varphi(1 \cdot x_0) = \varphi(1) \varphi(x_0). \end{align*} Dividing by $\varphi(x_0) \ne 0$ gives $\varphi(1) = 1$. [/step]

[step:Prove the pointwise bound $|\varphi(x)| \le \|x\|$ for all $x \in A$]Suppose for contradiction that there exists $x \in A$ with $|\varphi(x)| > \|x\|$. Note this forces $\varphi(x) \ne 0$. Define \begin{align*} y := \frac{x}{\varphi(x)} \in A. \end{align*} Then $\varphi(y) = \varphi(x)/\varphi(x) = 1$ and \begin{align*} \|y\| = \frac{\|x\|}{|\varphi(x)|} < 1. \end{align*} Set $a := 1 - y$, so $\|1 - a\| = \|y\| < 1$. By [Invertibility Near the Identity](/theorems/2667) — whose hypothesis $\|1 - a\| < 1$ is just verified — $a = 1 - y$ is invertible in $A$. Let $z := (1 - y)^{-1} \in A$. Apply $\varphi$ to the identity $z (1 - y) = 1$, using the homomorphism property and $\varphi(1) = 1$ from Step 2: \begin{align*} 1 = \varphi(1) = \varphi(z (1 - y)) = \varphi(z) \cdot \varphi(1 - y) = \varphi(z) \cdot (\varphi(1) - \varphi(y)) = \varphi(z) \cdot (1 - 1) = 0, \end{align*} a contradiction. Therefore $|\varphi(x)| \le \|x\|$ for all $x \in A$.[/step]

[guided]The pointwise bound $|\varphi(x)| \le \|x\|$ is the heart of the proof. Why is it true? Suppose to the contrary that $\varphi$ ever exceeds the norm: $|\varphi(x)| > \|x\|$ for some $x$. The trick is to **rescale to land at a point of norm strictly less than $1$ where $\varphi$ takes the value $1$**: set $y := x/\varphi(x)$, so $\varphi(y) = 1$ and $\|y\| < 1$. Now we use a fundamental fact about Banach algebras: elements close to the identity are invertible. Specifically, [Invertibility Near the Identity](/theorems/2667) says that if $\|1 - a\| < 1$, then $a$ is invertible. Setting $a := 1 - y$, we have $\|1 - a\| = \|y\| < 1$, so $1 - y$ is invertible. But wait: a character preserves invertibility — if $u \in A$ is invertible with inverse $u^{-1}$, then $\varphi(u) \varphi(u^{-1}) = \varphi(u u^{-1}) = \varphi(1) = 1$, so $\varphi(u) \ne 0$. Applying this with $u := 1 - y$: $\varphi(1 - y) = \varphi(1) - \varphi(y) = 1 - 1 = 0$, but $1 - y$ is invertible, so $\varphi(1 - y) \ne 0$. This contradicts $\varphi(1 - y) = 0$. The contradiction shows the supposition $|\varphi(x)| > \|x\|$ was false. Hence $|\varphi(x)| \le \|x\|$ for all $x$. What is the geometric meaning? The kernel $\ker \varphi$ is a maximal ideal (codimension 1 since $\varphi$ is a non-zero linear functional onto $\mathbb{C}$), and the bound says this ideal is "far from the identity" — specifically, the open unit ball around $1$ does not meet $\ker \varphi$. This is consistent with $\ker \varphi$ avoiding $G(A)$ (proper ideals miss invertibles), since a unit ball around $1$ lies in $G(A)$ by theorem 2667.[/guided]

[step:Conclude continuity and compute the norm] By Step 3, for every $x \in A$ with $\|x\| \le 1$ we have $|\varphi(x)| \le \|x\| \le 1$, so \begin{align*} \|\varphi\|_{A^*} = \sup_{\|x\| \le 1} |\varphi(x)| \le 1. \end{align*} In particular, $\varphi$ is bounded as a linear map $A \to \mathbb{C}$, hence continuous, so $\varphi \in A^* = \mathcal{L}(A, \mathbb{C})$. When $A$ is unital, the lower bound follows from Step 2: \begin{align*} \|\varphi\|_{A^*} = \sup_{\|x\| \le 1} |\varphi(x)| \ge \frac{|\varphi(1)|}{\|1\|}. \end{align*} Since $\|1\| = 1$ in any unital BA (by convention, see also Step 1) and $\varphi(1) = 1$, this gives $\|\varphi\|_{A^*} \ge 1$. Combining with the upper bound, $\|\varphi\|_{A^*} = 1$. In the non-unital case, the reduction in Step 1 gives only $\|\varphi\|_{A^*} \le 1$ — equality may fail since $A$ has no $1$ to evaluate $\varphi$ at. [/step]