Characters on a unital Banach algebra are continuous

Let $A$ be a nonzero unital complex Banach algebra (Unital Banach algebra) and let $\chi$ be a character on $A$ (Character and maximal ideal space). Then:

1. $\chi$ is unital: $\chi (1)=1$;
2. $\chi (a)\in {\sigma }_A(a)$ for every $a\in A$ (Spectrum and resolvent set in a Banach algebra);
3. $\mathrm{\mid }\chi (a)\mathrm{\mid }\le \mathrm{\parallel }a\mathrm{\parallel }$ for every $a\in A$; consequently $\chi$ is bounded linear with $\mathrm{\parallel }\chi \mathrm{\parallel }=1$, and in particular $\chi$ is continuous.

No Hahn–Banach theorem, no spectral-radius formula and no existence of characters is used: the argument runs on the Neumann series alone and is a theorem of ZF.

• The unit hypothesis streamlines the proof, but continuity survives without it. For a nonzero unital Banach algebra the computation $\chi (1{)}^2=\chi (1)$ forces unitality. A character on a nonunital Banach algebra $B$ extends to the algebraic sum-norm unitization $\mathbb{C}\oplus B$, with product $(\lambda ,a)(\mu ,b)=(\lambda \mu ,\lambda b+\mu a+ab)$, by ${\chi }^{+}(\lambda ,a)=\lambda +\chi (a)$. This is a unital character on a unital Banach algebra, so the theorem applied to ${\chi }^{+}$ shows that the original character is continuous as well.
• Choice-free. Steps 1.1–2.2 use only the algebra axioms, the definition of the spectrum and the Neumann series; no selection from nonempty sets and no separation theorem occurs.
• Where the bound is used. Part 3 is what puts $\mathrm{\Delta }(A)$ inside the dual unit ball and identifies the pointwise-evaluation topology with the weak-star subspace topology; this is the standard automatic-continuity statement for characters.