An abelian Banach algebra without characters Can one give an example of an abelian Banach algebra with empty character space? Such algebra must be necessarily non-unital.
I couldn't find any examples of such algebras.
Thanks!
 A: The Volterra algebra $V$ is an example of a commutative Banach algebra without maximal ideals (hence with empty character space). See also Definition 4.7.38 in 

H. G. Dales, Banach algebras and automatic continuity, London Math. Soc.
  Monographs, Volume 24, Clarendon Press, Oxford, 2000.

Okay, let me prove this claim. This relies on three facts:
The algebra $V$ has a bounded approximate identity, e.g. $(n\cdot \mathbf{1}_{\big[0, \tfrac{1}{n}\big]})_{n=1}^\infty$, hence by the Cohen factorisation theorem, $V = V^2$. Consequently, all maximal ideals of $V$ (if exist) are closed. 
Now apply a result of Dixmier which tells you that no prime ideal of $V$ is closed. Of course, maximal ideals are prime so, the conclusion follows. You will find the proof of Dixmier's result in the above-mentioned book by Dales (Theorem 4.7.58).
A: A related comment: I think I have an example of a commutative unital complex algebra $A$ without 
any nontrivial complex homomorphisms. Take 
$A$ to be all rational functions with complex coefficients, that is 
$$
A= \{[p/q]: p,q \text{ complex polynomials and }q \text{ not identically } 0\}.
$$
Here $[p/q]$ denotes the equivalence class of $(p,q)$ under the relation
$(p,q)\sim(r,s)$ if $ps=qr$. Then $A$ is a commutative unital complex algebra (which is also a field, and the only maximal ideal is $0$). But there is no nontrivial complex homomorphism $\varphi:A \rightarrow \mathbb{C}$, because if $\varphi(z)=\alpha$, then 
$$
\varphi\Big(\frac{1}{z-\alpha}\Big)=\frac{1}{\varphi(z-\alpha)}=\frac{1}{0},
$$ 
a contradiction.
