Let $A$ be a commutative Banach algebra (not necessarily with a unit) and set $\Delta(A) := \{\varphi: A \rightarrow \mathbb{C} : \varphi \text{ is an homomorphism with } \varphi \not\equiv 0\}$.
If $A$ is an arbitrary Banach space equipped with the trivial multiplication ($xy := 0$ for $x, y \in A$), then $A$ is a Banach algebra with $\Delta(A) = \emptyset$, since $\varphi(x)^2 = \varphi(x^2) = \varphi(0) = 0$ for all $x \in A$ and all homomorphisms $\varphi: A \rightarrow \mathbb{C}$.
Question: Are there any more Banach algebras $A$ with $\Delta(A) = \emptyset$? If yes, can we characterize all these algebras?
My “work” so far:
- If $x^2 = 0$ for all $x \in A$, then $\Delta(A) = 0$ by the same reasoning as above. Edit: As expanding $(x+y)^2$ shows, this is only true for the trivial multiplication (as $A$ is commutative).
- If $A$ has a unit, then $\Delta(A) \neq \emptyset$.
Somewhat more generally, if all elements are nilpotent (i.e. for each $x \in A$ there is $n$ such that $x^n = 0$), then $\Delta(A) = \emptyset$.
Or still more generally, if for all $x \in A$, $\lim_{n \to \infty} \left\|x^n\right\|^{1/n} = 0$.
EDIT: To see this, note that if $\varphi \in \Delta(A)$, $|\varphi(x)| = |\varphi(x^n)|^{1/n} \le \left(\|\varphi\|\left\|x^n\right\|\right)^{1/n}$
This condition has a name: $x \in A$ is quasinilpotent if $\lim_{n \to \infty} \left\|x^n\right\|^{1/n} = 0$. The Banach algebra $A$ is radical (or topologically nil) if every element is quasinilpotent.
For a nontrivial example of a radical commutative Banach algebra, take $L^1(0,1)$ with the product $(fg)(x) = \int_0^x f(x-s) g(s)\; ds$.