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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 $\Delta(A) \neq \emptyset$.
user505379
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Keba
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1 Answers1

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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 $(fg)(x) = \int_0^x f(x-s) g(s)\; ds$.

Robert Israel
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  • Thanks for your answer. I see why the nilpotent case works and can think of some examples (like certain sub algebras of $\mathbb{C}{n \times n}$). – Keba Sep 02 '16 at 19:08
  • Can you give an example for an algebra fulfilling $|x^n|^\frac1n = 0$ and containing a non nilpotent element? Also, I am not sure why $|x^n|^\frac1n = 0$ is already enough. – Keba Sep 02 '16 at 19:22
  • @Keba, see also this post http://math.stackexchange.com/a/1893799/17929 – Tomasz Kania Sep 11 '16 at 17:06