7
$\begingroup$

Let $L$ be the Banach algebra of $L^1$-functions from $\mathbb{R}$ to $\mathbb{C}$ with $L^1$-norm and convolution as algebra multiplication. Assume that we knew that the homomorphisms from $L$ to $\mathbb{C}$ are the zero map and evaluation of the Fourier transform at individual real numbers: $f \mapsto \int_{\mathbb R} f(t)e^{it\alpha}dt$ for some real $\alpha$. We may add a unit $e$ to $L$ artificially by considering the new Banach algebra $A:=L\oplus \mathbb{C}\cdot e$ with natural operations. Then the fact that any $L^1$-function whose Fourier transform is zero must be zero itself may be rephrased algebraically: the algebra $A$ is semisimple (as maximal ideals of unital Banach algebras correspond to homomorphisms to $\mathbb{C}$ by the Gelfand-Mazur theorem).

My question is whether this may be proved a priori and independently (and maybe for some wide class of commutative unital Banach algebras).

$\endgroup$
5
  • $\begingroup$ Could you clarify exactly what results/theory/tools you are trying to avoid using? $\endgroup$ Commented Oct 17, 2011 at 3:59
  • $\begingroup$ @Christopher: the goal is rather not avoiding any theory, but understanding this fact in Banach algebras context. For example, unital $C^{*}$-algebras are semisimple, but this algebra does not seem to have star (or am I blind here?). But maybe some Banach algebras with clear algebraic property or additional structure are also always semisimple, and $A$ is in this class by not so hard to check reasons. $\endgroup$ Commented Oct 17, 2011 at 5:39
  • 1
    $\begingroup$ L^1(G) is a Banach *-algebra, see planetmath.org/encyclopedia/GroupAlgebra2.html or en.wikipedia.org/wiki/… $\endgroup$ Commented Nov 4, 2012 at 18:09
  • 1
    $\begingroup$ @Dmitri. It is, but this star does not make it C*-algebra (since $\|ff^{*}\|\ne \|f\|^2$ in general). $\endgroup$ Commented Nov 8, 2012 at 8:57
  • $\begingroup$ @Dmitri: Banach-star algebras need not be semisimple, and can have rather unpleasant properties in general $\endgroup$
    – Yemon Choi
    Commented Nov 22, 2012 at 9:49

0

You must log in to answer this question.

Browse other questions tagged .