# injectivity of Fourier transform: is there algebraic proof?

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).

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Could you clarify exactly what results/theory/tools you are trying to avoid using? –  Christopher A. Wong Oct 17 '11 at 3:59
@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. –  Fedor Petrov Oct 17 '11 at 5:39
L^1(G) is a Banach *-algebra, see planetmath.org/encyclopedia/GroupAlgebra2.html or en.wikipedia.org/wiki/… –  Dmitri Pavlov Nov 4 '12 at 18:09
@Dmitri. It is, but this star does not make it C*-algebra (since $\|ff^{*}\|\ne \|f\|^2$ in general). –  Fedor Petrov Nov 8 '12 at 8:57
@Dmitri: Banach-star algebras need not be semisimple, and can have rather unpleasant properties in general –  Yemon Choi Nov 22 '12 at 9:49