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I've been trying to understand something about the Gelfand transform. As I understand it, for a complex Banach algebra $A$, we consider the set $$\Delta = \{\phi \in A^* : \phi(x*y) = \phi(x)\phi(y)\},$$ where $A^*$ denotes the continuous dual of $A$, i.e., the space of continuous linear functionals on $A$. The elements of $\Delta$ are in one-to-one correspondence with the maximal ideals of $A$.

Now, for any locally compact abelian Hausdorff group $G$, $L^1(G)$ is Banach space, and with multiplication $*$ defined as convolution, it is a Banach algebra. It turns out in fact that every complex homomorphism of $L^1(G)$ is convolution with a character -- I'll ask about that later. Let's take the case $G = \mathbb{R}$ and $A = L^1(\mathbb{R})$. The complex homomorphisms turn out to be $$\phi: f \rightarrow \int_\mathbb{R} f(x) e^{-ix\xi} dx,$$ which we know as the Fourier transform. What is the kernel of such a homomorphism? According to Gelfand theory, it should be a maximal ideal of $A = L^1(G)$. I don't see how the set of functions $f$ for which $$\int_\mathbb{R} f(x) e^{-ix \xi} dx = 0$$ forms an ideal or maximal ideal. In my mind, the maximal ideals of $\mathbb{R}$ when multiplication is defined by pointwise multiplication (i.e. $(f*g)(x) = f(x)g(x)$, not convolution) should be the functions which vanish at a single point. Where am I going wrong?

The references I'm using are: Rudin's Functional analysis, p. 277, 280, Katznelson's Introduction to Harmonic Analysis p. 202, Theorem 2.8 (2.9?), Green Rudin p. 191.

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up vote 2 down vote accepted

The kernel of $\phi$ is the set of functions $f \in L^1(G)$ whose Fourier transform $\hat{f}$ vanishes at the point $\xi$. Since convolution in $L^1(G)$ corresponds to pointwise multiplication of the Fourier transforms, this is indeed an ideal.

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I am retarded and don't know why I didn't see that. Thank you. Please close. – snarski Dec 4 '12 at 0:39

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