# Convolution algebra - help understanding a worked example

1. Jun 13, 2010

### Machu Picchu

The latex code here is doing all sorts of odd things... :( ... anyway,

The convolution algebra is $$l_1(\mathbb{Z},\mathbb{C})$$, the set of functions $$f:\mathbb{Z}\rightarrow\mathbb{C}$$ which satisfy

$$||f||:=\sum_{n=-\infty}^{\infty}|f(n)|<\infty$$

with pointwise addition and scalar multiplication, and multiplication of functions defined by

$$f*g(n)=\sum_{m=-\infty}^{\infty}f(n-m)g(m)$$

(this is a commutative Banach algebra).

For $$z\in\mathbb{T}$$, the unit circle in the complex plane, the functional $$\psi_z:l_1(\mathbb{Z},\mathbb{C})\rightarrow\mathbb{C}$$ is defined by

$$\psi_z(f)=\sum_{n=-\infty}^{infty}f(n)z^n$$

.$$\psi_z$$ is a non-zero homomorphism (in fact the set of all of these is the set of all non-zero complex homomorphisms).

For a function f in the algebra, the gelfand transform is

$$\hat{f}(\psi_z)=\psi_z(f)=\sum_{n=-\infty}^{\infty}f(n)z^n$$

The example I'm trying to understand shows how to find the inverse of a particular function f.

Part of the working says that "hat" is injective, so [at this stage the latex code is being absolutely ridiculous - priting something I had in a previous sentence that I subsequently deleted, and the actual thing I want is nowhere to be found :( ... I want to write that you can interchange the order of "inversing" and "hatting" f]. This I don't understand... it's probably quite simple, but something's not clicking for me unfortunately.

Thanks for any help.

Last edited: Jun 13, 2010
2. Jun 13, 2010

### JSuarez

As the Gelfand transform is linear, what must be f, so that

$$\hat{f}(\psi_z)=0$$

, for all $\psi_z$?

3. Jun 14, 2010

### DrRocket

First your LaTex problem. I had the same problem and was informed that there is a bug in the software. The workaround is to click the "refresh" button on your browser before you click on " Preview Post". That seems to make LaTex behave.

Now, for the mathematics. Your specific question is not absolutely clear to me, but perhaps it would help to point out that what you are dealing with is nothing more and nothing less than the ordinary theory of Fourier series. The Gelfand transform on $$L^1$$ of a locally compact abelian group is just the Fourier transform. The only thing here that may look at bit unfamiliar is that rather than starting with a function on the torus you are starting with one on the integers. But the dual group of the torus is the integers and vice versa. So, when you take the Gelfand transform of an $$l^1$$ series you get a function on the torus and the Fourier series of that function is essentially the series that you started with (modulo a reflection).