Convolution algebra - help understanding a worked example

In summary, the conversation discusses the convolution algebra l_1(\mathbb{Z},\mathbb{C}), the Gelfand transform, and the inverse of a particular function in the algebra. The Gelfand transform is essentially the Fourier transform and the inverse function can be found by interchanging the order of "inversing" and "hatting" the function.
  • #1
Machu Picchu
1
0
The latex code here is doing all sorts of odd things... :( ... anyway,

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

[tex]||f||:=\sum_{n=-\infty}^{\infty}|f(n)|<\infty[/tex]

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

[tex]f*g(n)=\sum_{m=-\infty}^{\infty}f(n-m)g(m)[/tex]

(this is a commutative Banach algebra).

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

[tex]\psi_z(f)=\sum_{n=-\infty}^{infty}f(n)z^n[/tex]


.[tex]\psi_z[/tex] 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


[tex]\hat{f}(\psi_z)=\psi_z(f)=\sum_{n=-\infty}^{\infty}f(n)z^n[/tex]


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.
 
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  • #2
As the Gelfand transform is linear, what must be f, so that

[tex]\hat{f}(\psi_z)=0[/tex]

, for all [itex]\psi_z[/itex]?
 
  • #3
Machu Picchu said:
The latex code here is doing all sorts of odd things... :( ... anyway,

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

[tex]||f||:=\sum_{n=-\infty}^{\infty}|f(n)|<\infty[/tex]

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

[tex]f*g(n)=\sum_{m=-\infty}^{\infty}f(n-m)g(m)[/tex]

(this is a commutative Banach algebra).

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

[tex]\psi_z(f)=\sum_{n=-\infty}^{infty}f(n)z^n[/tex]


.[tex]\psi_z[/tex] 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


[tex]\hat{f}(\psi_z)=\psi_z(f)=\sum_{n=-\infty}^{\infty}f(n)z^n[/tex]


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.

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 [tex] L^1[/tex] 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 [tex] l^1[/tex] 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).
 

Related to Convolution algebra - help understanding a worked example

1. What is convolution algebra?

Convolution algebra is a mathematical concept that involves the combination of two functions to produce a third function. It is a fundamental operation in signal processing and is used to model systems with multiple inputs and outputs.

2. How is convolution algebra used in real-world applications?

Convolution algebra is used in various fields such as engineering, physics, and computer science to analyze and manipulate signals and data. It is commonly used in image and audio processing, as well as in the study of differential equations and linear systems.

3. Can you provide an example of convolution algebra?

Sure, let's say we have two functions f(x) and g(x) that represent the input and output of a system. The convolution of these two functions, denoted by f * g, is a third function that describes the output of the system when the input is convolved with a unit impulse function.

4. How does convolution algebra relate to Fourier transforms?

Convolution algebra is closely related to Fourier transforms, as the convolution of two functions in the time domain is equivalent to the multiplication of their Fourier transforms in the frequency domain. This relationship is used in many applications, such as signal filtering and spectral analysis.

5. Are there any limitations to convolution algebra?

While convolution algebra is a powerful tool, it does have some limitations. One of the main limitations is that it assumes the system being modeled is linear and time-invariant. Additionally, the accuracy of the convolution operation depends on the resolution of the input functions, so it may not always provide an exact representation of the output.

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