Convolution Inverse: Family of Functions Explained

In summary, the convolution inverse for a function is a function that takes the inverse of the convolution. This inverse is only defined for functions in the Laplace domain, and for measures that have one-point support.
  • #1
mnb96
715
5
Hello,
I noticed that it is possible to define an inverse for the convolution operator so that a function f convolved by its convolution-inverse [tex]f^{\ast-1}[/tex] gives the delta-function: [tex]f \ast f^{\ast-1} = \delta[/tex]
http://en.wikipedia.org/wiki/Convolution#Convolution_inverse

Which is the family of functions that admits this convolution inverse?
 
Physics news on Phys.org
  • #2
mnb96 said:
Hello,
I noticed that it is possible to define an inverse for the convolution operator so that a function f convolved by its convolution-inverse [tex]f^{\ast-1}[/tex] gives the delta-function: [tex]f \ast f^{\ast-1} = \delta[/tex]
http://en.wikipedia.org/wiki/Convolution#Convolution_inverse

Which is the family of functions that admits this convolution inverse?

As long as the Laplace transform is defined for the function then you can calculate the convolution inverse by simply taking the inverse in the Laplace domain. Of course for a causal function the convolution inverse will be anti-causal.
 
  • #3
...ok, thanks for the hint.
could we generalize more? Namely, for which family of functions is the Laplace transform defined? for all the functions in [tex]L^2(\Re)[/tex]?
 
  • #4
I am sorry but I have to resume this thread.

The answer I received was not clear, and I don't see what the Laplace Transform has to do with the problem. I am besides interested in the whole real-axis, where the Laplace transform is not even defined.

Can anyone actually show how to obtain a convolution inverse for a function f, such that [tex]f \ast f^{\ast-1} = \delta[/tex]?
 
  • #5
If [tex]f[/tex] and [tex]g[/tex] are functions of a real variable, then so is their convolution. Since [tex]\delta[/tex] is not (it is a Schwarz distribution, not a function of a real variable), the "convolution inverse" of a function of a real variable is never a function of a real variable.

Non-function example: The distribution [tex]\delta(x-c)[/tex] is called a "unit mass at [tex]c[/tex]". The convolution inverse of the unit mass at [tex]c[/tex] is the unit mass at [tex]-c[/tex].

Further non-examples: if [tex]\mu[/tex] and [tex]\nu[/tex] are measures on the line, then the support of the convolution is the sum of the supports (added as sets). So in order for the convolution of two measures to be [tex]\delta[/tex], both components must have one-point support. So even in this case, the only examples are essentially the ones given above.

We could try to go to more complicated distributions, dipoles and so on. But why?
 
  • #6
OK, so what happens if I write:

[tex]f\ast f' = \delta[/tex]

[tex]\mathcal{F}\{f\ \ast f'\} = \mathcal{F}\{\delta\}[/tex]

[tex]\mathcal{F}\{f\} \mathcal{F}\{f'\} = 1[/tex]

[tex]\mathcal{F}\{f'\} = 1 / \mathcal{F}\{f\}[/tex]

[tex]f' = \mathcal{F}^{-1}\{ 1 / \mathcal{F}\{f\} \}[/tex]

Obviously we have to assume that [tex]\mathcal{F}\{f\}[/tex] is never zero, which for exaple happens when f is a Gaussian.
Why things seem to work in the Fourier domain but not in the time domain? Shouldn't the Fourier Transform define an isomorphism?
 
  • #7
mnb96 said:
OK, so what happens if I write:

[tex]f\ast f' = \delta[/tex]

[tex]\mathcal{F}\{f\ \ast f'\} = \mathcal{F}\{\delta\}[/tex]

[tex]\mathcal{F}\{f\} \mathcal{F}\{f'\} = 1[/tex]

[tex]\mathcal{F}\{f'\} = 1 / \mathcal{F}\{f\}[/tex]

[tex]f' = \mathcal{F}^{-1}\{ 1 / \mathcal{F}\{f\} \}[/tex]

Obviously we have to assume that [tex]\mathcal{F}\{f\}[/tex] is never zero, which for exaple happens when f is a Gaussian.
Why things seem to work in the Fourier domain but not in the time domain? Shouldn't the Fourier Transform define an isomorphism?

if [tex]G = \mathcal{F}(g)[/tex] is the Fourier transform of a function, then [tex]1/G[/tex] is not the Fourier transform of a function.

For example: [tex]G[/tex] goes to [tex]0[/tex] at [tex]\infty[/tex], but [tex]1/G[/tex] doesn't.
 
  • #8
Be very careful, you are going too fast.

Functions on [tex]\mathcal{L}_2[/tex] need not have to have an Fourier transform in the usual sense. You have to comfort yourself with the extension of [tex]\mathcal{L}_1\cap \mathcal{L}_2[/tex]

Things are not that trivial when it comes to these things. I would suggest you dig into deeper instead of fast conclusions
 

1. What is the definition of "Convolution Inverse"?

"Convolution Inverse" refers to a mathematical operation that reverses the effects of convolution, which is a process of combining two functions to produce a third function that expresses how the shape of one is modified by the other.

2. How is the Convolution Inverse related to the Family of Functions?

The Convolution Inverse is a mathematical operation that is used to find the inverse of a function in the Family of Functions. This allows us to find the original function that was modified by convolution.

3. What is the significance of the Convolution Inverse in real-world applications?

The Convolution Inverse has many practical applications, such as in signal processing, image processing, and data analysis. It allows us to reverse the effects of convolution and retrieve important information from processed data.

4. Are there any limitations to using the Convolution Inverse in mathematical calculations?

Yes, there are limitations to using the Convolution Inverse, as it can only be applied to certain types of functions. Additionally, it may be computationally intensive and require advanced mathematical techniques for its application.

5. How can the Convolution Inverse be visualized and understood?

The Convolution Inverse can be visualized and understood through graphical representations. It is also helpful to understand the concept of convolution and how it modifies functions. Additionally, studying mathematical proofs and examples can aid in understanding the Convolution Inverse.

Similar threads

  • Linear and Abstract Algebra
Replies
1
Views
1K
  • Linear and Abstract Algebra
Replies
3
Views
885
  • Linear and Abstract Algebra
Replies
4
Views
725
  • Calculus and Beyond Homework Help
Replies
1
Views
733
  • Programming and Computer Science
Replies
1
Views
794
  • Calculus and Beyond Homework Help
Replies
13
Views
1K
Replies
13
Views
1K
Replies
1
Views
1K
  • Linear and Abstract Algebra
Replies
7
Views
28K
  • Linear and Abstract Algebra
2
Replies
43
Views
5K
Back
Top