# Convolution inverse

1. Mar 2, 2009

### mnb96

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 $$f^{\ast-1}$$ gives the delta-function: $$f \ast f^{\ast-1} = \delta$$
http://en.wikipedia.org/wiki/Convolution#Convolution_inverse

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

2. Mar 2, 2009

### John Creighto

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. Mar 3, 2009

### mnb96

...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 $$L^2(\Re)$$?

4. Jul 2, 2009

### mnb96

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 $$f \ast f^{\ast-1} = \delta$$?

5. Jul 2, 2009

### g_edgar

If $$f$$ and $$g$$ are functions of a real variable, then so is their convolution. Since $$\delta$$ 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 $$\delta(x-c)$$ is called a "unit mass at $$c$$". The convolution inverse of the unit mass at $$c$$ is the unit mass at $$-c$$.

Further non-examples: if $$\mu$$ and $$\nu$$ 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 $$\delta$$, 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. Jul 2, 2009

### mnb96

OK, so what happens if I write:

$$f\ast f' = \delta$$

$$\mathcal{F}\{f\ \ast f'\} = \mathcal{F}\{\delta\}$$

$$\mathcal{F}\{f\} \mathcal{F}\{f'\} = 1$$

$$\mathcal{F}\{f'\} = 1 / \mathcal{F}\{f\}$$

$$f' = \mathcal{F}^{-1}\{ 1 / \mathcal{F}\{f\} \}$$

Obviously we have to assume that $$\mathcal{F}\{f\}$$ 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. Jul 2, 2009

### g_edgar

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

For example: $$G$$ goes to $$0$$ at $$\infty$$, but $$1/G$$ doesn't.

8. Jul 4, 2009

### trambolin

Be very careful, you are going too fast.

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

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