Strange convolution equation

[exmachina]

I've arrived at the following equation involving the convolution of two functions:

$f(x) = \int_{-\infty}^{\infty} f(t) g(t-x) dt = f(x) \ast g(x)$

Where:

$g(z) = e^{-z^2/2}$

In other words, a function convoluted with a Gaussian pdf results in the same function.

I've tried taking Fourier transforms, realizing that the FT of a gaussian results in another Gaussian:

$F[f(x)] = F[f(x) \ast g(x)] = F[f(x)] \cdot F[g(x)]$

But this results in the $F[f(x)]$ cancelling out, leaving me with just:

$1 = F[g(x)] = e^{-w^2/2}$

Any suggestions?

 

[micromass]

What if f=0?

 

[exmachina]

Yes that is the trivial solution.

Perhaps this can be casted as an eigenvalue problem - as it seems to imply that the convolution operator (wrt to the gaussian) may have certain eigenvalues and corresponding eigenfunctions f(x) being one of them

 

[exmachina]

edit - doh - this obviously implies that f(x) must be equal to 0 (no other solution satisfies:

f(x)=f(x)g(x) unless g(x) = 1, which in this case, it isn't)

 

[blue_raver22]

I would suggest checking your calculations and making sure that you are using the correct properties and definitions of the Fourier transform. It is possible that there may be a mistake in your calculations that is leading to the cancellation of terms.

Additionally, it is important to note that the convolution of two functions is not always equal to one of the functions. In this case, it seems that the convolution of a Gaussian with itself results in a Gaussian, but this may not always be the case.

Another possibility is to consider using other mathematical tools, such as Laplace transforms or z-transforms, to solve the convolution equation. These may provide a different perspective and potentially lead to a solution.

Lastly, it may also be helpful to consult with a colleague or mentor who has experience with convolution and Fourier transforms to get their insights and suggestions on how to approach this problem. Collaboration and discussion can often lead to a better understanding and resolution of complex equations.