What Happens When a Function is Convolved with a Gaussian PDF?

[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)