Convolution question

I'm not sure I can make sense out of your question. Over what interval are you taking the convolution? What happens if f(x) is a constant or f(x)= x?

It is true that the convolution, over [itex]-\infty[/itex] to [itex]\infty[/itex], of two Gaussians is a Gaussian so this may be a "fixed point" theorem.

 

What makes you think this? In raising a function to a very high power, the parts with |f(x)|>1 get increasingly larger while those with |f(x)|<1 vanish. So, for example, if the function has several narrow peaks whose height is greater than one, its very high powers will have spikes where each of those peaks were and vanish everywhere else.

 

I don't think indigojoker meant "high power". I think he meant applying the convolution a large number of times.

 

Right, but then you noted that a convolution in, say, the time domain corresponds to a product in the frequency domain, so convolving a function with itself n times corresponds to raising its fourier transform to the nth power. This is correct, but it doesn't follow that you'll always end up with gaussian like functions.

 

I don't see why anything that isn't a guassian should "approach" a gaussian. And this is impossible to tell anyway unless you say exactly what you mean by "approach", ie, how do you measure how similar two functions are?

 

Think about what convolution means in the frequency domain, and it's clear that almost any function will not end up as a gaussian.