Is there a function [math] g: R \rightarrow R [/math] such that [math] g^2=g*g [/math], where * denotes convolution?We can take g to be integrable and bounded. This problem is hard because it’s both nonlinear and non-local. Obvious strategies all fail, with simple inequalities failing to rule out a solution and Fourier transforming getting you back to where you started. Attempting to do power series or other forms of expansion end poorly because what is well behaved for the square term isn’t for the autoconvolution, and vice versa. Simple inequalities show that no such solution can exist which is supported on [math] (1/e, \infty) [/math], even over the shared domain of [math] g^2 [/math] and [math] g*g [/math]. In general, no finite support works because the convolution will double it. It’s worth noting that a solution to the problem can be horizontally rescaled to a solution for [math] g^2= c g*g [/math]. This isn’t the case if we define the problem over the domain (0,1) and take the convolution to be periodic (modulo 1). In that case, Jensen immediately rules out solutions to [math] g^2= c g*g [/math] unless [math] c>1 [/math]. Beyond that I have no idea. Another way to solve the problem would be to find a function who is its own Fourier transform, and whose square is its own Fourier transform as well. This is evidently difficult. Any new strategies would be greatly appreciated.
Obvious strategies do not all fail. Convolution in the Fourier domain is multiplication, so in the Fourier domain, you are looking for functions fixed by squaring. These functions are the ones that take values only in {0,1}.The simplest function of the kind you are seeking is the Fourier transform of a rectangle, which is the sinc function (f(0) = 1, f(x) =sin(x)/x).
>>16296365g=0
>>16296377Sorry, misread your question.
>>16296378Yes this is true. Given the homogeneity of the problem, I should restate "can you find a [math] ||g|| = 1 [/math] with [math] g^2 = g*g [/math].
>>16296365I think a Gaussian might satisfy this. The square of a Gaussian is proportional to a Gaussian with half the variance. The FT of a Gaussian is also a Gaussian with variance scaling as the inverse of the original's variance.You just need to (hopefully) find some scaling that it makes it work.
>>16296387Doesn't work, squaring gives half the variance but convolution gives twice the variance.
>>16296387I've tried my best to get this to work. Problem is that convolution stretches Gaussians, and squaring shrinks them.You could even try a linear superposition of all possible widths of Gaussians centered at the origin. I.e. for an amplitude function [math]S(\sigma) [/math], try [math] g(x) = \int d\sigma \hspace{.05cm} S(\sigma) \mathbb{N}(0,\sigma)([/math]. This unfortunately leads nowhere, as you are still stuck trying to solve a non-local non-linear integral equation.
>>16296387Convolution adds variances, it doesn't multiply them.>>16296398Yes, convolution grows variances, squaring shrinks them and Fourier transforming inverts them. The variance side of it works out. I'm not sure about the amplitudes though.
>>16296390Sorry misreplied again. You need twice the variance because you're in the Fourier domain, (in which scales are inverted somehow)
>>16296365No there's not
relevant stack exchange post with more detail: https://math.stackexchange.com/questions/4768288/is-there-a-function-whose-autoconvolution-is-its-square-g2x-gg-x
