For which $f$ does $f*f=\sqrt{2\pi}f$ hold?

Question

Im facing the following problem: Find all Elements $f$ of $L^1(\mathbb{T})$ with $f*f=\sqrt{2\pi}f$.

$\mathbb{T}$ being the unit circle. $*$ stands for convolution.

So my first idea was to use the convolution theorem: The convolution theorem states that

$\mathcal{F}\{f * g\} = k\cdot \mathcal{F}\{f\}\cdot \mathcal{F}\{g\}$

where $\mathcal{F}\{f\}$ denotes the Fourier transform of $f$, and $k$ is a constant that depends on the specific Normalizing constant of the Fourier transform.

But I cannot get rid of the Fourier transform. In fact I have no idea what the necessary condition should look like. Any ideas on this?

EDIT: I found this to be somekind of similar problem: http://sebastiaanjanssens.nl/math/calculating-an-auto-convolution-integral-by-fourier-transforms/

Here the result is $f*f=\pi f$. But in fact I need to work "backwords" in order to get my $f$.

Answer 1

Essentially, after applying Fourier transform you face the problem $$g^2 = A g, \quad \text{ A is a fixed constant}$$ with $g$ being a Fourier image of a trigonometric polynomial $f$.

Now what can you conclude on possible values taken by $g$?

Answer 2

With $f(\theta) = f_0 + f_1\cos\theta + f_2\cos(2\theta) + \cdots$, and assuming appropriate convergence for exchange of the sums and integral, \begin{align*} (f * f)(\theta) &= \int_0^{2\pi}\, f(\theta - \phi)f(\phi)\, d\phi \\ &= \int_0^{2\pi}\sum_{m=0}^\infty \sum_{n=0}^\infty\, f_m f_n\cos(m(\theta - \phi)) \cos(n\phi)\, d\phi \\ &= \sum_{m=0}^\infty \sum_{n=0}^\infty\, f_m f_n\int_0^{2\pi}\cos(m(\theta - \phi)) \cos(n\phi)\, d\phi \\ &= 2\pi f_0^2 + \sum_{m=1}^\infty \sum_{n=1}^\infty\, \pi f_m^2 \cos(m\theta). \end{align*} Equate the coefficients to see what's required for $f*f = Cf$.