Convolution theorem for bernoulli polynomials

By | June 13, 2018

The convolution theorem states that the fourier transform $\mathcal{F}$ of the convolution $*$ of two functions $f,g$ is equal to the pointwise product $\cdot$ of the fourier transform of each function.
Or

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

Now the fourier coefficients of the $n$th bernoulli polynomial $B_n$ is $\frac{1}{k^n}$ for the fourier basis $\exp(2\pi i k x)$. The pointwise product of the fourier coefficients of two bernoulli polynomials $B_m$ and $B_n$ is $$\frac{1}{k^n}\cdot \frac{1}{k^m}=\frac{1}{k^{m+n}}$$ for the basis function $\exp(2\pi i k y)$ (here $i$ is the imaginary unit) and the product of the basis function as it appears in the definition of convolution works out as $$\exp(2\pi i k (x - y))\cdot \exp(2\pi i k y)=\exp(2\pi i k x)\textrm{.}$$ So it appears that the convolution of the bernoulli polynomials is a simple example of the convolution theorem if the fourier expansion of the bernoulli polynomials is used.
Unfortuately i am shaky with the details of the fourier transform (difference between fourier series and fourier transform, finite interval vs. lower integration bound from $-\infty$ to $+\infty$) and the details of the convolution theorem for that matter. For that reason i am looking for someone to help me with the details using the bernoulli polynomials as an example.