Discrete Fourier Transform and Opposite Convolution Theorem

Question

I am reading the Wiki for DFT. There is a part for circular convolution theorem which sounds a bit odd saying:

$$ \mathcal{F} \left \{ \mathbf{x\cdot y} \right \}_k \ \stackrel{\mathrm{def}}{=} \sum_{n=0}^{N-1} x_n \cdot y_n \cdot e^{-\frac{2\pi i}{N} k n} =\frac{1}{N} (\mathbf{X * Y_N})_k, \,$$   which is the circular convolution of $\mathbf{X}$ and $\mathbf{Y}$.

I feel it's incorrect because writing $y_n$ in Fourier basis, and multiplying it by $e^{-\frac{2\pi i}{N} k n}$ will result in a shift in index by $k$, and that means the result should be the cross-corrolation of two signals.

And secondly if either this or my argument is correct why this does not hold for continuous signals.

P.S. The question has been asked here as well since I was unaware of the most proper domain for my question.

Answer 1

I'm not sure I understand your argument, but here's a proof which should make the relationship clear:

$$IDFT\left\{\frac{1}{N}(X*Y_N)_k\right\}=\frac{1}{N^2}\sum_{k=0}^{N-1}(X*Y_N)_ke^{i2\pi nk/N}=\\ =\frac{1}{N^2}\sum_{k=0}^{N-1}\sum_{l=0}^{N-1}X_lY_{(k-l)_N}e^{i2\pi nk/N}= \frac{1}{N^2}\sum_{l=0}^{N-1}X_l\sum_{k=0}^{N-1}Y_{(k-l)_N}e^{i2\pi nk/N}=\\ \frac{1}{N^2}\sum_{l=0}^{N-1}X_le^{i2\pi nl/N}\sum_{k=0}^{N-1}Y_{(k-l)_N}e^{i2\pi n(k-l)/N}=\\=\frac{1}{N}\sum_{l=0}^{N-1}X_le^{i2\pi nl/N}\cdot\frac{1}{N}\sum_{k=0}^{N-1}Y_{k}e^{i2\pi nk/N}=x_ny_n$$

where $(k-l)_N$ means $(k-l)$ modulo $N$. I hope this helps.