Prove Discrete Time Fourier Series Multiplication property

Question

Note: This is not a homework problem. I'm just stalled at a point because I think I might be interpreting the duality property incorrectly.

If $x_1[n]$ and $x_2[n]$ are periodic with period N, then if I periodically convolve them in time domain as follows:

$$y[n]=\sum_{m=0}^{N-1} x_1[m]x_2[n-m] \iff Y[k]=X_1[k]X_2[k] $$

Then I want to prove the corresponding property wherein I multiply in time domain using duality property.

As per duality property if $x[n]$ is periodic in the time domain with period N with Fourier series coefficients as X[k]

$$x[n] \iff X[k]$$

then

$$X[n] \iff Nx[-k]$$

Then if I multiply in the time domain this is what I get

$$x_1[n]x_2[n] \iff N \sum_{m=0}^{N-1} X_1[m]X_2[k-m]$$

But textbook says

$$x_1[n]x_2[n] \iff {\frac{1}{N}} \sum_{m=0}^{N-1} X_1[m]X_2[k-m]$$

where am I going wrong?

Answer 1

well, the thrid equation should be $X[n] \iff Nx[-k]$

The DFS of $x_1[n]x_2[n]$ is $$ \mathrm{DFS}\{x_1[n]x_2[n]\} = \sum_{n=0}^{N-1} x_1[n]x_2[n]W_N^{nk} \tag{1} $$ where $W_N^{nk}=e^{-j\frac{2\pi}{N}nk}$.

Now recall the definition of IDFS $$ x_1[n] = \mathrm{IDFS}\{X_1[m]\} = \frac{1}{N}\sum_{m=0}^{N-1}X_1[m]W_{N}^{-mn} \tag{2} $$ Substituting Eq. (2) into Eq. (1) and swap the order of summation, we have $$ \begin{aligned} \mathrm{DFS}\{x_1[n]x_2[n]\} &= \sum_{n=0}^{N-1} \left(\frac{1}{N}\sum_{m=0}^{N-1}X_1[m]W_{N}^{-mn}\right)x_2[n]W_N^{nk}\\ &=\frac{1}{N}\sum_{m=0}^{N-1}X_1[m] \left( \sum_{n=0}^{N-1}x_2[n]W_N^{n(k-m)} \right)\\ &=\frac{1}{N}\sum_{m=0}^{N-1}X_1[m] X_2[k-m] \end{aligned}\tag{3} $$