Discrete Time Fourior Transfrom Properties and Proofs - Rhea

$ = \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} e^{-j\omega 2\pi} $
$ = e^{-j\omega 2\pi} \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} $
$ = (1)\chi(\omega) = \chi(\omega) $
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$ \sum_{n=-\infty}^{\infty}ax_{1}[n]e^{-j\omega n} + \sum_{n=-\infty}^{\infty}bx_{2}[n]e^{-j\omega n} $
$ =a\chi_{1}(\omega) + b\chi_{2}(\omega) $
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2) $ e^{-j{\omega}_{o}n}x[n] \rightarrow \chi[\omega - \omega_{o}] $

let $ m = n - n_{o} $
$ \sum_{m=-\infty}^{\infty}x[m]e^{-j\omega m + n_{o}} $
$ = e^{-j\omega n_{o}}\sum_{m=-\infty}^{\infty}x[m] $
$ = e^{-j\omega n_{o}}\chi(\omega) $
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$ = \sum_{n=-\infty}^{\infty}x[n][cos(\omega n) + jsin(\omega n)] $
$ = \sum_{n=-\infty}^{\infty}x[n]cos(\omega n) + \sum_{n=-\infty}^{\infty}x[n]jsin(\omega n) $
$ = \sum_{n=-\infty}^{\infty}x[n]\frac{1}{2}[e^{j\omega n} + e^{-j\omega n}] + \sum_{n=-\infty}^{\infty}x[n]j\frac{1}{2j}[e^{j\omega n} - e^{-j\omega n}] $
Things Cancel out and you are left with..
$ = \sum_{n=-\infty}^{\infty}x[n]e^{j\omega n} $
$ = \chi^{*}(-\omega) $
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$ = \frac{1}{2\pi}\int_{0}^{2\pi}\chi(\omega)[\sum_{n=-\infty}^{\infty}x[n](\frac{1}{2\pi}e^{j\omega n}]d\omega $
$ = \frac{1}{2\pi}\int_{0}^{2\pi}\chi(\omega)[\chi(-\omega)]d\omega $
$ = \frac{1}{2\pi }\int_{0}^{2\pi}\left | \chi (\omega) \right |^{2}d\omega $
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$ \mathfrak{F}(x[n]*y[n]) = \sum_{n=-\infty}^{\infty}[\sum_{k=-\infty}^{\infty}x[k]*y[n-k]]e^{-j\omega n} $
We can replace $ e^{-j\omega n} $ by $ e^{-j\omega n} = e^{-j\omega n + j\omega k -j\omega k }= e^{-j\omega( n - k ) - j\omega k }= e^{-j\omega( n - k )} e^{-j\omega k } $
So..
$ = \sum_{n=-\infty}^{\infty}[\sum_{k=-\infty}^{\infty}x[k]*y[n-k]]e^{-j\omega( n - k )} e^{-j\omega k} $
$ = \sum_{k=-\infty}^{\infty}e^{-j\omega k }[\sum_{n=-\infty}^{\infty}x[k]*y[n-k]]e^{-j\omega( n - k )}] $
$ = \chi(\omega)\gamma(\omega) $
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