Prove that $\sum_{k=-\infty}^\infty e^{-j2\pi f k T}=\sum_{k=-\infty}^\infty\delta(f-\frac{k}{T})$ This is part of a proof itself.
$\sum_{k=-\infty}^\infty e^{-j2\pi f k T}=\sum_{k=-\infty}^\infty\delta(f-\frac{k}{T})$
$\delta$ is Dirac function.
It's been a while I am thinking about this part but I have no clue.
Edit:
Sorry it took a long time to answer.
The lecture can be found on this page ECE5660  SPREAD SPECTRUM COMMUNICATIONS
and in lecture 3, chapter 3 section. Here is the link to the file. The part I'm asking is on top of the page 8.
 A: This is essentially the Poisson summation formula (cf. http://en.wikipedia.org/wiki/Poisson_summation_formula ), but written in the notation of tempered distributions.
Note that the left hand side does not make sense as an ordinary function, but can be interpreted as a tempered distribution by
$$
\left\langle \sum_{k=-\infty}^{\infty}e^{-2\pi ixkT},f\right\rangle :=\sum_{k=-\infty}^{\infty}\int_{\mathbb{R}}f\left(x\right)\cdot e^{-2\pi ixkT}\, dx=\sum_{k=-\infty}^{\infty}\widehat{f}\left(kT\right).
$$
Now consider
$$
g\left(x\right):=\sum_{k=-\infty}^{\infty}f\left(x+\frac{k}{T}\right).
$$
Then $g$ is $\frac{1}{T}$-periodic and we have (here, $\widehat{g}\left(\ell\right)$
denotes the $\ell$-th Fourier coefficient w.r.t the orthonormal basis
$\left(x\mapsto \sqrt{T}\cdot e^{2\pi i\ell Tx}\right)_{\ell\in\mathbb{Z}}$
of $L^{2}\left(\left[0,\frac{1}{T}\right]\right)$):
\begin{eqnarray*}
\widehat{g}\left(\ell\right) & = & \sum_{k=-\infty}^{\infty}\int_{0}^{\frac{1}{T}}f\left(x+\frac{k}{T}\right)\cdot \sqrt{T}\cdot e^{-2\pi i\ell Tx}\, dx\\
 & = & \sum_{k=-\infty}^{\infty}\int_{0}^{\frac{1}{T}}f\left(x+\frac{k}{T}\right)\cdot \sqrt{T}\cdot e^{-2\pi i\ell T\left(x+\frac{k}{T}\right)}\, dx\\
 & = & \sum_{k=-\infty}^{\infty}\int_{\frac{k}{T}}^{\frac{k+1}{T}}f\left(y\right)\cdot \sqrt{T}\cdot e^{-2\pi i\ell Ty}\, dy\\
 & = & \sqrt{T}\cdot\int_{\mathbb{R}}f\left(y\right)\cdot e^{-2\pi i\ell Ty}\, dy=\sqrt{T}\cdot\widehat{f}\left(T\ell\right).
\end{eqnarray*}
If $f$ is a Schwartz function, the same is true of $\widehat{f}$,
which implies that the series
$$
\sum_{\ell=-\infty}^{\infty}\widehat{g}\left(\ell\right)\cdot \sqrt{T}\cdot e^{2\pi i\ell Tx}=T\cdot\sum_{\ell=-\infty}^{\infty}\widehat{f}\left(T\ell\right)\cdot e^{2\pi i\ell Tx}
$$
converges (locally) uniformly on $\mathbb{R}$ because of the rapid
decay of $\widehat{f}\left(T\ell\right)$.
By injectivity of the Fourier transform on $L^{1}\left(\left[0,\frac{1}{T}\right]\right)$,
conclude
$$
g=T\cdot\sum_{\ell=-\infty}^{\infty}\widehat{f}\left(T\ell\right)\cdot e^{2\pi i\ell Tx}
$$
almost everywhere and then everywhere, because both sides are continuous
(because $f$ is Schwartz, the series defining $g$ also converges
uniformly).
Plugging in $x=0$ yields
$$
\left\langle \sum_{k=-\infty}^{\infty}\delta\left(\cdot-\frac{k}{T}\right),f\right\rangle =\sum_{k=-\infty}^{\infty}f\left(\frac{k}{T}\right)=g\left(0\right)=T\cdot\sum_{\ell=-\infty}^{\infty}\widehat{f}\left(T\ell\right)=T\cdot\left\langle \sum_{k=-\infty}^{\infty}e^{-2\pi ixkT},f\right\rangle ,
$$
where I used the "physicist" notation for the delta distribution
on the left hand side.
The above also shows that your target equation is only correct up to a factor of $T$.
