Definition
Let $f : \mathbb R \to \mathbb C$ be a function with period $2l$ and Lebesgue integrable on $[-l, l].$
Then, define $S_N[f] : \mathbb R \to \mathbb C$ by $S_N [f](x)=\displaystyle\sum_{k=-N}^N c_k (f) e^{ik\omega x} ,$ where $\omega=\dfrac{\pi}{l}$ and $c_k(f)=\dfrac{1}{2l}\displaystyle\int_{-l}^l f(s) e^{-ik\omega s} \ \ ds$.
Then, $f$ is Fourier series expandable in $L^2([-l,l]) $ $\underset{\mathrm{def}}\iff$ $\| S_N[f]-f \|_{L^2([-l,l])} \to 0 $ as $N \to \infty$
Then, I want to prove that all trigonometric polynomials are Fourier series expandable.
Let $g$ be a trigonometric polynomial.
Then, there exists $M \in \mathbb N, \{a_j \}_{j=1}^M \subset \mathbb C, \{ k_j \}_{j=1}^M \subset \mathbb Z$
such that $k_j \neq k_l$ for $j\neq l$ and $g(x)=\displaystyle\sum_{j=1}^M a_j e^{ik_j\ \omega x}$.
First, I consider $S_N [g](x)=\displaystyle\sum_{k=-N}^N c_k (g) e^{ik\omega x}$.
\begin{align} c_k(g) &=\dfrac{1}{2l}\int_{-l}^l g(s) e^{-ik\omega s} \ ds \\ &=\sum_{j=1}^M a_j \cdot \dfrac{1}{2l} \int_{-l}^l e^{i(k_j-k) \omega s}\ \ ds. \end{align}
Thus, \begin{align} S_N [g](x) &=\displaystyle\sum_{k=-N}^N c_k (g) e^{ik\omega x} \\ &=\sum_{k=-N}^N \ \Bigg[ \sum_{j=1}^M a_j \cdot \dfrac{1}{2l} \int_{-l}^l e^{i(k_j-k) \omega s}\ \ ds \cdot e^{ik\omega x}\ \Bigg]. \cdots (\ast) \end{align}
And \begin{align} \| S_N[g]-g \|_{L^2([-l,l])} &=\int_{-l}^l |S_N[g](x)-g(x)|^2 \ dx \end{align} So I have to consider $|S_N[g](x)-g(x)| =\left| \displaystyle\sum_{k=-N}^N \ \Bigg[ \sum_{j=1}^M a_j \cdot \dfrac{1}{2l} \int_{-l}^l e^{i(k_j-k) \omega s}\ \ ds \cdot e^{ik\omega x}\ \Bigg] - \displaystyle\sum_{j=1}^M a_j e^{ik_j\ \omega x} \right|$
but this is too complicated and I don't know how I can proceed. I think that in order to simplify $(\ast)$, I can use $$\dfrac{1}{2l} \int_{-l}^l e^{i(p-q) \omega x}\ = \begin{cases} 0\ \mathrm{if} \ p\neq q \\ 1\ \mathrm{if} \ p=q \end{cases}$$ but I have no idea how I could process $(\ast)$ using this.
Thanks for your help.