Let
$$f_n(t)=\sum_{j=1}^n \chi_{A_j}(t)$$
be a piecewise continuous function, where $A_j(t)=(t_j-t_{j-1})$ is the indicator function. A picture of the intervals of the indicator functions are given here, in terms of $\xi$ (Xi on the image) instead of $t$:
Also, a picture of such a function is:
Now, let the Fourier series of the function be given by:
$$f_n(t)=\frac{\alpha_0}{2}+\sum_{k=1}^\infty \alpha_k\cos\omega k t+\beta_k\sin\omega kt $$
Let us consider the $j^{th}$ indicator function. Here its coefficients are, where $\chi$ is the respective value of f(t) on the specific interval of the $j^{th}$ indicator function:
\begin{equation} \alpha_0^j=\frac{1}{a}\int_{-a}^a\chi_{A_j}(t)\text{d}t \end{equation}
\begin{equation} \alpha_k^j=\frac{1}{a}\int_{-a}^a\chi_{A_j}(t)\cos\bigg(\frac{\pi k}{a}t\bigg)\text{d}t \end{equation}
\begin{equation} \beta_k^j=\frac{1}{a}\int_{-a}^a\chi_{A_j}(t)\sin\bigg(\frac{\pi k}{a}t\bigg)\text{d}t \end{equation}
Here comes the major clue: The number of intervals of the piece-wise function, dividing (-a,a), which is $n$ (at the top), are related to the number of the given Fourier coefficients.
So if $n=4$, then we should have 4 coefficient for each type, so $\{\alpha_0^1, \alpha_0^2, \alpha_0^3, \alpha_0^4\}$, and $\{\alpha_k^1, \alpha_k^2, \alpha_k^3, \alpha_k^4\}$ and $\{\beta_k^1, \beta_k^2, \beta_k^3, \beta_k^4\}$.
Ok, that is fine. 4 indicator functions make up the piecewise function, and give 4 coefficients of each of the three types given above. If we sum these four coefficient together, then we get the whole coefficient for the entire piecewise function at the top.
But at this stage I should prove that we can uniquely determine the piece-wise constant function (numbers for $\chi_j, \ j=1,2,\cdots, n $) on a given grid of points $t_j, \ j=0,1,\cdots, n $ solely from the coefficients. This means in other words that we can find $\chi_1$ from $\alpha_0^1$ for instance, or $\chi_2$ from $\alpha_k^2$ for that matter.
But where should I start?
Any hints are appreciated
Thanks