Prove the independence of "a" of a Fourier coefficient of a Fourier series calculated over the interval $[-a,a]$

28 Views Asked by At

I need to prove that the Fourier coefficients of a Fourier series (say for instance $\alpha_0$) calculated for a function on the interval $[-a,a]$ is independent on $a$ itself.

Define the stepwise function:

\begin{equation} \chi_{A_j}(t)=\begin{cases} 1, \ \ \ \ \xi_{j-1}<t<\xi_j\\ 0, \ \ \ \ \ \text{elsewhere}, \end{cases} \end{equation}

Define the Fourier coefficient:

$$\alpha_0=\frac{1}{2a}\int_{-a}^a\chi_{A_j}(t)$$

Claim: The Fourier coefficient is independent on $a$. Sounds weird to me, but apparently this is true.

Let

\begin{equation} \alpha_0^j=\frac{1}{2a}\int_{-a}^a\chi_{A_j}(t)\text{d}t=\frac{1}{2a}\int_{\xi_j-1}^{\xi_j}\text{d}t=\frac{1}{2a}(\xi_j-\xi_{j-1}) \end{equation} with $(\xi_j-\xi_{j-1})=h$ we get:

\begin{equation} \alpha_0^j=\frac{h}{2a} \end{equation} Since $h=\frac{2a}{n}$ we get:

\begin{equation}\label{a01} \alpha_{0}^j=\frac{1}{n}, \end{equation}

Clearly, $\alpha_0$ is independent on $a$. But this is not sufficient I think,

Any ideas?

Thanks