I need some help with this exercise:
Given $A>0$, let $L_{A}^2(\mathbb{R})$ the subspace of $L^2(\mathbb{R})$ of the functions $f$ that satisfy $\hat{f}=\chi_{[\frac{-A}{2},\frac{A}{2}]}\hat{f}$. Prove that the set: $\left\{\sqrt{A}\frac{\sin(\pi(At-k))}{\pi(At-k)}/ k\in{\mathbb{Z}}\right\}$ is an orthonormal basis of $L^2_A(\mathbb{R})$.
This is an exercise of a Fourier series course, and I really don't know how to lead with it... Thanks a lot for any help.
Expanding $\sin(2\pi x)$ yields $$ \int_{-\infty}^\infty\frac{\sin(2\pi x)}{\pi x}e^{-2\pi ix\xi}\,\mathrm{d}x =\int_{-\infty}^\infty\frac{e^{2\pi ix(1-\xi)}-e^{-2\pi ix(1+\xi)}}{2\pi ix}\,\mathrm{d}x\tag{1} $$ Using the contour, $\gamma$, from $-\infty-i\epsilon$ to $+\infty-i\epsilon$, then circling back counterclockwise through the upper half-plane for positive $\alpha$, or clockwise through the lower half-plane for negative $\alpha$, we get $$ \int_\gamma\frac{e^{2\pi ix\alpha}}{2\pi ix}\,\mathrm{d}x =\left\{\begin{array}{cl} 1&\text{if }\alpha\ge0\\ 0&\text{if }\alpha\lt0 \end{array}\right.\tag{2} $$ Thus, using $(1)$ and $(2)$, we have $$ \int_{-\infty}^\infty\frac{\sin(2\pi x)}{\pi x}e^{-2\pi ix\xi}\,\mathrm{d}x =\left\{\begin{array}{cl} 1&\text{if }-1\le\xi\le1\\ 0&\text{otherwise} \end{array}\right.\tag{3} $$ Usual scaling of the Fourier Transform yields $$ \sqrt{A}\int_{-\infty}^\infty\frac{\sin(\pi Ax)}{\pi Ax}e^{-2\pi ix\xi}\,\mathrm{d}x =\left\{\begin{array}{cl} \frac1{\sqrt{A}}&\text{if }-\frac A2\le\xi\le\frac A2\\ 0&\text{otherwise} \end{array}\right.\tag{4} $$ Usual translation of the Fourier Transform yields $$ \sqrt{A}\int_{-\infty}^\infty\frac{\sin(\pi(Ax-k))}{\pi(Ax-k)}e^{-2\pi ix\xi}\,\mathrm{d}x =\left\{\begin{array}{cl} \frac1{\sqrt{A}}e^{-2\pi i\xi k/A}&\text{if }-\frac A2\le\xi\le\frac A2\\ 0&\text{otherwise} \end{array}\right.\tag{5} $$ By the standard $L^2$ duality of the Fourier Transform, since the functions on the right hand side of $(5)$ are orthonormal basis elements of $L^2\left[-\frac A2,\frac A2\right]$, so are $\sqrt{A}\frac{\sin(\pi(Ax-k))}{\pi(Ax-k)}$ an orthonormal basis of $L_A^2(\mathbb{R})$.