Let $H_a=\{f \in L^2: \hat{f}(\xi)=0 \text { for } |\xi| > a\}$. Define $\operatorname{sinc} x=\frac{\sin \pi x}{\pi x}$
Now $$f_k(x)=\sqrt{2a}\operatorname{sinc}(2ax-k)=\sqrt{2a}\tau_{\frac{k}{2a}}(\operatorname{sinc}(2ax))=\frac{1}{\sqrt{2a}}\tau_{\frac{k}{2a}}(\check{\chi}_{[-a,a]})(x)$$
So $$\hat{f}_k(x)=\frac{1}{\sqrt{2a}}e^{-2i\frac{k}{2a}\pi x} \hat{\check{\chi}}_{[-a,a]}(x) = \frac 1 {\sqrt{2a}} e^{i\frac{k}{a}\pi x} \chi_{[-a,a]}(x)$$
Thus, $f_k \in H_a$.
Moreover $$\langle f_k,f_\ell \rangle=\langle \hat{f}_k,\hat{f}_\ell \rangle = \int_{-a}^a \frac 1 {2a} e^{i\pi x(\frac{k-\ell} a)} \, dx$$
So $\{f_k: k \in \mathbb{Z}\}$ form an orthonormal set.
Now suppose that there exists a $g \in H_a$ such that $\langle g,f_k \rangle=0$ for all $k \in \mathbb{Z}$. Then $$\langle g,f_k \rangle=\langle \hat{g},\hat{f}_k \rangle = \int_{-a}^a \hat{g}(x)e^{\frac{-ikx \pi}{a}} \, dx = 0$$ for all $k$.
Let $M$ be the closed span of $\{e^{\frac{-ikx \pi}{a}}: k \in \mathbb{Z}\}$. By an application of Stone Weirstrass Theorem, we see that $M=L^2[-a,a]$. Thus $\hat{g}=0$. This implies that $g=0$.
This is what I think should happen for the above problem. I have a few doubts though. First of all, when I am calculating the fourier inversion of $\chi_{[-a,a]}$, a priori I don't know if I can do that or not. Secondly, while showing the orthogonality of the $f_k$'s,I am using Plancherel's Theorem but I am not sure, why I can use that.
Thanks for the help!