Let $d >0.$ Prove that $\{ e^{i k x / d} \}_{k \in \mathbb{Z}}$ is a frame for $L^2(-R,R)$ if and only if $0 < R \leq d \pi.$
The frame condition implies that for $\{ f_k \}_{k=1}^{\infty}$ $$A \cdot \| f \|^2 \leq \sum_{k=1}^m \left\vert\langle f, f_k \rangle\right\vert^2 \leq B \cdot \| f \|^2, ~ \forall f \in L^2(-R,R).$$ How to incorporate the square integrable to function space to prove the given result. I'm struggling with this problem. Any help is much appreciated.
Let $e_k(x) = \frac{1}{\sqrt{2r}}e^{i \pi k x/ r}$. The Fourier series main theorem is that $\{e_k\}_{k \in \mathbb{Z}}$ is an orthonormal (unitary) basis of $L^2(-r,r)$.
If and only if $ R \le r$, then $f \in L^2(-R,R)$ can be seen as $f1_{|x| < R} \in L^2(-r,r)$,
so that $$f1_{|x| < R} = \underbrace{\sum_{k\in \mathbb{Z}}}_{\text{in } L^2(-r,r)} \langle f,e_k\rangle e_k$$ And Parseval's theorem gives $$\|f 1_{|x| < R}\|^2_{L^2(-r,r)} = \|f\|^2_{L^2(-R,R)} = \sum_{k\in \mathbb{Z}} |\langle f,e_k\rangle|^2$$
Now if $\forall x, f(x) \in \mathbb{R}$, then $|\langle f,e_k\rangle| = |\langle f,e_{-k}\rangle|$, and you also have $\frac{1}{2}\|f\|^2 <\sum_{k=0}^\infty |\langle f,e_k\rangle|^2 < \|f\|^2$