Apologies if this is not the correct way of asking a previously-asked but still couldn't solve problem. I tried editing the question and getting a response too. I'm not here to get any negative vote. Just need help. Here's the link
Let $\{ \lambda_k \}_{k \in \mathbb{Z}}$ be a sequence in $\mathbb{R}$ and assume that $\{ e^{i \lambda_k x} \}_{k \in \mathbb{Z}}$ is a Bessel sequence in $L^2(-\pi, \pi).$ Define $\phi$ by $\hat{\phi}=\chi_{[-\pi,\pi]}.$ How to prove that $\Phi:=\{ \phi( \cdot - \lambda_k/{(2 \pi)}), k \in \mathbb{Z} \}$ is a Bessel sequence in $L^2(\mathbb{R}).$ Further what can we say about the system $\Phi$ if $\{ e^{i \lambda_k x} \}_{k \in \mathbb{Z}}$ is a frame for $L^2(-\pi, \pi) ?$
I'm still struggling to come up with a solution. If anyone can explain me how to do this problem, it's a big help. Thank you.
I am assuming that the Fourier transform is defined as $$\hat{f}(\xi)=\int_\mathbb{R}f(x)e^{-2\pi i x\xi}dx.$$ Then it is easy to observe that $$\widehat{f(\cdot-t)}(\xi)=e^{-2\pi i t\xi}\hat{f}(\xi).$$
Now let $f\in L^2(\mathbb{R})$, it follows from Plancherel's theorem and the fact that $\{e^{i\lambda_kx}\}$ is a Bessel sequence that $$\begin{aligned}\sum_{k\in\mathbb{Z}}\left|\int_\mathbb{R}f(x)\overline{\phi(x-\frac{\lambda_k}{2\pi})}dx\right|^2&=\sum_{k\in\mathbb{Z}}\left|\int_\mathbb{R}\hat{f}(x)\overline{\widehat{\phi(\cdot-\frac{\lambda_k}{2\pi})}(x)}dx\right|^2\\ &=\sum_{k\in\mathbb{Z}}\left|\int_\mathbb{R}\hat{f}(x)\left[e^{i\lambda_kx}\overline{\hat{\phi}(x)}\right]dx\right|^2\\ &=\sum_{k\in\mathbb{Z}}\left|\int_\mathbb{R}\hat{f}(x)\overline{\hat{\phi}(x)}e^{i\lambda_kx}dx\right|^2\\ &\leq C\|\hat{f}\overline{\hat{\phi}}\|_2^2\\ &\leq C\|\hat{f}\|_2^2\\ &=C\|f\|_2^2 \end{aligned}$$