Reconstruction after Unitary Operator on Paley-Wiener space

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Background:

According to the Sampling Theorem for Reproducing Kernel Hilbert Spaces (RKHS), for any RKHS $H_{rk}$ with reproducing kernel $k(\cdot, \cdot)$, if I can find a set of points $\{t_{n}\}_{n \in \mathbb{Z}} \in \mathbb{R}$ such that $\{k(\cdot, t_{n}) \}_{n \in \mathbb{Z}}$ is an orthogonal basis for the space, I can reproduce any $g \in H_{rk}$ via kernel interpolation at $\{t_{n}\}_{n \in \mathbb{Z}}$.

So now suppose I have a function $f \in PW_{\pi}$, where $PW_{\pi}$ is the Paley-Weiner space with Fourier support $[-\pi, \pi]$. I then apply a unitary operator $U$ to the function and get $g = Uf$. It is well known that $PW_{\pi}$ is a RKHS, with $\{sinc_{\pi}(t-n)\}_{n \in \mathbb{Z}}$ as both the orthogonal basis and the reproducing kernel.


My Questions:

1.Is $UPW_{\pi}$ an RKHS?

2.If it is, is it possible to reconstruct $f$ via interpolation from the samples of $g$? If so, can we find the samples of $f$ that are required to reconstruct $f$ from the samples of $g$?


My Answers:

  1. Yes. My intuition is that since open sets on $L^{2}[\mathbb{R}]$ are unit circles, and since $U$ is unitary, it is just rotating these balls, which means $UPW_{\pi} = PW_{\pi}$. I am not sure how to prove this though, or if this is even the way to do so.

  2. Yes. All we have to do is interpolate $g$ and then map it back to $f$ via $f = U^{-1}g$. So if we want to find a sample of $f$, we first find all the needed samples of $g$, interpolate $g$ via the sampling theorem, and then resample $U^{-1}g$ at the points $\{U^{-1}t_{n}\}_{n \in \mathbb{Z}}$. These points $\{U^{-1}t_{n}\}_{n \in \mathbb{Z}}$ will be the corresponding sampling points of $f$. Any idea on how to start proving this?


Apologies:

My functional analysis is very weak. I am just trying to piece things together as I go, so I may be missing some fundamental concepts or parts here. If anybody has any suggestions on things to study to help me start to understand how to prove this, I would be grateful.