I would like to show that
$$\frac{1}{2\pi}\int_{-\pi}^{\pi} \left( \sum_{j \in \mathbb{Z}} e^{-ikj} f(\varepsilon j)\right) e^{ikx} \, dk \to f(\varepsilon x)$$ as $\varepsilon \to 0$ in $L^2$.
The left expression can be thought of as an interpolation operator on the sequnce $f(\varepsilon j)$.
Using a basic change of variables with $u = k/\varepsilon$ one finds that $$\frac{1}{2\pi}\int_{-\pi}^\pi \left(\sum_{j \in \mathbb{Z}} e^{-ikj} f(\varepsilon j)\right) e^{ikx} \, dk = \frac{1}{2\pi}\int_{-\pi/\varepsilon}^{\pi/\varepsilon} \left(\sum_{j \in \mathbb{Z}} \varepsilon e^{-iu\varepsilon j}f(\varepsilon j)\right) e^{iu\varepsilon x} \, du.$$
Now we have it in a form where if we could pretend limits always turned out as they should and cheat
$$\lim_{\epsilon \to 0}\frac{1}{2\pi}\int_{-\pi/\epsilon}^{\pi/\epsilon}(\sum_{j \in \mathbb{Z}}\epsilon e^{-iu\epsilon j}f(\epsilon j))e^{iu\epsilon x}du =\frac{1}{2\pi}\int_{-\infty}^{\infty}(\int_{-\infty}^{\infty}e^{-ius}f(s)ds)e^{iu\epsilon x}du=f(\epsilon x).$$
The last equality follows from Fourier inversion. But of course the limit is not this easy and more so since the limit needs to be in the $L^2$ sense. One approach I have tried is to swap the integral and the sum. One comes up with $$\frac{1}{2\pi}\int_{-\pi/\epsilon}^{\pi/\epsilon}(\sum_{j \in \mathbb{Z}}\epsilon e^{-iu\epsilon j}f(\epsilon j))e^{iu\epsilon x}du=\sum_{j\in\mathbb{Z}}f(\epsilon j)\operatorname{sinc}(x-j).$$ Here $\operatorname{sinc}(x)$ is normalized. So now can it be shown that $$\sum_{j\in\mathbb{Z}}f(\epsilon j)\operatorname{sinc}(x-j) \to f(\epsilon x)$$ in $L^2$?
Maybe if the $\{\operatorname{sinc}(x-j)\}_{j \in \mathbb{Z}}$ form an Orthonormal basis in $L^2$ then I can write $f(\epsilon x)=\sum_{j \in \mathbb{Z}}c^\epsilon_j\operatorname{sinc}(x-j)$. In this case I could try to prove that $\sum_{j \in \mathbb{Z}}c^\epsilon_j\operatorname{sinc}(x-j) \to \sum_{j\in\mathbb{Z}} f(\epsilon j)\operatorname{sinc}(x-j)$ in $L^2$ where $c^\epsilon_j=\int_{-\infty}^{\infty}f(\epsilon x)\operatorname{sinc}(x-j)dx$. I am pretty sure that if $\epsilon$ is very small that $c_j^\epsilon\approx f(\epsilon j)$ with a change of variables it can be seen that $\operatorname{sinc}$ approaches a Dirac delta, but even if this can be rigorously shown, one still must sum $c_j^{\epsilon}-f(\epsilon j)$. At this point I feel like I am going in circles and would appreciate some insight through any means into attaining the first limit. I don't completely care where the $f$ live. Preferably they are $L^2$, but they may need to bandlimited i.e. their Fourier transform has compact support? It would even be helpful first step to me if one could show that $$\left\|\frac{1}{2\pi}\int_{-\pi}^\pi \left( \sum_{j \in \mathbb{Z}} e^{-ikj}f(\varepsilon j)\right) e^{ikx} \, dk -f(\varepsilon x)\right\|_{L^2}<C$$ whenever $\varepsilon <1$ where $C$ does not depend on $\varepsilon$.