For a finite vector space we know that $\|x\|_{\infty} \leq \|x\|_{2}.$
I am looking for something similar for the topological Paley-Wiener space (bandlimited and $L^2(\mathbb{R})$ ). Specifically, I need something like $$\|f\|_{\infty}\leq C \| f\|_{2},$$ for some constant $C$. I will formalise the problem as follows.
For simplicity we assume the bandwidth to be $\pi$. We know that a function $f$ in this space is written as $f(t)=\sum_{k\in\mathbb{Z}}c_k \frac{\mathrm{sin\pi(t-k)}}{\pi(t-k)}$. We also know that, because $\frac{\mathrm{sin\pi(t-k)}}{\pi(t-k)}$ is an orthonormal basis, we get $$\|c\|_{\ell^2}=\|f\|_{L^2}.$$
Moreover, we know that, because $\frac{\mathrm{sin\pi(t-k)}}{\pi(n-k)}=\delta_{nk}$, where $\delta$ is the Kronecker delta, we get $f(k)=c_k$ and thus $$\|f(k)\|_{\infty}=\|c\|_{\infty}\leq\|c\|_{\ell^2}=\|f\|_{L^2}.$$
The issue that arises here is that $f$ can be higher than its samples $\|f(k)\|_{\infty}$. Also, it feels that maybe this bound is getting a bit too loose. Does anyone know any way to get the bound I need? If you work in the field and give any insight, even without the solution, it would be very helpful. Thank you!
Use the Cauchy-Schwarz inequality
$$ \left|f\left(t\right)\right| \leq \lVert c\rVert_{\ell^{2}}\sqrt{\sum_{k \in \mathbb{Z}}\left(\frac{\sin\pi\left(t-k\right)}{\pi(t-k)}\right)^{2}} = \lVert f\rVert_{L^{2}}\sqrt{\sum_{k \in \mathbb{Z}}\left(\frac{\sin\pi\left(t-k\right)}{\pi(t-k)}\right)^{2}}. $$
The sum in the square root you can use the bound $\lceil t\rceil - k \geq t-k \geq \lfloor t\rfloor - k$ and bound the terms where $k$ is closest to $t$ with
$$ \left|\frac{\sin\pi\left(t-k\right)}{\pi(t-k)}\right| \leq 1 $$
so you get something like
$$ \sqrt{\sum_{k \in \mathbb{Z}}\left(\frac{\sin\pi\left(t-k\right)}{\pi(t-k)}\right)^{2}} \leq \sqrt{2\left(1+\sum_{k=1}^{\infty}\frac{1}{k^{2}}\right)} = \sqrt{2\left(1+\frac{\pi^{2}}{6}\right)}. $$