Unable to prove the $L^2$ version of Nyquist–Shannon sampling theorem without an additional assumption I want to prove the following version of Nyquist–Shannon sampling theorem for square-integrable functions:

Let $f \in L^2(\mathbb{R})$ and $B > 0$. If $\hat{f}(\xi) = 0$ for
almost every $|\xi| > B$, then $f$ agrees almost everywhere with a
continuous function $g \in C_0(\mathbb{R})$ such that the series
$$ \sum_{n \in \mathbb{Z}} g\left( \frac{n}{2B} \right) \frac{\sin(\pi
(2Bx - n))}{\pi (2Bx - n)} $$ converges to $f$ in $L^2(\mathbb{R})$.

I was able to prove that $f$ agrees amost everywhere with $g \in C_0(\mathbb{R})$ defined by
$$ g(x) = \int_{-B}^B \hat{f}(\xi) e^{2\pi i \xi x} d\xi $$
which satisfies the following identity for every $x \in \mathbb{R}$:
$$ g(x) = \sum_{n \in \mathbb{Z}} g\left( \frac{n}{2B} \right) \frac{\sin(\pi
(2Bx - n))}{\pi (2Bx - n)} $$
What I failed to prove is that the series on the RHS converges in $L^2(\mathbb{R})$:
$$ \sum_{n \in \mathbb{Z}} \lVert g\left( \frac{n}{2B} \right) \frac{\sin(\pi
(2Bx - n))}{\pi (2Bx - n)} \rVert_{L^2(\mathbb{R})} = \frac{1}{\sqrt{2B}} \sum_{n \in \mathbb{Z}} \left| g\left( \frac{n}{2B} \right) \right| $$
which is finite if and only if $ \{g(n/2B)\} \in l^1(\mathbb{Z}) $. But I was only able to prove $ \{g(n/2B)\} \in l^2 $, not in $l^1$.
 A: Okay, I've figured it out. We indeed cannot conclude that the series converges in $L^2$. We can only conclude that the series converges absolutely and uniformly. The revised version of the theorem is as follow:

Let $f \in L^2(\mathbb{R})$ and $B > 0$. If $\hat{f}(\xi) = 0$ for
almost every $|\xi| > B$, then $f$ agrees almost everywhere with a
continuous function $g \in C_0(\mathbb{R})$ that satisfies
$$ g(x) = \sum_{n \in \mathbb{Z}} g\left( \frac{n}{2B} \right) \frac{\sin(\pi
(2Bx - n))}{\pi (2Bx - n)} \quad \forall x \in \mathbb{R}$$
Moreover, the series on RHS converges absolutely and uniformly.

Below I would give a relatively short proof using Parseval identity.
Proof:
Since $\hat{f} \in L^2(\mathbb{R})$ and $\hat{f}(\xi) = 0$ for almost every $|\xi| > B$, the restriction $\hat{f}|_{[-B, B]}$ has a $2B$-periodic extension $\phi \in L^2(\mathbb{T}_{2B})$ such that $\hat{f} = \phi \chi_{[-B, B]}$ a.e. Moreover, we actually have $\hat{f} \in L^1(\mathbb{R}) \cap L^2(\mathbb{R})$, so $f$ agrees almost everywhere with the continuous function $g \in C_0(\mathbb{R}^d)$ defined by
$$
g(x) = \int_{-\infty}^\infty \hat{f}(\xi) e^{2\pi i x \xi} d\xi = \int_{-B}^B \phi(\xi) e^{2\pi i \xi x} d\xi
$$Consider the Hilbert space $L^2(\mathbb{T}_{2B})$ with the inner product $ \langle \varphi, \psi \rangle = \frac{1}{2B} \int_{-B}^B \varphi(\xi) \overline{\psi(\xi)} d\xi $. For each $x \in \mathbb{R}$, we can write $ g(x) = 2B \langle \phi, e^{-2\pi i \xi x} \rangle $. But then by the Parseval identity with respect to the orthornormal basis $\{e^{-2\pi i \xi n / 2B} : n \in \mathbb{Z}\}$, we have for each $x \in \mathbb{R}$:
$$
g(x)
= 2B \sum_n \langle \phi, e^{-2\pi i \xi n / 2B} \rangle \overline{\langle  e^{-2\pi i \xi x}, e^{-2\pi i \xi n / 2B} \rangle}
= \sum_n g\left( \frac{n}{2B} \right) \frac{\sin(\pi (2Bx - n))}{\pi (2Bx - n)}
$$
Lastly, the absolute and uniform convergence of RHS follows from the Cauchy-Schwartz inequality and the Parseval identity again:
\begin{align*}
\sum_n \left| g\left( \frac{n}{2B} \right) \frac{\sin(\pi (2Bx - n))}{\pi (2Bx - n)} \right|
&\leq 2B \sqrt{ \sum_n | \langle \phi, e^{-2\pi i \xi n / 2B} \rangle |^2 }
\sqrt{ \sum_n | \langle  e^{-2\pi i \xi x}, e^{-2\pi i \xi n / 2B} \rangle |^2 } \\
&= 2B \lVert \phi \rVert_{L^2(\mathbb{T}_{2B})} \lVert e^{-2\pi i \xi x} \rVert_{L^2(\mathbb{T}_{2B})}
= 2B \lVert \phi \rVert_{L^2(\mathbb{T}_{2B})} < \infty
\end{align*}
