= Solution
Use $\widehat f(\lambda)=\int f(t)e^{-i\lambda t}\,dt$ and interpret the regularity assumption to include $f\in L^2(\mathbb R)$ with the continuous representative furnished by <Fourier inversion>. Write $F=\widehat f$. The bandwidth assumption implies $F\in L^1$ as well as $L^2$, by the <Cauchy-Schwarz inequality> on its finite support, so
$$
f(t)=\frac1{2\pi}\int_{-\pi}^{\pi}F(\lambda)e^{it\lambda}\,d\lambda
$$
is continuous and defined at every $t$. The integer samples are precisely the <Fourier coefficients> of $F$ with the sign convention appropriate to the basis $e^{-in\lambda}$:
$$
f(n)=\frac1{2\pi}\int_{-\pi}^{\pi}F(\lambda)e^{in\lambda}\,d\lambda.
$$
The completeness and <Parseval identity> for the exponential <orthonormal basis> on $[-\pi,\pi]$ give $F=\sum_{n\in\mathbb Z}f(n)e^{-in\lambda}$ in $L^2$ and
$$
\sum_{n\in\mathbb Z}|f(n)|^2=\frac1{2\pi}\|F\|_2^2=\|f\|_2^2.
$$
The last equality is the <Plancherel theorem>. Apply the inverse integral to the finite symmetric <Fourier partial sums>. Since
$$
\frac1{2\pi}\int_{-\pi}^{\pi}e^{i(t-n)\lambda}\,d\lambda=\frac{\sin\pi(t-n)}{\pi(t-n)},
$$
we obtain the <Nyquist–Shannon sampling theorem> in this normalization:
$$
\boxed{f(t)=\sum_{n\in\mathbb Z}f(n)\frac{\sin\pi(t-n)}{\pi(t-n)}\quad(t\in\mathbb R),}
$$
where the quotient is one at $t=n$. This is a <sampling expansion by periodic Fourier projection>, not merely an almost-everywhere inversion. Indeed, the inverse integral of an error $G\in L^2[-\pi,\pi]$ is bounded, uniformly in $t$, by $\|G\|_2/\sqrt{2\pi}$. Hence the symmetric finite sampling sums converge uniformly on all of $\mathbb R$. Moreover, the <Parseval identity> applied to $e^{it\lambda}$ gives
$$
\sum_{n\in\mathbb Z}\left|\frac{\sin\pi(t-n)}{\pi(t-n)}\right|^2=1.
$$
The <Cauchy-Schwarz inequality> therefore makes the sampling series absolutely convergent at every $t$, with its absolute tail bounded uniformly by $(\sum_{|n|>N}|f(n)|^2)^{1/2}$. Thus the integer samples recover the function everywhere.
For the nonuniqueness beyond this bandwidth, fix $0<\epsilon_0<\min(\epsilon,\pi)$ and a nonzero smooth function $\Phi$ compactly supported in $(-\epsilon_0,\epsilon_0)$. Let $h$ be its inverse <Fourier transform> and put
$$
g(t)=\sin(\pi t)h(t),\qquad \widehat g(\lambda)=\frac{\Phi(\lambda-\pi)-\Phi(\lambda+\pi)}{2i}.
$$
The function $h$, and hence $g$, is a <Schwartz function>. The two shifted spectral supports are disjoint, so $\widehat g\ne0$, but its support is contained strictly inside $(-\pi-\epsilon,\pi+\epsilon)$. Every integer sample of $g$ is zero. Therefore \b[$g$ and the zero function have identical samples but are distinct], providing a <strictly supercritical sampling alias counterexample> for every positive bandwidth enlargement.
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