Nyquist–Shannon sampling theorem

Remarks

The difference $h = f - g$ is band-limited and vanishes at all points $k/(2B)$. On $[-B, B]$, $\hat h$ has the Fourier coefficients $\frac{1}{2B} \int_{-B}^{B} \hat h(\xi) e^{-2\pi i k \xi / (2B)}\, d\xi = \frac{1}{2B} h(-\frac{k}{2B})$ by Fourier inversion, and they are all zero. So $\hat h = 0$ in $L^2$, and $h = 0$. The Whittaker–Shannon interpolation formula gives $f$ explicitly from its samples.

The rate $2B$ samples per unit length is called the Nyquist rate. It cannot be lowered: if the spacing is $h > 1/(2B)$, then $f(x) = \sin(\pi x / h)\, \operatorname{sinc}(2\delta x)^2$ with $\delta = \frac12 (B - \frac{1}{2h}) > 0$ is a nonzero function, band-limited with bandwidth $B$, in $L^2(\R)$, that vanishes at every point $kh$. Here $\operatorname{sinc}(t) = \sin(\pi t)/(\pi t)$, and $\operatorname{sinc}(2\delta x)^2$ has transform supported in $[-2\delta, 2\delta]$.

The theorem is named after H. Nyquist and C. E. Shannon; E. T. Whittaker and V. A. Kotelnikov found equivalent results.

Going deeper

For a student

The proof is a duality between the line and the circle. The samples of $f$ are the Fourier coefficients of $\hat f$, viewed as a periodic function on an interval of length $2B$. Equivalently, multiplying $f$ by a Dirac comb of spacing $1/(2B)$ periodizes $\hat f$ with period $2B$, and the copies do not overlap exactly when the band fits. Landau's density theorem extends the necessity of the Nyquist rate to irregular sampling sets.