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The Nyquist–Shannon sampling theorem says that a bandlimited function can be recovered exactly from equally spaced samples, provided the sampling rate is high enough. It is the bridge between a continuous waveform and a sequence of numbers: the missing values between samples are determined by the restriction on its frequencies.

Here is a precise finite-energy version. Let $f\in L^2(\mathbb{R})$, meaning ${\int }_{\mathbb{R}}|f(t){|}^{2}dt<\infty$, and suppose its Fourier transform, defined through the Plancherel theorem, is zero outside $[-2\pi B,2\pi B]$. Thus $B>0$ is the largest allowed ordinary frequency, in cycles per second, rather than an angular frequency. Set $T=1/f_s$, where $f_s$ is the number of samples per second. If $f_s\ge 2B$, the continuous representative of $f$ is recovered byThe quotient is the normalized sinc function, with value one when its argument is zero. Each shifted sinc function is one at its own sampling time and zero at every other sampling time. Its coefficient is therefore exactly the required sample value. For this L2 space formulation, the reconstruction converges in the L2 space norm and uniformly in $t$. The critical rate $f_s=2B$ is allowed for this class; choosing $f_s>2B$ leaves a gap between the occupied frequencies and the sampling boundary.

For a numerical example, a bandlimited function with no frequencies above $3$ cycles per second can be reconstructed from its samples every $0.1$ seconds: $f_s=10>6$. Exact reconstruction uses the whole sequence indexed by $\mathbb{Z}$. A finite recording and a finite sum of sinc functions introduce truncation error; joining adjacent samples by straight lines is generally a different reconstruction.

The obstruction is aliasing. At sampling rate $f_s=10$, the two distinct waveformshave identical samples at $t=n/10$, becauseThe samples alone cannot tell whether the waveform oscillates twice or eight times per second. If we know in advance that all frequencies are below $5$, the $8$-cycle waveform is excluded and the ambiguity disappears. These pure cosines are an illustrative aliasing example, rather than members of the finite-energy L2 space class in the stated theorem.

Exactly at a sampling boundary, the function class matters. At rate $f_s$, the sine $\sin (2\pi (f_s/2)t)$ has zero values at every $n/f_s$. It is a nonzero waveform indistinguishable from zero on that grid, but has infinite energy and is excluded by the finite-energy hypothesis above. This explains why a rule intended to cover arbitrary pure sinusoids should demand a strict inequality. For spatial data, replace seconds by distance: the Nyquist spatial frequency is $1/(2p)$ for sample pitch $p$.

The proof idea is short: the samples are Fourier coefficients of the Fourier transform restricted to a sampling-sized frequency interval. Because a bandlimited function has no contribution outside that interval, its Fourier series determines the whole Fourier transform; Fourier inversion then gives the sinc function expansion. See sampling expansion by periodic Fourier projection for this connection.

Claude E. Shannon's Communication in the Presence of Noise (1949), Section II, Theorem 1 and equation (7), gives the sampling statement and reconstruction using shifted sinc functions in a communication setting. Shannon explicitly notes earlier mathematical forms and credits Nyquist with emphasizing the sampling interval in telegraphy.^[ref]

Use $\hat{f}(\lambda )=\int f(t)e^{-i\lambda t}dt$ and interpret the regularity assumption to include $f\in L^2(\mathbb{R})$ with the continuous representative furnished by Fourier inversion. Write $F=\hat{f}$. The bandwidth assumption implies $F\in L^1$ as well as $L^2$, by the Cauchy-Schwarz inequality on its finite support, sois continuous and defined at every $t$. The integer samples are precisely the Fourier coefficients of $F$ with the sign convention appropriate to the basis $e^{-in\lambda }$: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 $L^2$ andThe last equality is the Plancherel theorem. Apply the inverse integral to the finite symmetric Fourier partial sums. Sincewe obtain the Nyquist–Shannon sampling theorem in this normalization: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 $\Vert G{\Vert }_2/\sqrt{2\pi }$. Hence the symmetric finite sampling sums converge uniformly on all of $\mathbb{R}$. Moreover, the Parseval identity applied to $e^{it\lambda }$ givesThe 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 $0<{ϵ}_0<\min (ϵ,\pi )$ and a nonzero smooth function $\Phi$ compactly supported in $(-{ϵ}_0,{ϵ}_0)$. Let $h$ be its inverse Fourier transform and putThe function $h$, and hence $g$, is a Schwartz function. The two shifted spectral supports are disjoint, so $\hat{g}\ne 0$, but its support is contained strictly inside $(-\pi -ϵ,\pi +ϵ)$. Every integer sample of $g$ is zero. Therefore $g$ and the zero function have identical samples but are distinct, providing a strictly supercritical sampling alias counterexample for every positive bandwidth enlargement.

Use the transform convention $F(\lambda )={\int }_{\mathbb{R}}f(t)e^{-i\lambda t}dt$. The given Fourier inversion theorem givesHere $F$ is continuous, vanishes at both endpoints and belongs to $L^2[-\pi ,\pi ]$. It therefore defines a continuous periodic function. Its Fourier coefficient at index $-n$ is $f(n)$. By part (i),Pair this convergence with $e^{it\lambda }$. Since that function has normalized $L^2$ norm one, the Cauchy-Schwarz inequality yields, uniformly in $t$,The elementary integral is the sinc function,Thus the sampling expansion by periodic Fourier projection isThere is no pointwise interchange with an unproved Fourier series: the calculation first uses finite sums and then an $L^2$ limit.

The convergence can also be made absolute. Parseval's identity gives ${\sum }_n|f(n){|}^2=\Vert F{\Vert }_2^2$, and Bessel's inequality applied to $e^{it\lambda }$ gives ${\sum }_n|D(t-n){|}^2\le 1$. Henceuniformly in $t$. At an integer argument, $D$ is one at zero and zero at the other integers, so the expansion interpolates the samples exactly.