Parseval's and Plancherel's Theorems

The isometry is quietly load-bearing across modern measurement. In gravitational-wave astronomy, LIGO and Virgo detect signals by matched filtering, and the optimal signal-to-noise ratio is computed in the frequency domain as $\rho^2=4\int_0^\infty |\tilde h(f)|^2/S_n(f)\,\mathrm{d}f$ — a noise-weighted Parseval integral. Doing the correlation in frequency space is legitimate only because Plancherel guarantees it agrees with the time-domain version, and it is far cheaper via the FFT; Wiener–Khinchin supplies the very power spectral density $S_n(f)$ that weights the integral. In magnetic-resonance imaging the scanner samples $k$-space and reconstructs by an inverse FFT, and the discrete Parseval theorem lets engineers track signal energy and noise identically in image space and $k$-space. Codecs use the same discrete identity to allocate bits by energy, and quantum-information calculations lean on the polarized form because state fidelity is a representation-independent inner product. The open frontier lies less in the theorem than in its fast discrete cousins — sparse and non-uniform FFTs, and quantum algorithms whose speedups rest on the same unitary Fourier structure.

Picture a function not as a curve but as an arrow in a space with one perpendicular axis per Fourier mode; resolving it onto those axes gives its Fourier coefficients, its coordinates. Coordinates change when you rotate the axes, but the length of the arrow does not. Parseval's identity is that invariance for the Fourier basis: the squared length is the sum of squared coordinates in any orthonormal frame. The Fourier transform is just the rotation carrying "physical-space" axes into "frequency" axes, and a rotation cannot stretch a vector. The first figure shows this with the smallest honest example, a right triangle.

Physically, $|f(x)|^2$ is an energy density in space and $|\tilde f(k)|^2$ an energy density per unit wavenumber. The total is the same however you slice it: a plucked string's energy can be tallied by integrating along its length or by summing its harmonics, and the two must agree to the last decimal — not by coincidence but because orthonormality forces it. The second figure makes this quantitative for the sawtooth $f(x)=x$, where adding up harmonic energies literally computes the Basel sum $\sum 1/n^2 = \pi^2/6$.

The polarized form $\langle f,g\rangle=\langle\tilde f,\tilde g\rangle$ is the real engine: $g=f$ gives the energy theorem, a shifted $g$ gives the Wiener–Khinchin power spectrum, and two quantum states $f,g$ show overlaps are representation-independent. And because nice functions are dense in $L^2$ while the transform preserves their norms, it extends by continuity to the whole space — rescuing the transform of $\operatorname{sinc}$-like functions that are square-integrable but not absolutely integrable.

Start with the periodic case in its complex form, where the exponential basis makes the algebra shortest. Expand $f$, multiply by its conjugate, integrate over one period, and orthonormality kills every cross term.

where: $f(x)$ — square-integrable $2\pi$-periodic function; $c_n=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-inx}\,\mathrm{d}x$ — complex Fourier coefficient; $n$ — integer mode index. Reading: the mean-square value of $f$ equals the total power carried by its modes. The same statement in the real trigonometric basis is what most textbooks quote. The factors $\tfrac14$ and $\tfrac12$ are the mean-square values of the constant, cosine, and sine building blocks.

where: $a_n,b_n$ — real cosine/sine coefficients over period $T$; $a_0$ — twice the mean of $f$; $T$ — fundamental period. Reading: the DC term contributes $a_0^2/4$ and each harmonic half its squared amplitude, because $\overline{\cos^2}=\overline{\sin^2}=\tfrac12$. Now the continuum. For a non-periodic $f$ the coefficient sum becomes an integral over a continuous frequency, and in the symmetric convention the equality is exact — no prefactor at all.

where: $\tilde f(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(x)e^{-ikx}\,\mathrm{d}x$ — symmetric Fourier transform; $k$ — angular wavenumber. Reading: this is Plancherel's theorem — the transform is unitary, and the missing $2\pi$ is the reward for splitting the kernel symmetrically. The most useful form is the polarized one, preserving the full inner product rather than just the norm — every other identity here is a special case of it.

where: $f,g$ — two square-integrable functions; $\overline{\,\cdot\,}$ — complex conjugation. Reading: the transform preserves angles as well as lengths. Put $g=f$ and equation (4) collapses to (3); this is why proving the polarized form once proves everything. Both identities descend from a single completeness statement — that the exponentials resolve the identity. In the continuum this is the delta-function normalization of the plane waves.

where: $\delta$ — Dirac delta distribution; $x,x'$ — two points in physical space. Reading: the plane waves form a "continuum orthonormal basis"; this is the continuum cousin of $\frac{1}{2\pi}\int e^{i(m-n)x}\,\mathrm{d}x=\delta_{mn}$ and collapses the double integral in Plancherel's proof. When the orthonormal set is incomplete, equality weakens to an inequality — you can never capture more energy than the function has.

where: $c_n$ — coefficients against an orthonormal (possibly incomplete) set. Reading: Bessel's inequality. Equality holds if and only if the set is complete, so a measured energy deficit is a direct diagnostic of missing modes — the basis of truncation-error estimates in spectral methods. Finally the corollary that dominates signal processing: the transform of the autocorrelation is the power spectrum.

where: $R_{ff}$ — autocorrelation function; $|\tilde f(k)|^2$ — power spectral density; $\tau$ — lag. Reading: Wiener–Khinchin — the polarized relation with $g(x)=f(x-\tau)$, turning a time-domain average into a frequency-domain magnitude.

How the identity specializes depends on your Fourier convention, on periodicity, and on whether the basis is complete:

1 · The Basel sum from a sawtooth (fully worked)

Apply Parseval to $f(x)=x$ on $(-\pi,\pi)$ and read off $\sum_{n=1}^\infty 1/n^2$.

Step 1. Find the coefficients. The sawtooth is odd, so $a_n=0$, and $b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}x\sin nx\,\mathrm{d}x=\frac{2(-1)^{n+1}}{n}$. Hence $b_n^2=4/n^2$ for every $n$, independent of the alternating sign. The mode energies are $4/n^2$: 4, 1, 0.444, 0.25, ….

Step 2. Compute the left side in the trigonometric normalization $\frac{1}{\pi}\int_{-\pi}^{\pi}f^2\,\mathrm{d}x=\frac{a_0^2}{2}+\sum(a_n^2+b_n^2)$. Direct integration gives $\frac{1}{\pi}\int_{-\pi}^{\pi}x^2\,\mathrm{d}x=\frac{1}{\pi}\cdot\frac{2\pi^3}{3}=\frac{2\pi^2}{3}\approx 6.5797$. Total energy $=2\pi^2/3$.

Step 3. Equate to the coefficient sum: $\frac{2\pi^2}{3}=\sum b_n^2=\sum\frac{4}{n^2}=4\sum\frac{1}{n^2}$, so $\sum\frac{1}{n^2}=\frac{\pi^2}{6}$. $\displaystyle\sum_{1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}\approx 1.6449$, Euler's Basel value, from pure energy conservation — the number the second figure builds bar by bar.

2 · One more power: $\sum 1/n^4$ from a parabola (fully worked)

Apply Parseval to $f(x)=x^2$ on $(-\pi,\pi)$.

Step 1. The coefficients of the even function $x^2$ are $a_0=\frac{1}{\pi}\int_{-\pi}^\pi x^2\,\mathrm{d}x=\frac{2\pi^2}{3}$ and $a_n=\frac{1}{\pi}\int_{-\pi}^\pi x^2\cos nx\,\mathrm{d}x=\frac{4(-1)^n}{n^2}$, with all $b_n=0$. So $a_n^2=16/n^4$.

Step 2. The left side is $\frac{1}{\pi}\int_{-\pi}^\pi x^4\,\mathrm{d}x=\frac{1}{\pi}\cdot\frac{2\pi^5}{5}=\frac{2\pi^4}{5}$. The DC contribution is $\frac{a_0^2}{2}=\frac12\left(\frac{2\pi^2}{3}\right)^2=\frac{2\pi^4}{9}$. Left side $=2\pi^4/5$, DC term $=2\pi^4/9$.

Step 3. Parseval reads $\frac{2\pi^4}{5}=\frac{2\pi^4}{9}+\sum_{n=1}^\infty\frac{16}{n^4}$. Then $16\sum\frac{1}{n^4}=2\pi^4\!\left(\frac15-\frac19\right)=2\pi^4\cdot\frac{4}{45}=\frac{8\pi^4}{45}$, giving $\sum_{n=1}^\infty\frac{1}{n^4}=\frac{\pi^4}{90}$. $\displaystyle\sum_{1}^{\infty}\frac{1}{n^4}=\frac{\pi^4}{90}\approx 1.0823$ — a value Euler also found, here dropping out of a single squared integral.

3 · Plancherel on a rectangular pulse (transform case)

Take $f(x)=1$ for $|x|<1$ and $0$ otherwise, and use Plancherel to evaluate $\int_{-\infty}^\infty \frac{\sin^2 k}{k^2}\,\mathrm{d}k$.

Step 1. The symmetric transform is $\tilde f(k)=\frac{1}{\sqrt{2\pi}}\int_{-1}^{1}e^{-ikx}\,\mathrm{d}x=\frac{1}{\sqrt{2\pi}}\cdot\frac{2\sin k}{k}=\sqrt{\frac{2}{\pi}}\,\frac{\sin k}{k}$, so $|\tilde f(k)|^2=\frac{2}{\pi}\frac{\sin^2 k}{k^2}$. The transform of a box is a sinc.

Step 2. The physical-space energy is trivial: $\int_{-\infty}^\infty|f|^2\,\mathrm{d}x=\int_{-1}^{1}1\,\mathrm{d}x=2$. Left side $=2$.

Step 3. Plancherel forces $2=\int|\tilde f|^2\,\mathrm{d}k=\frac{2}{\pi}\int\frac{\sin^2 k}{k^2}\,\mathrm{d}k$, so $\int\frac{\sin^2 k}{k^2}\,\mathrm{d}k=\pi$. $\displaystyle\int_{-\infty}^{\infty}\frac{\sin^2 k}{k^2}\,\mathrm{d}k=\pi$, an awkward integral handed over for free — the continuum sibling of the two series sums above.

These theorems carry two names because they were proved a century apart: an eighteenth-century formal identity about series, and a twentieth-century rigorous theorem about the Hilbert space $L^2$. The portrait belongs to the man who closed the gap.

Texts. Stone & Goldbart, Mathematics for Physics (Cambridge, 2009), Ch. 2 (function spaces) and Ch. 9 (integral transforms) · Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., §§19.2, 20.2 · Hassani, Mathematical Physics, 2nd ed., Ch. 6 (Hilbert spaces) and Ch. 8 · Byron & Fuller, Mathematics of Classical and Quantum Physics, Ch. 5 · Riley, Hobson & Bence, Mathematical Methods for Physics and Engineering, 3rd ed., §§12.8, 13.1 · Reed & Simon, Methods of Modern Mathematical Physics I: Functional Analysis, §IX.1 (Plancherel theorem).

Provenance. The complex and trigonometric Parseval forms are checked against Riley–Hobson–Bence §12.8 and Arfken §19.2; the Plancherel derivation (completeness collapse and Fubini swap) against Stone & Goldbart Ch. 9 and Reed & Simon §IX.1. Numerical checks recomputed here: the Basel sum $\sum 1/n^2=\pi^2/6\approx1.6449$ and $\sum 1/n^4=\pi^4/90\approx1.0823$ from Parseval on $x$ and $x^2$, and $\int_{-\infty}^\infty(\sin k/k)^2\,\mathrm{d}k=\pi$ from Plancherel on a rectangular pulse. Convention prefactors ($1$, $1/2\pi$, or none for ordinary frequency) follow directly from the kernel normalization and should be re-derived by inserting each inverse-transform definition rather than quoted from memory.

Prerequisites. Fourier Series and Periodic Functions · The Fourier Transform · The Dirac Delta Function. Leads to. The Convolution Theorem · Tempered Distributions and the Fourier Transform · Properties of the Fourier Transform.

Ask the tutor: Ask me to apply Parseval to any function of your choosing — say the square wave or a triangle pulse — and watch a stubborn series or integral fall out of energy conservation.