Parseval's Identity | Real Analysis

The function's total square equals the coefficients' total square, as in the formula below. Adding modes pushes the captured share of energy toward all of it. Nothing is lost and nothing is invented.

For example, count the 💰 coins in a 🐷 piggy bank by denomination and add them up. Changing how you count never changes the total.

Examples & Exercises

Verify Parseval's identity for the square wave $f(x) = \operatorname{sgn}(x)$ (on $[-\pi, \pi]$ ): first check numerically in the scene that $S_N$ approaches 2, then prove rigorously from the Basel problem $\sum_{n\ge 1} \frac{1}{n^2} = \frac{\pi^2}{6}$ that $\sum_{\text{odd } n} \frac{16}{\pi^2 n^2} = 2$ .

\frac{1}{\pi}\int_{-\pi}^{\pi} f^2 = 2 = \sum_{\text{odd } n} \frac{16}{\pi^2 n^2}, \qquad \sum_{\text{odd } n} \frac{1}{n^2} = \frac{\pi^2}{8}

Square wave,

N = 40

S_{40} = \sum_{\text{odd } n \le 40} \frac{16}{\pi^2 n^2} \approx 1.9798

— about 99% of the target 2.

Show proof

Since $f^2 = 1$ , the $L^2$ norm is $\frac{1}{\pi}\int_{-\pi}^{\pi} f^2\,dx = \frac{2\pi}{\pi} = 2$ . As $f$ is odd, $a_n = 0$ , and $b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\sin(nx)\,dx = \frac{2}{\pi}\int_0^{\pi} \sin(nx)\,dx = \frac{2(1 - (-1)^n)}{n\pi}$ — $\frac{4}{n\pi}$ for odd $n$ , 0 for even $n$ .
Hence Parseval's right side is $\frac{a_0^2}{2} + \sum_{n\ge 1} b_n^2 = \sum_{\text{odd } n} \frac{16}{\pi^2 n^2} = \frac{16}{\pi^2}\sum_{\text{odd } n} \frac{1}{n^2}$ . From the Basel problem $\sum_{n\ge 1} \frac{1}{n^2} = \frac{\pi^2}{6}$ ; the even terms alone give $\sum_{k\ge 1} \frac{1}{(2k)^2} = \frac{1}{4}\cdot\frac{\pi^2}{6} = \frac{\pi^2}{24}$ , so the odd terms sum to $\frac{\pi^2}{6} - \frac{\pi^2}{24} = \frac{\pi^2}{8}$ .
Substituting gives $\frac{16}{\pi^2}\cdot\frac{\pi^2}{8} = 2$ — exactly the $L^2$ norm 2 on the left.
Parseval's identity is proved in this example.
The scene's bars pressing against 2 at $N = 40$ are the partial sums of this infinite series.

For $f$ Riemann integrable on $[-\pi, \pi]$ , using the orthogonality of the trigonometric system, expand $\|f - S_N\|_2^2 \ge 0$ to prove completely Bessel's inequality $\frac{a_0^2}{2} + \sum_{n=1}^{N}(a_n^2 + b_n^2) \le \frac{1}{\pi}\int_{-\pi}^{\pi} f^2$ (the finite core of Parseval).

0 \le \|f - S_N\|_2^2 = \|f\|_2^2 - \left(\frac{a_0^2}{2} + \sum_{n=1}^{N}(a_n^2 + b_n^2)\right)

Sawtooth,

N = 20

S_{20} = \sum_{n\le 20} \frac{4}{n^2} \approx 6.38

sits below the

L^2

norm

\frac{2\pi^2}{3} \approx 6.5797

— Bessel's inequality itself.

Show proof

State the hypotheses: assume $f$ is Riemann integrable on $[-\pi, \pi]$ .
Then $f^2$ is integrable as well, and the Fourier coefficients $a_n, b_n$ and $\int_{-\pi}^{\pi} f^2$ are all well-defined.
With the inner product $\langle g, h\rangle = \frac{1}{\pi}\int_{-\pi}^{\pi} g(x)h(x)\,dx$ , the trigonometric system $\{1/\sqrt{2}, \cos(nx), \sin(nx)\}$ is orthonormal.
Writing the Fourier partial sum as $S_N = \frac{a_0}{2} + \sum_{n=1}^{N}(a_n\cos(nx) + b_n\sin(nx))$ , we get $\|f - S_N\|_2^2 = \langle f - S_N, f - S_N\rangle = \|f\|_2^2 - 2\langle f, S_N\rangle + \|S_N\|_2^2$ — always nonnegative.
The coefficient definitions are exactly $\langle f, \cos(nx)\rangle = a_n$ , $\langle f, \sin(nx)\rangle = b_n$ , $\langle f, 1\rangle = a_0$ , so $\langle f, S_N\rangle = \frac{a_0^2}{2} + \sum_{n=1}^{N}(a_n^2 + b_n^2)$ . By orthogonality, $\|S_N\|_2^2 = \frac{a_0^2}{2} + \sum_{n=1}^{N}(a_n^2 + b_n^2)$ as well (every cross term is 0).
Substituting gives $0 \le \|f - S_N\|_2^2 = \|f\|_2^2 - 2E_N + E_N = \|f\|_2^2 - E_N$ (with $E_N = \frac{a_0^2}{2} + \sum_{n=1}^{N}(a_n^2 + b_n^2)$ ) — i.e. $E_N \le \|f\|_2^2$ .
That is Bessel's inequality, and why the scene's bars never cross the dashed line.
The $N \to \infty$ limit is Parseval's identity (problem 3 below).

Prove that if Parseval's identity $\frac{a_0^2}{2} + \sum_{n\ge 1}(a_n^2 + b_n^2) = \|f\|_2^2$ holds, then the Fourier partial sums $S_N$ converge to $f$ in $L^2$ .

Mention the completeness (density of trigonometric polynomials) only as the bridge — that step is idea-level.

\|f - S_N\|_2^2 = \|f\|_2^2 - \left(\frac{a_0^2}{2} + \sum_{n=1}^{N}(a_n^2 + b_n^2)\right) \;\xrightarrow[N \to \infty]{}\; 0

Parabola,

N = 12

: the energy enters as

1/n^4

, so

S_{12}

already captures over 99.99% of the

L^2

norm

\frac{2\pi^4}{5} \approx 38.96

—

L^2

convergence happening before your eyes.

Show proof

Problem 2's expansion is an identity, not just an inequality: keeping $\|f - S_N\|_2^2$ instead of replacing it by $0$ gives $\|f - S_N\|_2^2 = \|f\|_2^2 - E_N$ (with $E_N = \frac{a_0^2}{2} + \sum_{n=1}^{N}(a_n^2 + b_n^2)$ ) — the cross terms vanish by orthogonality, so this holds exactly.
Parseval's identity says $E_N \to \|f\|_2^2$ as $N \to \infty$ .
Hence $\|f - S_N\|_2^2 = \|f\|_2^2 - E_N \to 0$ — i.e. $S_N \to f$ in $L^2$ .
That one step, from inequality to equality, is the whole argument: the implication itself is completely proved.
The bridge (idea level): Parseval's identity itself needs the trigonometric system to be "complete" — finite trigonometric polynomials must approximate $f$ in $L^2$ to any accuracy.
Caution: every trigonometric polynomial satisfies $T(-\pi) = T(\pi)$ , so uniform approximation of a continuous $f$ needs the hypothesis $f(-\pi) = f(\pi)$ (in that case the Stone-Weierstrass theorem for trigonometric polynomials applies).
For a general $f$ the endpoints are measure zero in $L^2$ , so this condition does not block $L^2$ approximation: first approximate $f$ in $L^2$ by continuous functions, adjust the endpoint values to match, then apply the approximation above.
The proof is omitted here.
Once the bridge stands, the implication above completes the $L^2$ convergence.