Parseval's Identity

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.

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

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. That density follows from the Weierstrass approximation theorem for continuous functions plus density of continuous functions in $L^2$ ; the proof is omitted here. Once the bridge stands, the implication above completes the $L^2$ convergence.

Examples & Exercises