Parseval's Theorem/Formulation 2

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Theorem

Let $f$ be a real function which is square-integrable over the interval $\openint {-\pi} \pi$.

Let $f$ be expressed by the Fourier series:

$\map f x = \ds \sum_{n \mathop = -\infty}^\infty c_n e^{i n x}$

where:

$c_n = \ds \frac 1 {2 \pi} \int_{-\pi}^\pi \map f t e^{-i n t} \rd t$

Then:

$\ds \frac 1 {2 \pi} \int_{-\pi}^\pi \size {\map f x}^2 \rd x = \sum_{n \mathop = -\infty}^\infty \size {c_n}^2$

Proof

$\ds \frac 1 {2 \pi} \int_{-\pi}^\pi \size {\map f x}^2 \rd x$

$=$

$\ds \frac 1 {2 \pi} \int_{-\pi}^\pi \map f x \overline {\map f x} \rd x$

Modulus in Terms of Conjugate

$\ds $

$=$

$\ds \frac 1 {2 \pi} \int_{-\pi}^\pi \sum_{n \mathop = -\infty}^\infty c_n e^{i n x} \overline {\sum_{m \mathop = -\infty}^\infty c_m e^{i m x} } \rd x$

$\ds $

$=$

$\ds \frac 1 {2 \pi} \int_{-\pi}^\pi \sum_{n \mathop = -\infty}^\infty c_n e^{i n x} \sum_{m \mathop = -\infty}^\infty \overline {c_m} e^{-i m x} \rd x$

Sum of Complex Conjugates

$\ds $

$=$

$\ds \frac 1 {2 \pi} \int_{-\pi}^\pi \sum_{n \mathop = -\infty}^\infty \sum_{m \mathop = -\infty}^\infty c_n \overline {c_m} e^{i x \paren {n - m} } \rd x$

$\ds $

$=$

$\ds \frac 1 {2 \pi} \sum_{n \mathop = -\infty}^\infty \sum_{m \mathop = -\infty}^\infty c_n \overline {c_m} \int_{-\pi}^\pi e^{i x \paren {n - m} } \rd x$

Fubini's Theorem

$\ds $

$=$

$\ds \frac 1 {2 \pi} \sum_{n \mathop = -\infty}^\infty \sum_{m \mathop = -\infty}^\infty c_n \overline {c_m} 2 \pi \delta_{n m}$

Integral over 2 pi of Exponential of i by n x

$\ds $

$=$

$\ds \frac {2 \pi} {2 \pi} \sum_{n \mathop = -\infty}^\infty c_n \overline {c_n}$

$\ds $

$=$

$\ds \sum_{n \mathop = -\infty}^\infty \size {c_n}^2$

$\blacksquare$

Source of Name

This entry was named for Marc-Antoine Parseval.