Bessel's Inequality and Parseval's Formula: The Energy Theorem

We have seen some types of approximations, such as Taylor and Fourier approximations. The type of convergence used changes depending on the nature of the approximation. One of the most useful for Fourier approximations is L²-convergence:

Let f (x) be an integrable function on the interval [- $\pi$ , $\pi$ ], such that

f (x) $\displaystyle \sim$ a₀ + $\displaystyle \sum_{n=1}^{\infty}$ $\displaystyle \Big($ a_ncos(nx) + b_nsin(nx) $\displaystyle \Big)$ .

f_N(x) = a₀ + $\displaystyle \sum_{n=1}^{n=N}$ $\displaystyle \Big($ a_ncos(nx) + b_nsin(nx) $\displaystyle \Big)$ .

$\displaystyle \int_{-\pi}^{\pi}$ (f (x) - f_N(x))²dx = $\displaystyle \int_{-\pi}^{\pi}$ (f²(x) - 2f (x)f_N(x) + f²_N(x))dx.

$\displaystyle \int_{-\pi}^{\pi}$ f²_N(x)dx = $\displaystyle \pi$ $\displaystyle \left(\vphantom{2 a^2_0 + \sum_{n=1}^{n=N}(a^2_n + b^2_n)}\right.$ 2a²₀ + $\displaystyle \sum_{n=1}^{n=N}$ (a²_n + b²_n) $\displaystyle \left.\vphantom{2 a^2_0 + \sum_{n=1}^{n=N}(a^2_n + b^2_n)}\right)$ .

$\displaystyle \int_{-\pi}^{\pi}$ f (x)f_N(x)dx = $\displaystyle \pi$ $\displaystyle \left(\vphantom{2 a^2_0 + \sum_{n=1}^{n=N}(a^2_n + b^2_n)}\right.$ 2a²₀ + $\displaystyle \sum_{n=1}^{n=N}$ (a²_n + b²_n) $\displaystyle \left.\vphantom{2 a^2_0 + \sum_{n=1}^{n=N}(a^2_n + b^2_n)}\right)$ .

$\displaystyle \int_{-\pi}^{\pi}$ (f (x) - f_N(x))²dx = $\displaystyle \int_{-\pi}^{\pi}$ f²(x)dx - $\displaystyle \pi$ $\displaystyle \left(\vphantom{2 a^2_0 + \sum_{n=1}^{n=N}(a^2_n + b^2_n)}\right.$ 2a²₀ + $\displaystyle \sum_{n=1}^{n=N}$ (a²_n + b²_n) $\displaystyle \left.\vphantom{2 a^2_0 + \sum_{n=1}^{n=N}(a^2_n + b^2_n)}\right)$ .

$\displaystyle \pi$ $\displaystyle \left(\vphantom{2 a^2_0 + \sum_{n=1}^{n=N}(a^2_n + b^2_n)}\right.$ 2a²₀ + $\displaystyle \sum_{n=1}^{n=N}$ (a²_n + b²_n) $\displaystyle \left.\vphantom{2 a^2_0 + \sum_{n=1}^{n=N}(a^2_n + b^2_n)}\right)$ $\displaystyle \leq$ $\displaystyle \int_{-\pi}^{\pi}$ f²(x)dx,

Theorem. Bessel's Inequality Let f (x) be a function defined on [- $\pi$ , $\pi$ ] such that f²(x) has a finite integral on [- $\pi$ , $\pi$ ]. If a_n and b_n are the Fourier coefficients of the function f (x), then we have

$\displaystyle \pi$ $\displaystyle \left(\vphantom{2a^2_0 + \sum_{n=1}^{\infty}(a^2_n + b^2_n)}\right.$ 2a²₀ + $\displaystyle \sum_{n=1}^{\infty}$ (a²_n + b²_n) $\displaystyle \left.\vphantom{2a^2_0 + \sum_{n=1}^{\infty}(a^2_n + b^2_n)}\right)$ $\displaystyle \leq$ $\displaystyle \int_{-\pi}^{\pi}$ f²(x)dx.

Remark. The quantity A_n = $\sqrt{a_n^2 + b^2_n}$ is called the amplitude of the n^th harmonic. The square of the amplitude has a useful interpretation. Indeed, borrowing terminology from the study of periodic waves, we define the energy E of a 2 $\pi$ -periodic function f (x) to be the number

$\displaystyle \left(\vphantom{2a^2_0 + \sum_{n=1}^{\infty}(a^2_n + b^2_n)}\right.$ 2a²₀ + $\displaystyle \sum_{n=1}^{\infty}$ (a²_n + b²_n) $\displaystyle \left.\vphantom{2a^2_0 + \sum_{n=1}^{\infty}(a^2_n + b^2_n)}\right)$ $\displaystyle \leq$ E.

You may ask the following question: when does the inequality become an equality?

Note that for Fourier polynomials, the inequality does indeed become an equality. Using this, one can show that the answer to the question is in the affirmative if and only if

$\displaystyle \lim_{N \rightarrow \infty}^{}$ $\displaystyle \int_{-\pi}^{\pi}$ (f (x) - f_N(x))²dx = 0.

$\displaystyle \pi$ (2a²₀ + $\displaystyle \sum_{n=1}^{\infty}$ (a²_n + b²_n)) = $\displaystyle \int_{-\pi}^{\pi}$ f²(x)dx.

Theorem. Parseval's Formula or the Energy Theorem. Let f (x) be a function defined on [- $\pi$ , $\pi$ ] such that f²(x) has a finite integral on [- $\pi$ , $\pi$ ]. If a_n and b_n are the Fourier coefficients of f (x), then we have

2a²₀ + $\displaystyle \sum_{n=1}^{\infty}$ (a²_n + b²_n) = $\displaystyle {\frac{1}{\pi}}$ $\displaystyle \int_{-\pi}^{\pi}$ f²(x)dx = E

$\displaystyle \lim_{N \rightarrow \infty}^{}$ $\displaystyle \int_{-\pi}^{\pi}$ (f (x) - f_N(x))²dx = 0.

Remark. One may wonder when does Parseval's Formula hold? This is the case, for example, for piecewise smooth functions. The reason behind is the uniform convergence of the Fourier partial sums to f (x), ie.

$\displaystyle \lim_{N \rightarrow \infty}^{}$ $\displaystyle \sup_{x \in [-\pi,\pi]}^{}$ | f (x) - f_N(x)| = 0.

f²(x) = a₀f (x) + $\displaystyle \sum_{n=1}^{\infty}$ (a_ncos(nx)f (x) + b_nsin(nx)f (x)).

$\displaystyle \int_{-\pi}^{\pi}$ f²(x)dx = a₀ $\displaystyle \int_{-\pi}^{\pi}$ f (x)dx + $\displaystyle \sum_{n=1}^{\infty}$ $\displaystyle \int_{-\pi}^{\pi}$ (a_ncos(nx)f (x) + b_nsin(nx)f (x))dx.

Application: Least Square Error.
One application of Parseval's Formula is the measure of the least square error $\sigma^{2}_{N}$ defined by

$\displaystyle \sigma^{2}_{N}$ = $\displaystyle {\frac{1}{2\pi}}$ $\displaystyle \int_{-\pi}^{\pi}$ (f (x) - f_N(x))²dx.

$\displaystyle \sigma^{2}_{N}$ = $\displaystyle {\textstyle\frac{1}{2}}$ $\displaystyle \sum_{n=N+1}^{\infty}$ (a²_n + b²_n).

Example. Let f (x) = | x| be defined on [- $\pi$ , $\pi$ ]. Find $\sigma^{2}_{N}$ and its asymptotic behavior when N gets large.
Answer. Since f (x) is even, we have b_n = 0. On the other hand, easy calculations give

$\displaystyle \sigma^{2}_{2N-1}$ = $\displaystyle \sigma^{2}_{2N}$ = $\displaystyle {\textstyle\frac{1}{2}}$ $\displaystyle \sum_{n=2N+1}^{\infty}$ $\displaystyle \left(\vphantom{\frac{4}{\pi (2n-1)^2}}\right.$ $\displaystyle {\frac{4}{\pi (2n-1)^2}}$ $\displaystyle \left.\vphantom{\frac{4}{\pi (2n-1)^2}}\right)^{2}_{}$ = $\displaystyle {\frac{8}{\pi^2}}$ $\displaystyle \sum_{n=2N+1}^{\infty}$ $\displaystyle {\frac{1}{(2n-1)^4}}$ .

$\displaystyle \int_{N}^{\infty}$ $\displaystyle {\frac{dx}{(2x-1)^4}}$ = $\displaystyle {\textstyle\frac{1}{6}}$ $\displaystyle {\frac{1}{(2N-1)^3}}$ ,

$\displaystyle \sigma^{2}_{N}$ = O $\displaystyle \left(\vphantom{\frac{1}{N^3}}\right.$ $\displaystyle {\frac{1}{N^3}}$ $\displaystyle \left.\vphantom{\frac{1}{N^3}}\right)$ , as N $\displaystyle \rightarrow$ $\displaystyle \infty$ .