Bessel's Inequality and Parseval Formula: The Energy Theorem

Bessel's Inequality and Parseval Formula: The Energy Theorem

Bessel's Inequality and Parseval Formula: The Energy Theorem

We have seen some types of approximations, such as Taylor and Fourier approximations. The type of convergence used may change depending on the nature of the approximation. One of the most useful c is the L^p-convergence.

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

$\begin{displaymath}\int_{-\pi}^{\pi} f^2(x) dx < \infty.\end{displaymath}$

We will say that f(x) is square integrable. Consider the associated Fourier series

$\begin{displaymath}f(x) \sim a_0 + \sum_{n=1}^{\infty}\Big(a_n\cos(nx) + b_n\sin(nx)\Big).\end{displaymath}$

Set

$\begin{displaymath}f_N(x) = a_0 + \sum_{n=1}^{n=N}\Big(a_n\cos(nx) + b_n\sin(nx)\Big).\end{displaymath}$

Hence

$\begin{displaymath}\int_{-\pi}^{\pi} (f(x) - f_N(x))^2 dx = \int_{-\pi}^{\pi} (f^2(x) - 2f(x)f_N(x) + f^2_N(x))dx.\end{displaymath}$

Easy calculations give

$\begin{displaymath}\int_{-\pi}^{\pi}f^2_N(x) dx = \pi \left(\frac{a^2_0}{2} + \sum_{n=1}^{n=N}(a^2_n + b^2_n)\right).\end{displaymath}$

We also have

$\begin{displaymath}\int_{-\pi}^{\pi}f(x)f_N(x)dx = \pi \left(\frac{a^2_0}{2} + \sum_{n=1}^{n=N}(a^2_n + b^2_n)\right).\end{displaymath}$

Therefore

$\begin{displaymath}\int_{-\pi}^{\pi} (f(x) - f_N(x))^2 dx = \int_{-\pi}^{\pi}f^2... ...\left(\frac{a^2_0}{2} + \sum_{n=1}^{n=N}(a^2_n + b^2_n)\right).\end{displaymath}$

Since $\int_{-\pi}^{\pi} (f(x) - f_N(x))^2 dx \geq 0$ , then

$\begin{displaymath}\pi \left(\frac{a^2_0}{2} + \sum_{n=1}^{n=N}(a^2_n + b^2_n)\right) \leq \int_{-\pi}^{\pi}f^2(x)dx,\end{displaymath}$

for any $N \geq 1$ . This clearly implies the following result:

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

$\begin{displaymath}\pi \left(2a^2_0 + \sum_{n=1}^{\infty}(a^2_n + b^2_n)\right) \leq \int_{-\pi}^{\pi}f^2(x)dx.\end{displaymath}$

In particular, the series $\sum_{n=1}^{\infty}(a^2_n + b^2_n))$ is convergent.

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

$\begin{displaymath}E = \frac{1}{\pi} \int_{-\pi}^{\pi}f^2(x)dx.\end{displaymath}$

So Bessel's inequality translates into:

$\begin{displaymath}\left(2a^2_0 + \sum_{n=1}^{\infty}(a^2_n + b^2_n)\right) \leq E.\end{displaymath}$

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

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

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

In this case, we have

$\begin{displaymath}\pi (\frac{a^2_0}{2} + \sum_{n=1}^{\infty}(a^2_n + b^2_n)) = \int_{-\pi}^{\pi}f^2(x)dx.\end{displaymath}$

Theorem. Parseval 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

$\begin{displaymath}\frac{a^2_0}{2} + \sum_{n=1}^{\infty}(a^2_n + b^2_n) = \frac{1}{\pi} \int_{-\pi}^{\pi}f^2(x)dx = E\end{displaymath}$

if and only if

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

Remark. One may wonder when does the Parseval 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.

$\begin{displaymath}\lim_{N \rightarrow \infty} \sup_{x \in [-\pi,\pi]} \vert f(x) - f_N(x)\vert = 0.\end{displaymath}$

The proof in this case is quite easy. Indeed, since f(x) is continuous then

$\begin{displaymath}f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty}(a_n\cos(nx) + b_n\sin(nx)).\end{displaymath}$

Hence

$\begin{displaymath}f^2(x) = \frac{a_0}{2}f(x) + \sum_{n=1}^{\infty}(a_n\cos(nx)f(x) + b_n\sin(nx)f(x)).\end{displaymath}$

The uniform convergence of this series enables us to integrate term-by-term to obtain

$\begin{displaymath}\int_{-\pi}^{\pi} f^2(x)dx = \frac{a_0}{2}\int_{-\pi}^{\pi}f(... ...\infty}\int_{-\pi}^{\pi}(a_n\cos(nx)f(x) + b_n\sin(nx)f(x))dx.\end{displaymath}$

By using the definition of the Fourier coefficients, we get the desired conclusion.

Application: Least Square Error.
One application of the Parseval formula is the measure of the least square error $\sigma^2_N$ defined by

$\begin{displaymath}\sigma^2_N = \frac{1}{2\pi}\int_{-\pi}^{\pi} (f(x) - f_N(x))^2dx.\end{displaymath}$

If the function f(x) satisfies the assumptions of the Energy Theorem, then we have

$\begin{displaymath}\sigma^2_N = \frac{1}{2} \sum_{n=N+1}^{\infty}(a^2_n + b^2_n).\end{displaymath}$

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

$\begin{displaymath}a_{2n} = 0, \;\;\mbox{and} \;\; a_{2n-1} = - \frac{4}{\pi (2n-1)^2}.\end{displaymath}$

Hence

$\begin{displaymath}\sigma^2_{2N-1} = \sigma^2_{2N} = \frac{1}{2} \sum_{n=2N+1}^{... ...2 = \frac{8}{\pi^2} \sum_{n=2N+1}^{\infty} \frac{1}{(2n-1)^4}.\end{displaymath}$

Using the equality

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

we get

$\begin{displaymath}\sigma^2_N = O\left(\frac{1}{N^3}\right),\;\;\; \mbox{as}\;\;N \rightarrow \infty.\end{displaymath}$

Recall that the sequences $\{u_n\}$ and $\{v_n\}$ satisfy

u_n = O(v_n)

if $\displaystyle \left\{\frac{u_n}{v_n}\right\}$ is a bounded squence.