📂Fourier Analysis

Fourier Series and Bessel's Inequality

Formulas

If a function $f$ defined on the interval $[-L,L)$ is Riemann integrable, the following inequality holds, known as the Bessel’s inequality.

$$\dfrac{1}{4}|a_{0}|^{2} +\dfrac{1}{2}\sum\limits_{n=1}^{\infty} \left(|a_{n}|^{2} + |b_{n}|^{2} \right) =\sum \limits_{n=-\infty}^{\infty} | c_{n} |^{2} \le \dfrac{1}{2L}\int_{-L}^{L} | f(t)|^{2} dt$$

Here, $a_{0},\ a_{n},\ b_{n}$ is the Fourier coefficient of $f$, and $c_{n}$ is the complex Fourier coefficient of $f$.

Proof

For any complex number $z$, since $|z|^{2}= z \overline{z}$,

$$\begin{align*} 0 &\le \left| f(t)-\sum\limits_{n=-N}^{N}c_{n} e^{i\frac{n\pi t}{L}} \right|^{2} \\ &=\left(f(t)-\sum\limits_{n=-N}^{N}c_{n} e^{i\frac{n\pi t}{L}} \right) \left(\overline{f(t)}-\sum\limits_{n=-N}^{N} \overline{c_{n}} e^{-i\frac{n\pi t}{L}} \right) \\ &= |f(t)|^{2} - \sum \limits_{n=-N}^{N} \left( f(t) \overline{c_{n}} e^{-i\frac{n\pi t}{L}} + \overline{f(t)}c_{n} e^{i\frac{n\pi t}{L}} \right) + \sum \limits_{m=-N}^{N}\sum\limits_{n=-N}^{N} |c_{n}|^{2}e^{i\frac{(m-n)\pi t}{L}} \end{align*}$$

As can be seen from the first line, the expression is greater than or equal to $0$, hence the integrated value is also greater than or equal to $0$. Therefore, the following inequality holds.

$$\begin{align*} 0& \le \dfrac{1}{2L}\int_{-L}^{L} |f(t)|^{2} dt - \sum \limits_{n=-N}^{N} \left(\overline{c_{n}} \dfrac{1}{2L}\int_{-L}^{L}f(t) e^{-i\frac{n\pi t}{L}}dt + c_{n}\dfrac{1}{2L}\int_{-L}^{L} \overline{f(t)} e^{i\frac{n\pi t}{L}}dt \right) \\ &\quad+ \sum \limits_{m=-N}^{N}\sum\limits_{n=-N}^{N} |c_{n}|^{2}\dfrac{1}{2L}\int_{-L}^{L}e^{i\frac{(m-n)\pi t}{L}}dt \end{align*}$$

Using the definition of complex Fourier coefficients for the second term, and the orthogonality of the exponential function for the third term, it simplifies to the following.

$$\begin{align*} 0 &\le \dfrac{1}{2L}\int_{-L}^{L} |f(t)|^{2} dt - \sum \limits_{n=-N}^{N} \left(\bar c_{n} \dfrac{1}{2L}\int_{-L}^{L}f(t) e^{-i\frac{n\pi t}{L}}dt + c_{n}\dfrac{1}{2L}\int_{-L}^{L} \bar f(t)e^{i\frac{n\pi t}{L}}dt \right) + \sum \limits_{m=-N}^{N}\sum\limits_{n=-N}^{N} |c_{n}|^{2}\dfrac{1}{2L}\int_{-L}^{L}e^{i\frac{(m-n)\pi}{L}t}dt \\ &= \dfrac{1}{2L}\int_{-L}^{L} |f(t)|^{2} dt - \sum \limits_{n=-N}^{N} \left(\bar c_{n} c_{n} + c_{n}\bar c_{n} \right) + \sum \limits_{m=-N}^{N}\sum\limits_{n=-N}^{N} |c_{n}|^{2}\delta_{mn} \\ &= \dfrac{1}{2L}\int_{-L}^{L} |f(t)|^{2} dt - \sum \limits_{n=-N}^{N}2 |c_{n}|^{2} + \sum \limits_{n=-N}^{N}|c_{n}|^{2} \\ &= \dfrac{1}{2L}\int_{-L}^{L} |f(t)|^{2} dt - \sum \limits_{n=-N}^{N} |c_{n}|^{2} \end{align*}$$

Taking the limit of $N \rightarrow \infty$ and moving the second term on the last expression yields the following.

$$\sum \limits_{n=-\infty}^{\infty} |c_{n}|^{2} \le \dfrac{1}{2L}\int_{-L}^{L} |f(t)|^{2}dt$$

According to the relationship between the Fourier coefficients and complex Fourier coefficients, the following equations are obtained.

$$\begin{align*} |a_{n}|^{2}+|b_{n}|^{2} &= a_{n} \bar a_{n} + b_{n} \bar b_{n} \\ &= (c_{n}+c_{-n})(\bar c_{n} + \bar c_{-n})+ i (c_{n}-c_{-n})(-i)(\bar c_{n} - \bar c_{-n} ) \\ &= 2c_{n}\bar c_{n} + 2c_{-n}\bar c_{-n} \\ &= 2\left( |c_{n}|^{2} + |c_{-n}|^{2} \right) \end{align*}$$

$$\implies \dfrac{1}{2} \sum \limits_{n=1}^{\infty} \left( |a_{n}|^{2}+|b_{n}|^{2} \right) = \sum \limits_{n=1}^{\infty} \left ( |c_{n}|^{2} + |c_{-n}|^{2} \right) = \sum \limits_{n=-\infty \\ n\ne 0}^{\infty} | c_n |^{2}$$

And since $|c_{0}|^{2} = \dfrac{1}{4}|a_{0}|^{2}$,

$$\dfrac{1}{4}|a_0|^{2} +\dfrac{1}{2}\sum\limits_{n=1}^{\infty} \left(|a_n|^{2} + |b_n|^{2} \right) =\sum \limits_{n=-\infty}^{\infty} | c_n |^{2} \le \dfrac{1}{2L}\int_{-L}^{L} | f(t)|^{2} dt$$

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