Parseval's Formula for the Sum of Coefficients of a Fourier Series

Recall from the Bessel's Inequality for the Sum of Coefficients of a Fourier Series page that if $\{ \varphi_0(x), \varphi_1(x), ... \}$ is an orthonormal system of functions on $I$, $f \in L^2(I)$, and $\displaystyle{f(x) \sim \sum_{n=0}^{\infty} c_n \varphi_n(x)}$ then we have that:

(1)

\begin{align} \quad \sum_{n=0}^{\infty} \mid c_n \mid^2 \leq \| f(x) \|^2 \end{align}

We would like to know when equality holds in Bessel's inequality, i.e., when $\displaystyle{\sum_{n=0}^{\infty} \mid c_n \mid^2 = \| f(x) \|^2}$ which is called Parseval's Formula. The following theorem tells us when Parseval's formula holds.

Theorem 1: Let $\{ \varphi_0(x), \varphi_1(x), ... \}$ be an orthonormal system of functions on $I$, $f \in L^2(I)$, $\displaystyle{f(x) \sim \sum_{n=0}^{\infty} c_n \varphi_n(x)}$, and $\displaystyle{s_n(x) = \sum_{k=0}^{n} c_k \varphi_k(x)}$. Then Parseval's formula holds, i.e., $\displaystyle{\sum_{n=0}^{\infty} \mid c_n \mid^2 = \| f(x) \|^2}$ if and only if $\displaystyle{\lim_{n \to \infty} \| f(x) - s_n(x) \| = 0}$.

• Proof: From the proof of Bessel's inequality (see above) we have:

(2)

\begin{align} \quad \sum_{k=0}^{n} \mid c_k \mid^2 + \| f(x) - s_n(x) \|^2 &= \| f(x) \|^2 \quad (*) \end{align}

• $\Rightarrow$ Suppose that Parseval's Inequality holds. Then from $(*)$ we must have that:

(3)

\begin{align} \quad \lim_{n \to \infty} \| f(x) - s_n(x) \|^2 = 0 \end{align}

• But then we must have that $\displaystyle{\lim_{n \to \infty} \| f(x) - s_n(x) \| = 0}$.

• $\Leftarrow$ Conversely, suppose that $\displaystyle{\lim_{n \to \infty} \| f(x) - s_n(x) \| = 0}$. Noting that $(\| f(x) - s_n(x) \|^2)_{n=0}^{\infty}$ is a sequence of real numbers, by taking the limit as $n \to \infty$ of $(*)$ we get:

(4)

\begin{align} \quad \lim_{n \to \infty} \left ( \sum_{k=0}^{n} \mid c_k \mid^2 + \| f(x) - s_n(x) \|^2 \right ) &= \| f(x) \|^2 \\ \quad \sum_{n=0}^{\infty} \mid c_n \mid^2 + \lim_{n \to \infty} \| f(x) - s_n(x) \|^2 &= \| f(x) \|^2 \\ \quad \sum_{n=0}^{\infty} \mid c_n \mid^2 &= \| f(x) \|^2 \quad \blacksquare \end{align}