The Riesz-Fischer Theorem for Fourier Series
 Table of Contents

# The Riesz-Fischer Theorem for 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 in $L^2(I)$ and if $f \in L^2(I)$ where $\displaystyle{f(x) \sim \sum_{n=0}^{\infty} c_n\varphi_n(x)}$ then $\displaystyle{\sum_{n=0}^{\infty} \mid c_n \mid^2 \leq \| f(x) \|^2}$.

Suppose now that we consider the same orthonormal system and we instead have a sequence of complex numbers $(c_n)_{n=0}^{\infty}$ such that $\displaystyle{\sum_{n=0}^{\infty} \mid c_n \mid^2}$ converges. The following theorem shows us that a function $f \in L^2(I)$ exists with those properties we desire from it.

 Theorem 1 (Riesz-Fischer Theorem for Fourier Series): Let $\mathcal S = \{ \varphi_0(x), \varphi_1(x), ... \}$ be an orthonormal system of functions in $L^2(I)$ and let $(c_n)_{n=0}^{\infty}$ be any sequence of complex numbers for which $\displaystyle{\sum_{n=0}^{\infty} \mid c_n \mid^2}$ converges. Then there exists a function $f \in L^2(I)$ with the following properties: a) $\displaystyle{\sum_{n=0}^{\infty} \mid c_n \mid^2 = \| f(x) \|^2}$. b) $c_k= (f, \varphi_k)$ for all $k \in \{ 0, 1, 2, ... \}$.

In the following proof we will use three important theorems. We will call upon the Riesz-Fischer theorem for sequences of Cauchy functions in $L^2(I)$, Parseval's formula, and the Cauchy-Schwarz inequality.

• Proof of a) Define a sequence of functions $(s_n(x))_{n=0}^{\infty}$ as follows:
(1)
\begin{align} \quad s_n(x) = \sum_{k=0}^{n} c_k\varphi_k(x) \end{align}
• Then this is a sequence of Lebesgue integrable functions since each $s_n$ is a finite linear combination of Lebesgue integrable functions which is also Lebesgue integrable.
• We claim that the sequence $(s_n(x))_{n=0}^{\infty}$ is also a Cauchy sequence. To show this, we note that since $\displaystyle{\sum_{n=0}^{\infty} \mid c_n \mid^2}$ converges that $\lim_{n \to \infty} \mid c_n \mid = 0$. So $(\mid c_n \mid)_{n=0}^{\infty}$ is a convergent sequence of positive real numbers approaching $0$, so, for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $\displaystyle{\sum_{k=n+1}^{p} \mid c_k \mid < \epsilon}$. Thus for $m, n \geq N$ we have that:
(2)
\begin{align} \quad \| s_m(x) - s_n(x) \| &= \biggr \| \sum_{k=0}^{m} c_k\varphi_k(x) - \sum_{k=0}^{n} c_k \varphi_k(x) \biggr \| \\ &= \biggr \| \sum_{k=n+1}^m c_k \varphi_k(x) \biggr \| \\ &= \sum_{k=n+1}^m \| c_k \varphi_k(x) \| \\ &= \sum_{k=n+1}^m \mid c_k \mid \\ &< \epsilon \end{align}
(3)
\begin{align} \quad \lim_{n \to \infty} \| f(x) - s_n(x) \| = 0 \quad (*) \end{align}
• Proof of b) Let $n \in \{0, 1, 2, ... \}$. Then $s_n(x) = \sum_{k=0}^{n} c_k \varphi_k(x)$. So for any $k \in \{ 0, 1, ..., n \}$ we have that:
(4)
\begin{align} \quad (s_n(x), \varphi_k(x)) &= \left ( \sum_{j=0}^{n} c_j \varphi_j(x), \varphi_k(x) \right ) \\ \quad &= \sum_{j=0}^{n} c_j (\varphi_j(x), \varphi_k(x)) \\ \quad &= c_k \end{align}
• So we will now show that $\mid c_k - (f(x), \varphi_k(x)) \mid < \epsilon$ for all $\epsilon > 0$. Notice that:
(5)
\begin{align} \quad \mid c_k - (f(x), \varphi_k(x)) \mid = \mid (s_n(x), \varphi_k(x)) - (f(x), \varphi_k(x)) \mid = \mid (s_n(x) - f(x), \varphi_k(x)) \mid \end{align}
• Using the Cauchy-Schwarz inequality gives us that:
(6)
\begin{align} \quad \mid c_k - (f(x), \varphi_k(x)) \mid = ... \leq \| s_n(x) - f(x) \| \| \varphi_k(x) \| = \| s_n(x) - f(x) \| \end{align}
• Since $\lim_{n \to \infty} \| s_n(x) - f(x) \| = 0$ from (a), for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $\| s_n(x) - f(x) \| < \epsilon$, i.e., $\mid c_k - (f(x), \varphi_k(x)) \mid < \epsilon$ for all $\epsilon > 0$ which shows that $c_k = (f(x), \varphi_k(x))$ for all $k \in \{0, 1, 2, ... \}$. $\blacksquare$
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License