Theorem 6.3.1.
The sequence given by
\begin{equation*}
e_n(x) = \frac{1}{\sqrt{2\pi}} e^{i n x}, \hspace{1cm} x \in (-\pi, \pi)
\end{equation*}
is a complete orthonormal sequence in \(L^2(-\pi, \pi)\text{.}\)
Section 6.3 Convergence of classical Fourier series
We now seek to establish that classical Fourier series are expansions in a complete orthonormal sequence (see Definition 5.1.1).
Proof.
That the functions \(e_n(x)\) are an orthonormal system in the standard inner product in \(L^2(-\pi, \pi)\) is left as an exercise.
We seek to invoke Theorem 5.4.3 to claim that \((e_n)\) is complete by arguing that
\begin{equation*}
\cl \spn \{e_n: n \in \Z\} = L^2(-\pi, \pi).
\end{equation*}
To perform this trick, we’ll actually work on a easier question by way of a result typically proved in graduate analysis that by necessity will have to be a black box. We claim that functions in \(L^2(-\pi, \pi)\) can be approximated uniformly by continuous \(2\pi\)-periodic functions - that is, the set of functions consisting of restrictions of \(2\pi\)-continuous functions to \((-\pi, \pi)\) is a dense subspace of \(L^2(-\pi,\pi)\text{.}\) Essentially then, the argument will be that any continuous function in \(L^2(-\pi, \pi)\) can be approximated uniformly by linear combinations of \(e_n\) and any \(L^2\) function can be approximated by continuous functions, which the standard setup for a triangle inequality argument.
So choose an arbitrary \(2\pi\)-continuous function \(f: (-\pi, \pi)\to \R\text{.}\) To show that \(f\in \cl \spn \{e_n\}\text{,}\) we need to find a sequence of functions in \(\spn \{e_n\}\) converging to \(f\text{.}\) Define the \(m\)th Fourier polynomial
\begin{equation*}
f_m = \sum_{n=-m}^m \ip{f}{e_n} e_n.
\end{equation*}
By construction, \(f_m\) is in the linear span of \((e_n)\text{,}\) and it would suffice to prove that \(f_m \to f\text{.}\) However, in order to set up the application of the Fejer kernel 6.2.1 results we established in the previous chapter, we’ll work with the Cesàro sum
1
en.wikipedia.org/wiki/Ces%C3%A0ro_summation
\begin{equation*}
F_m = \frac{1}{m+1} (f_0 + f_1 + \cdots + f_m),
\end{equation*}
and show that \(F_m \to f\) (this is also sufficient, as each \(F_m \in \spn\{e_n\}\)).
In \(L^2(-\pi,\pi)\text{,}\) the inner product \(\ip{f}{en}\) is given by
\begin{equation*}
\ip{f, e_n} = \frac{1}{\sqrt{2\pi}} \int_\pi^\pi f(x) e^{-inx}\, dx
\end{equation*}
and so for the Fourier polynomial \(f_m\text{,}\) we have
\begin{align*}
f_m(y) \amp= \sum_{n=-m}^m \ip{f}{e_n} e_n(y)\\
\amp= \frac{1}{2\pi} \sum_{n=-m}^{m} \left(\int_{-\pi}^\pi f(x) e^{-inx}\, dx\right)e^{iny}\\
\amp= \frac{1}{2\pi} \int_{-\pi}^\pi f(x) \sum_{n=-m}^m e^{in(y-x)}\, dx.
\end{align*}
Plugging this into our averaged function \(F_m\text{,}\) we get
\begin{align*}
F_m(y) \amp= \frac{1}{m+1} \sum_{j=0}^m f_j(y)\\
\amp= \frac{1}{m+1} \sum_{j=0}^m \frac{1}{2\pi} \int_{-\pi}^\pi f(x) \sum_{n=-m}^m e^{in(y-x)}\, dx\\
\amp= \frac{1}{2\pi} \int_{-\pi}^\pi f(x) \left(\frac{1}{m+1} \sum_{j=0}^m \sum_{n=-j}^j e^{in(y-x)}\right)\, dx.\\
\amp= \frac{1}{2\pi} \int_{-\pi}^\pi f(x) K_m(y-x)\, dx,
\end{align*}
where \(K_m(y-x)\) is the Fejér kernel 6.2.1.
We’re all set up to apply the technical lemmas. Essentially, the idea is going to be to show that \(F_m \to f\text{.}\) We’ll start by constructing an integral form of this difference to study. Let \(y \in [-\pi, \pi]\text{.}\) By Lemma 6.2.3, the substitution \(t = y-x\) gives
\begin{equation*}
\int_{y-\pi}^{y-\pi} K_m(y-x)\, dx = 2\pi,
\end{equation*}
which gives
\begin{equation*}
f(y) \int_{y-\pi}^{y-\pi} K_m(y-x)\, dx = f(y) 2\pi,
\end{equation*}
or more usefully,
\begin{equation}
f(y) =\frac{1}{2\pi} \int_{y - \pi}^{y+\pi} f(x) K_m(y-x)\, dx.\tag{6.3.1}
\end{equation}
Since \(K_m\) is \(2\pi\)-periodic, and \(f\) is \(2\pi\)-periodic, so is the function \(f(x)K_m(y-x)\) for a fixed \(y\text{.}\) Once consequence is that the integral of \(f(x)K_m(y-x)\) will evaluate to the same quantity on any interval of length \(2\pi\text{,}\) and so in particular, we can write
\begin{equation}
F_m(y) = \frac{1}{2\pi} \int_{y - \pi}^{y+\pi} f(x) K_m(y-x)\, dx.\tag{6.3.2}
\end{equation}
\begin{equation}
F_m(y) - f(y) = \frac{1}{2\pi} \int_{y-\pi}^{y + \pi} (f(x) - f(y)) K_m(y - x)\, dx,\tag{6.3.3}
\end{equation}
and now we can see what the game is. We will show that this difference is uniformly small for a large enough \(N \in \mathbb{N}\text{.}\)
Let \(\eps \gt 0\) be given. Since \(f\) is continuous on \([-\pi,\pi]\text{,}\) it is also bounded, say by \(M\) so that \(\abs{f(x)} \leq M\) for all \(x\in \R\) (since \(f\) is \(2\pi\)-periodic). Because it is continuous on a compact set, \(f\) is also uniformly continuous. So choose \(\delta \gt 0\) such that \(y \in [-\pi,\pi]\) and \(\abs{y - x} \lt \delta\) give
\begin{equation*}
\abs{f(x) - f(y)} \lt \frac{\eps}{2}.
\end{equation*}
So when \(y\) is near \(x\text{,}\) \(F_m\) and \(f\) are close. What if \(x, y\) are far apart? By Lemma 6.2.3, for our given \(\delta\text{,}\) we can find \(N \in \mathbb{N}\) such that
\begin{equation*}
\int_{-\pi}^{-\delta} + \int_{\delta}^\pi K_m(t) \, dt \lt \frac{\pi \eps}{2M}.
\end{equation*}
