The Riemann Localization Theorem

# The Riemann Localization Theorem

Recall from the Dirichlet's Kernel Representation of the Partial Sums of a Fourier Series page that if $f \in L([0, 2\pi])$ is a $2\pi$-periodic function and if $\displaystyle{s_n(x) = \frac{a_0}{2} + \sum_{k=1}^{n} (a_k \cos kx + b_k \sin kx)}$ is the $n^{\mathrm{th}}$ partial sum of the Fourier series generated by $f$ (with respect to the trigonometric system) then $s_n$ can be represented as an integral by:

(1)
\begin{align} \quad s_n(x) = \frac{2}{\pi} \int_0^{\pi} \frac{f(x + t) + f(x - t)}{2} D_n(t) \: dt \end{align}

The collection of functions $D_n$ are called Dirichlet's Kernel and are given by:

(2)
\begin{align} \quad D_n(x) = \frac{1}{2} + \sum_{k=1}^{n} \cos kx = \left\{\begin{matrix} n + \frac{1}{2} & \mathrm{if} \: t = 2m\pi, m \in \mathbb{Z} \\ \frac{\sin \left ( \left ( n + \frac{1}{2} \right ) t \right )}{2 \sin \left ( \frac{t}{2} \right )} & \mathrm{if} \: t \neq 2m\pi, m \in \mathbb{Z}\\ \end{matrix}\right. \end{align}

We will now state and prove a very important useful theorem known as The Riemann Localization Theorem which will tell us when a Fourier series generated by $f$ (with respect to the trigonometric system) converges at a point $x$.

 Theorem 1 (The Riemann Localization Theorem): Let $f \in L([0, 2\pi])$ be a $2\pi$-periodic function. Then the Fourier series generated by $f$ will converge at a point $x$ if and only if there exists a $b \in \mathbb{R}$ with $0 < b \leq \pi$ such that $\displaystyle{\lim_{n \to \infty} \frac{2}{\pi} \int_0^{b} \frac{f(x + t) + f(x - t)}{2} \frac{\sin \left (\left ( n + \frac{1}{2} \right )t \right )}{t} \: dt}$ exists in which case the Fourier series generated by $f$ will converge at $x$ to this limit.
• Proof: For each $x$, the Fourier series generated by $f$ converges at $x$ if and only if the sequence of partial sums converge at $x$, that is, the following limit exists:
(3)
\begin{align} \quad \lim_{n \to \infty} s_n(x) &= \lim_{n \to \infty} \frac{2}{\pi} \int_0^{\pi} \frac{f(x + t) + f(x - t)}{2} D_n(t) \: dt \\ \end{align}
• Since the value of an integral does not change if a countable collection of points in the domain are removed, we can replace $D_n(t)$ by $\displaystyle{\frac{\sin \left ( \left ( n + \frac{1}{2}\right ) t \right )}{2 \sin \left (\frac{t}{2}\right )}}$ to get that the Fourier series generated by $f$ converges at $x$ if and only if the following limit exists:
(4)
\begin{align} \quad \lim_{n \to \infty} s_n(x) &= \lim_{n \to \infty} \frac{2}{\pi} \int_0^{\pi} \frac{f(x + t) + f(x - t)}{2} \frac{\sin \left ( \left ( n + \frac{1}{2}\right ) t \right )}{2 \sin \left (\frac{t}{2}\right )} \: dt \\ \end{align}
• Now consider the following limit:
(5)
\begin{align} \quad \lim_{n \to \infty} \frac{2}{\pi} \int_0^{\pi} \frac{f(x + t) + f(x - t)}{2} \frac{\sin \left ( n + \frac{1}{2} \right )t}{t} \: dt - \lim_{n \to \infty} \frac{2}{\pi} \int_0^{\pi} \frac{f(x + t) + f(x - t)}{2} \frac{\sin \left ( n + \frac{1}{2} \right )t}{2 \sin \left ( \frac{t}{2} \right )} \: dt \\ = \lim_{n \to \infty} \frac{2}{\pi} \int_0^{\pi} \left ( \frac{1}{t} - \frac{1}{2\sin \left ( \frac{t}{2} \right ) } \right ) \frac{f(x + t) + f(x - t)}{2} \sin \left ( \left ( n + \frac{1}{2} \right ) t \right ) \: dt \end{align}
• Since $f \in L([0, 2\pi])$ and is $2\pi$-periodic we have that $f(x + t)$ and $f(x - t)$ are Lebesgue integrable on $[0, 2\pi]$ as well, and thus, $\displaystyle{\frac{f(x + t) + f(x - t)}{2}}$ is Lebesgue integrable on $[0, 2\pi]$. Furthermore, $\displaystyle{\frac{1}{t} - \frac{1}{2\sin \left ( \frac{t}{2} \right )}}$ is continuous and of bounded variation on $[0, \pi]$ if at $t = 0$ we define this function to equal $0$, and so $\displaystyle{\left (\frac{1}{t} - \frac{1}{2\sin \left ( \frac{t}{2} \right )} \right ) \frac{f(x + t) + f(x - t)}{2}}$ is Lebesgue integrable on $[0, 2\pi]$. So by The Riemann-Lebesgue Lemma we see that:
(6)
\begin{align} \quad \lim_{n \to \infty} \frac{2}{\pi} \int_0^{\pi} \left ( \frac{1}{t} - \frac{1}{2\sin \left ( \frac{t}{2} \right ) } \right ) \frac{f(x + t) + f(x - t)}{2} \sin \left ( \left ( n + \frac{1}{2} \right ) t \right ) \: dt = 0 \end{align}
• Which implies that:
(7)
\begin{align} \quad \lim_{n \to \infty} s_n(x) = \lim_{n \to \infty} \frac{2}{\pi} \int_0^{\pi} \frac{f(x + t) + f(x - t)}{2} \frac{\sin \left ( n + \frac{1}{2} \right )t}{t} \: dt = \lim_{n \to \infty} \frac{2}{\pi} \int_0^{\pi} \frac{f(x + t) + f(x - t)}{2} \frac{\sin \left ( n + \frac{1}{2} \right )t}{2 \sin \left ( \frac{t}{2} \right )} \: dt \\ \end{align}
• Now for any $b \in \mathbb{R}$ such that $0 < b \leq \pi$ we use the Riemann-Lebesgue Lemma again by noting that $\frac{1}{t}$ is continuous and of bounded variation on $[b, \pi]$ for all $0 < b \leq \pi$ and we see that:
(8)
\begin{align} \quad \lim_{n \to \infty} s_n(x) & = \lim_{n \to \infty} \frac{2}{\pi} \int_0^{b} \frac{f(x + t) + f(x - t)}{2} \frac{\sin \left ( n + \frac{1}{2} \right ) t}{t} \: dt + \underbrace{\lim_{n \to \infty} \int_b^{\pi} \frac{f(x + t) + f(x - t)}{2} \frac{1}{t} \sin \left (\left ( n + \frac{1}{2} \right )t \right ) \: dt}_{=0 \: \mathrm{by \: the \: Riemann-Lebesgue \: Lemma}} \\ \quad \lim_{n \to \infty} s_n(x) &= \lim_{n \to \infty} \frac{2}{\pi} \int_0^{b} \frac{f(x + t) + f(x - t)}{2} \frac{\sin \left ( n + \frac{1}{2} \right ) t}{t} \: dt \quad (*) \end{align}
• $\Rightarrow$ Suppose that the Fourier series generated by $f$ converges at $x$ then $\displaystyle{\lim_{n \to \infty} s_n(x)}$ converges and from $(*)$ above there exists a $b \in \mathbb{R}$ with $0 < b \leq \pi$ such that the following limit exists and for which the following limit is equal to the sum of the Fourier series generated by $f$ at $x$:
(9)
\begin{align} \quad \lim_{n \to \infty} \frac{2}{\pi} \int_0^{b} \frac{f(x + t) + f(x - t)}{2} \frac{\sin \left ( n + \frac{1}{2} \right ) t}{t} \: dt \end{align}
• $\Leftarrow$ Conversely, suppose that $\displaystyle{\lim_{n \to \infty} \frac{2}{\pi} \int_0^{b} \frac{f(x + t) + f(x - t)}{2} \frac{\sin \left ( n + \frac{1}{2} \right ) t}{t} \: dt }$ exists for some $b \in \mathbb{R}$ with $0 < b \leq \pi$. Then by $(*)$ we have that $\displaystyle{\lim_{n \to \infty} s_n(x)}$ equals this limit so the sequence of partial sums for the Fourier series generated by $f$ converges at $x$ which implies that the Fourier series generated by $f$ converges at $x$. $\blacksquare$