Fejer's Theorem

# Fejer's Theorem

Recall from the Fejer's Kernel Representation of the Arithmetic Means of the Partial Sums of a Fourier Series page that if $f \in L([0, 2\pi])$ is a $2\pi$-periodic function then the arithmetic means of the partial sums of the trigonometric Fourier series of $f$, $\displaystyle{\sigma_n(x) = \frac{s_1(x) + s_2(x) + ... + s_n(x)}{n}}$ can be expressed as an integral with:

(1)
\begin{align} \quad \sigma_n(x) = \frac{1}{n \pi} \int_0^{\pi} \frac{f(x + t) + f(x - t)}{2} \frac{\sin^2 \left ( \frac{1}{2} n t \right )}{\sin^2 \left ( \frac{1}{2} t \right )} \: dt = \frac{1}{\pi} \int_0^{\pi} \frac{f(x + t) + f(x - t)}{2} F_n(t) \: dt \end{align}

Where the functions $\displaystyle{F_n(t) = \frac{1}{n} \frac{\sin^2 \left ( \frac{1}{2} nt \right )}{\sin^2 \left ( \frac{1}{2} t \right )}}$ are called Fejer's kernel.

We will now look at a remarkable theorem called Fejer's theorem.

 Theorem 1 (Fejer's Theorem): Let $f \in L([0, 2\pi])$ be a $2\pi$-periodic function. Let $\displaystyle{s(x) = \lim_{t \to 0+} \frac{f(x + t) + f(x - t)}{2}}$. Then for each $x$ for which $s(x)$ is well-defined, the trigonometric Fourier series of $f$ at $x$ is Cesáro summable with sum $s(x)$, i.e., $\sigma_n(x) \to s(x)$ as $n \to \infty$. Furthermore, if $f$ is continuous on $[0, 2\pi]$ then $\sigma_n(x) \to f(x)$ uniformly as $n \to \infty$ on $[0, 2\pi]$.
• Proof: For each fixed $x$ for which $s(x)$ is defined, let:
(2)
\begin{align} \quad g_x(t) = \frac{f(x + t) + f(x - t)}{2} - s(x) \end{align}
• Then notice that:
(3)
\begin{align} \quad \lim_{t \to 0+} g_x(t) = \lim_{t \to 0+} \left [ \frac{f(x + t) + f(x - t)}{2} - s(x) \right ] = 0 \end{align}
• So for all $\epsilon > 0$ there exists a $\delta > 0$ with $\delta < \pi$ for which if $0 < t < \delta$ then $\mid g_x(t) \mid < \epsilon$. Now by Fejer's integral representation for
(4)
\begin{align} \quad \mid \sigma_n(x) - s(x) \mid = \biggr \lvert \frac{1}{n\pi} \int_0^{\pi} \frac{f(x + t) + f(x - t)}{2} \frac{\sin^2 \left ( \frac{1}{2} nt \right )}{\sin^2 \left ( \frac{1}{2} t \right )} \: dt - s(x) \biggr \rvert \end{align}
• On the Fejer's Kernel Examples 1 we proved that $\displaystyle{\frac{1}{n\pi} \int_0^{\pi} \frac{\sin^2 \left ( \frac{1}{2} nt \right )}{\sin^2 \left ( \frac{1}{2}t \right )} \: dt = 1}$, so $\displaystyle{\frac{1}{n\pi} \int_0^{\pi} s(x) \frac{\sin^2 \left ( \frac{1}{2} nt \right )}{\sin^2 \left ( \frac{1}{2}t \right )} \: dt = s(x)}$. Substituting this into the equation above gives us for $t$ with $0 < t < \delta$:
(5)
\begin{align} \quad \mid \sigma_n(x) - s(x) \mid &= \biggr \lvert \frac{1}{n\pi} \int_0^{\pi} \frac{f(x + t) + f(x - t)}{2} \frac{\sin^2 \left ( \frac{1}{2} nt \right )}{\sin^2 \left ( \frac{1}{2} t \right )} \: dt - \frac{1}{n\pi} \int_0^{\pi} s(x) \frac{\sin^2 \left ( \frac{1}{2} nt \right )}{\sin^2 \left ( \frac{1}{2}t \right )} \: dt \biggr \rvert \\ &= \frac{1}{n \pi} \biggr \lvert \int_0^{\pi} \left [ \frac{f(x + t) + f(x - t)}{2} - s(x) \right ] \frac{\sin^2 \left ( \frac{1}{2} nt \right )}{\sin^2 \left ( \frac{1}{2} t \right )} \: dt \biggr \rvert \\ &= \frac{1}{n \pi} \biggr \lvert g_x(t) \frac{\sin^2 \left ( \frac{1}{2} nt \right )}{\sin^2 \left ( \frac{1}{2} t \right )} \: dt \biggr \rvert \\ & \leq \frac{1}{n\pi} \int_0^{\pi} \mid g_x(t) \mid \frac{\sin^2 \left ( \frac{1}{2} nt \right )}{\sin^2 \left ( \frac{1}{2} t \right )} \: dt \\ & < \frac{1}{n\pi} \int_0^{\pi} \epsilon \frac{\sin^2 \left ( \frac{1}{2} nt \right )}{\sin^2 \left ( \frac{1}{2} t \right )} \: dt \\ & < \frac{\epsilon}{n\pi} \int_0^{\pi} \frac{\sin^2 \left ( \frac{1}{2} nt \right )}{\sin^2 \left ( \frac{1}{2} t \right )} \: dt \\ & < \frac{\epsilon}{n\pi} \\ & < \epsilon \end{align}
• Therefore $\sigma_n(x) \to s(x)$ as $n \to \infty$.
• If $f$ is continuous on $[0, 2\pi]$ then $f$ is bounded on $[0, 2\pi]$ so there exists an $M \in \mathbb{R}$, $M > 0$ such that $\mid g_x(t) \mid \leq M$ for all $t \in [0, 2\pi]$. So above, replacing "$\epsilon$" with "$M$" gives us that $\displaystyle{\mid \sigma_n(x) - s(x) \mid \leq \frac{M}{n\pi}}$, and taking the limit as $n \to \infty$ shows that $\sigma_n(x)$ converges to $s(x)$ uniformly on $[0, 2\pi]$. But since $f$ is continuous on $[0, 2\pi]$ we see that $\displaystyle{s(x) = \lim_{t \to 0} \frac{f(x + t) + f(x - t)}{2} = \frac{2f(x)}{2} = f(x)}$, so $\sigma_n(x) \to f(x)$ uniformly as $n \to \infty$ on $[0, 2\pi]$. $\blacksquare$
 Theorem 2: Let $f$ be a continuous function on $[0, 2\pi]$ that is $2\pi$-periodic with trigonometric Fourier series $\displaystyle{f(x) \sim \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n\cos nx + b_n \sin nx)}$. Then: a) $\displaystyle{\lim_{n \to \infty} \| s_n(x) - f(x) \| = 0}$ on $[0, 2\pi]$. b) $\displaystyle{\frac{1}{\pi} \int_0^{2\pi} \mid f(x) \mid^2 \: dx = \frac{a_0^2}{2} + \sum_{n=1}^{\infty} (a_n^2 + b_n^2)}$ (Parseval's Identity). c) $\displaystyle{\int_0^x f(t) \: dt = \frac{a_0x}{2} + \sum_{n=1}^{\infty} \int_0^x (a_n \cos nt + b_n \sin nt) \: dt}$. d) If $s_n(x)$ converges at $x$ then it converges to $f(x)$.