Fejer's Kernel Examples 1

# Fejer's Kernel Examples 1

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 and if $\sigma_n$ denotes the arithmetic means of the partial sums of the trigonometric Fourier series (i.e., $\displaystyle{\sigma_n = \frac{s_1 + s_2 + ... + s_n}{n}}$) then $\sigma_n$ has an integral representation 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} nt \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 $\displaystyle{F_n(t) = \frac{1}{n} \frac{\sin^2 \left ( \frac{1}{2} n t \right )}{\sin^2 \left ( \frac{1}{2} t \right )}}$ is called Fejer's kernel.

We will now look at some example problems regarding this representation and Fejer's kernel.

## Example 1

Prove that for all $n \in \mathbb{N}$ that $\displaystyle{\frac{1}{n \pi} \int_0^{\pi} \frac{\sin^2 \left ( \frac{1}{2} nt \right )}{\sin^2 \left ( \frac{1}{2} t \right )} \: dt = 1}$.

Consider the function $f(x) = 1$. Then $f \in L([0, 2\pi)$ and is (somewhat trivially) a $2\pi$-periodic function. So we can apply the theorem stated above. Its very easy to show that the Fourier series of $f$ is simply $f(x) = 1$, and $s_n(x) = 1$ for all $n \in \mathbb{N}$. Therefore, for all $n \in \mathbb{N}$ we have that:

(2)
\begin{align} \quad \sigma_n(x) = \frac{s_1 + s_2 + ... + s_n}{n} = \frac{1 + 1 + ... + 1}{n} = \frac{n}{n} = 1 \end{align}

By the Theorem stated above, this means that for all $n \in \mathbb{N}$:

(3)
\begin{align} \quad 1 = \sigma_n(x) = \frac{1}{n\pi} \int_0^{\pi} \frac{1 + 1}{2} \frac{\sin^2 \left ( \frac{1}{2} n t \right )}{\sin^2 \left ( \frac{1}{2} t \right )} \: dt = \frac{1}{n \pi} \int_0^{\pi} \frac{\sin^2 \left ( \frac{1}{2} n t \right )}{\sin^2 \left ( \frac{1}{2} t \right )} \: dt \end{align}

## Example 2

Let $f \in L([0, 2\pi])$ be a $2\pi$-periodic function. Prove that $\displaystyle{\lim_{n \to \infty} \sigma_n(x) = s}$ if and only if $\displaystyle{\lim_{n \to \infty} \frac{1}{n\pi} \int_0^{\pi} \left [ \frac{f(x + t) + f(x - t)}{2} - s \right ] \frac{\sin^2 \left ( \frac{1}{2} nt \right )}{\sin^2 \left ( \frac{1}{2} t \right )} \: dt = 0}$ by using Example 1 above.

By Example 1 above we have that for all $n \in \mathbb{N}$ that:

(4)
\begin{align} \quad 1 = \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 \end{align}

So we see that:

(5)
\begin{align} \quad \lim_{n \to \infty} \frac{1}{n\pi} \int_0^{\pi} \left [ \frac{f(x + t) + f(x - t)}{2} - s \right ] \frac{\sin^2 \left ( \frac{1}{2} nt \right )}{\sin^2 \left ( \frac{1}{2} t \right )} \: dt &= \lim_{n \to \infty} \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 - \lim_{n \to \infty} s \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 \\ &= \sigma_n(x) - s \end{align}

$\Rightarrow$ If $\displaystyle{\lim_{n \to \infty} \sigma_n(x) = s}$ then [[$\displaystyle{\lim_{n \to \infty} \sima_n(x) - s = 0}$ and the formula above shows that $\displaystyle{\lim_{n \to \infty} \frac{1}{n\pi} \int_0^{\pi} \left [ \frac{f(x + t) + f(x - t)}{2} - s \right ] \frac{\sin^2 \left ( \frac{1}{2} nt \right )}{\sin^2 \left ( \frac{1}{2} t \right )} \: dt = 0}$.

$\Leftarrow$ If $\displaystyle{\lim_{n \to \infty} \frac{1}{n\pi} \int_0^{\pi} \left [ \frac{f(x + t) + f(x - t)}{2} - s \right ] \frac{\sin^2 \left ( \frac{1}{2} nt \right )}{\sin^2 \left ( \frac{1}{2} t \right )} \: dt = 0}$ then $\sigma_n(x) - s \to 0$ as $n \to \infty$, i.e., $\displaystyle{\lim_{n \to \infty} \sigma_n(x) = s}$.