Dirichlet's Kernel Representation of the Partial Sums of a Fourier Series

# Dirichlet's Kernel Representation of the Partial Sums of a Fourier Series

Recall from the Dirichlet's Kernel page that Dirichlet's kernel is the collection of functions $D_n$ where for each $n \in \mathbb{N}$:

(1)
\begin{align} \quad D_n(t) = \frac{1}{2} + \sum_{k=1}^{n} \cos kt = \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 also looked at some properties of the functions in Dirichlet's kernel. We saw that for each $n \in \mathbb{N}$, $D_n$ is even, $2\pi$-periodic, and $\displaystyle{\int_0^{2\pi} D_n(t) \: dt = \pi}$. We will now look at a nice theorem which tells us that any function $f \in L([0, 2\pi])$ that is $2\pi$-periodic can have the partial sums of the Fourier series generated by $f$ represented in terms of an integral involving a function in Dirichlet's kernel.

 Theorem 1: Let $f \in L([0, 2\pi])$ be a $2\pi$-periodic function. For each $n \in \mathbb{N}$ let $\displaystyle{s_n(x) = \frac{a_0}{2} + \sum_{k=1}^{n} (a_k \cos kx + b_k \sin kx)}$ denote the partial sum of the Fourier series generated by $f$. Then for each $n \in \mathbb{N}$, $\displaystyle{s_n(x) = \frac{2}{\pi} \int_0^{\pi} \frac{f(x + t) + f(x - t)}{2} D_n(t) \: dt}$.

Recall that if $f \in L([0, 2\pi])$ then the Fourier series "generated" by $f$ is the Fourier series obtained with respect to the trigonometric system $\displaystyle{\left \{ \frac{1}{\sqrt{2\pi}}, \frac{\cos x}{\sqrt{\pi}}, \frac{\sin x}{\sqrt{\pi}}, \frac{\cos 2x}{\sqrt{\pi}}, \frac{\sin 2x}{\sqrt{\pi}}, ... \right \}}$ and $\displaystyle{f(x) \sim \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos nx + b_n \sin nx)}$ where the $a_n$'s and $b_n$'s are the Fourier coefficients of the Fourier series given by $\displaystyle{a_n = \frac{1}{\pi} \int_0^{2\pi} f(t) \cos nt \: dt}$ and $\displaystyle{b_n = \frac{1}{\pi} \int_0^{2\pi} f(t) \sin nt \: dt}$. Be sure to review the content on the The Fourier Series of Functions Relative to an Orthonormal System page as a refresher!

• Proof: We know that for all $n \in \mathbb{N}$ that $\displaystyle{s_n(x) = \frac{a_0}{2} + \sum_{k=1}^{n} (a_k \cos kx + b_k \sin kx)}$. Compacting this and we get:
(2)
\begin{align} \quad s_n(x) &= \frac{a_0}{2} + \sum_{k=1}^{n} (a_k \cos k x + b_k \sin kx ) \\ &= \frac{1}{\pi} \frac{1}{2} \int_0^{2\pi} f(t) \: dt + \sum_{k=1}^{n} \left ( \frac{1}{\pi} \int_0^{2\pi} f(t) \cos kt \cos kx \: dt + \frac{1}{\pi} \int_0^{2\pi} f(t) \sin kt \sin kx \: dt \right ) \\ &= \frac{1}{\pi} \int_0^{2\pi} f(t) \left [ \frac{1}{2} + \sum_{k=1}^{n} (\cos kt \cos kx + \sin kt \sin kx) \right ] \: dt \\ \end{align}
• We now use the following trigonometric identity:
(3)
\begin{align} \quad \cos (a - b) = \cos a \cos b + \sin a \sin b \end{align}
• Therefore:
(4)
\begin{align} \quad s_n(x) &= \frac{1}{\pi} \int_0^{2\pi} f(t) \left [ \frac{1}{2} + \sum_{k=1}^{n} \cos (kt - kx) \right ] \: dt \\ &= \frac{1}{\pi} \int_0^{2\pi} f(t) \left [ \frac{1}{2} + \sum_{k=1}^{n} \cos (k(t - x)) \right ] \: dt \\ &= \frac{1}{\pi} \int_0^{2\pi} f(t) D_n(t - x) \: dt \end{align}
• Notice that $f$ and $D_n$ are both $2\pi$ periodic. Therefore we have that:
(5)
\begin{align} \quad s_n(x) &= \frac{1}{\pi} \int_{x - \pi}^{x + \pi} f(t) D_n(t - x) \: dt \end{align}
• Let $u = t - x$. Then $du =dt$ and thus:
(6)
\begin{align} \quad s_n(x) &= \frac{1}{\pi} \int_{-\pi}^{\pi} f(u + x) D_n(u) \: du \\ &= \frac{1}{\pi} \int_{-\pi}^0 f(x + u) D_n(u) \: dt + \frac{1}{\pi} \int_0^{\pi} f(x + u) D_n(u) \: du \end{align}
• Since $D_n$ is an even function and $[-\pi, \pi]$ is centered at the origin we have that:
(7)
\begin{align} \quad s_n(x) &= \frac{1}{\pi}\int_0^{\pi} f(x - u) D_n(u) \: du + \frac{1}{\pi} \int_0^{\pi} f(x + u) D_n(u) \: du \\ &= \frac{1}{\pi} \int_0^{\pi} [f(x + u) + f(x - u)]D_n(u) \: du \\ &= \frac{2}{\pi} \int_0^{\pi} \frac{f(x + u) + f(x - u)}{2} D_n(u) \: du \end{align}
• Replacing the dummy variable $u$ with $t$ yields:
(8)
\begin{align} \quad s_n(x) &= \frac{2}{\pi} \int_0^{\pi} \frac{f(x + t) + f(x - t)}{2} D_n(t) \: dt \quad \blacksquare \end{align}