Kernels arising from the Fourier series

Here, we present a different way of looking at the Fourier series using kernels. We see that the partial Fourier sum of a function $f(x)$ can be expressed as the convolution between $f$ and the Dirichlet kernel. We have already seen that the partial Fourier sum converges uniformly to $f$ if $f(x)$ is continuously differentiable. Similarly, we show that the convolution of the function $f$ and the Fejér kernel converges uniformly to $f$, but this time, if $f(x)$ is continuous. These kernels are used to prove the $L^2$ convergence of the Fourier series.

Definition (Kernel function)

Let $X$ be a set, and $K:X\times X\to \mathbb{R}$ be a two-variable function. $K(x,y)$ is said to be a kernel function (or integral kernel or nucleus) if it satisfies the following conditions:

1. (symmetry) For any $x,y\in X$, $$K(x,y)=K(y,x).$$
2.  (positive semi-definiteness) For any $n\in \mathbb{N}$, $\{x_1,x_2,\cdots ,x_n\}\subset X$, $\{c_1,c_2,\cdots ,c_n\}\subset \mathbb{R}$, $$\sum _{i=1}^n\sum _{j=1}^n{c}_i{c}_{j}K(x_i,x_j)\ge 0.$$

We now consider kernel representations of the Fourier series.

The Dirichlet kernel

Consider the partial sum $S_n(x)$ of $S[f]$:

$$\begin{array}{rcl}{S}_n(x)& =& \sum _{|k|\le n}{c}_k{e}^{ikx}\\ & =& \sum _{|k|\le n}(\frac{1}{2\pi }{\int }_{-\pi }^{\pi }f(y)e^{-iky}dy)e^{ikx}\\ & =& {\int }_{-\pi }^{\pi }\frac{1}{2\pi }\sum _{k=-n}^n{e}^{ik(x-y)}f(y)dy.\end{array}$$

Let us define $D_n(\theta )(\theta \in \mathbb{R})$ by

$$\begin{array}{}\text{(eq:Ddef)}& D_n(\theta )=\frac{1}{2\pi }\sum _{k=-n}^n{e}^{ik\theta }.\end{array}$$

Noting

$$\begin{array}{rcl}\sum _{k=-n}^n{e}^{ik\theta }& =& \frac{e^{-in\theta }-e^{i(n+1)\theta }}{1-e^{i\theta }}\\ & =& \frac{e^{i(n+1/2)\theta }-e^{-i(n+1/2)\theta }}{e^{i\theta /2}-e^{-i\theta /2}}\\ & =& \frac{\sin (n+\frac{1}{2})\theta }{\sin \frac{\theta }{2}},\end{array}$$

we have

$$\begin{array}{}\text{(eq:Dfconv)}& S_n[f](x)=S_n(x)={\int }_{-\pi }^{\pi }{D}_n(x-y)f(y)dy\end{array}$$

where

$$D_n(x-y)=\frac{1}{2\pi }\frac{\sin (n+\frac{1}{2})(x-y)}{\sin \frac{x-y}{2}}(n=0,1,\cdots )$$

is called the Dirichlet kernel. As the name suggests, it is a kernel function. $D_n(\theta )$ is an even function of $\theta$, and has a period of $2\pi$. Thus, $D_n(x-y)$ is symmetric with respect to $x$ and $y$. Using (eq:Ddef), we can show that it is also positive semi-definite (exercise!).

For the constant function $f(x)=1$,  $c_n=0$ for $n\ne 0$ and $c_0=1$ so that $S_n[f](x)=1$ for all $n$. Therefore

$${\int }_{-\pi }^{\pi }{D}_n(x-y)dy=1(n=0,1,\cdots ).$$

The following figure shows the Dirichlet kernels with $n=0,1,5,20$.

The integral of the form of (Eq:Dfconv) is so common that it has a name.

Definition (Convolution)

Let $f,g\in {\mathcal{P}}_{2\pi }$ . The convolution of $f$ and $g$, often denoted as $f\ast g$, is defined as

$$(f\ast g)(x)={\int }_{-\pi }^{\pi }f(x-y)g(y)dy.$$

See also: Convolution (Wikipedia)

Thus, we have

$$S_n(x)=(D_n\ast f)(x).$$

The Fejér kernel

Given the sequence $\{a_n{\}}_{n=0}^{\mathrm{\infty }}$, define another sequence $\{b_n\}$ by

$$b_n=\frac{1}{n+1}\sum _{k=0}^n{a}_k=\frac{a_0+a_1+\cdots +a_n}{n+1}.$$

It is known that, if $\{a_n\}$ converges, then $\{b_n\}$ converges to the same value and the convergence is generally ``faster'' than $\{a_n\}$. Based on this observation, we introduce another kernel representation of the Fourier series. For the partial sum $S_n$ of the Fourier series of $f$, we define Fejér's partial sum ${\sigma }_n[f](x)$ by

$$\begin{array}{}\text{(eq:fejer)}& {\sigma }_n[f](x)={\sigma }_n(x)=\frac{1}{n+1}\sum _{k=0}^n{S}_k(x)(n=0,1,\cdots ).\end{array}$$

The integral representation of ${\sigma }_n(x)$ can be obtained as follows.

$$\begin{array}{rcl}{\sigma }_n(x)& =& \frac{1}{n+1}\sum _{k=0}^n{\int }_{-\pi }^{\pi }{D}_k(x-y)f(y)dy\\ & =& {\int }_{-\pi }^{\pi }(\frac{1}{n+1}\sum _{k=0}^n{D}_k(x-y))f(y)dy\end{array}$$

where $D_n$ is the Dirichlet kernel. Noting

$$\begin{array}{rcl}\sum _{k=0}^n\sin (k+\frac{1}{2})\theta & =& \frac{1}{\sin \frac{\theta }{2}}\sum _{k=0}^n\sin (k+\frac{1}{2})\theta \cdot \sin \frac{\theta }{2}\\ & =& \frac{1}{2\sin \frac{\theta }{2}}\sum _{k=0}^n(\cos k\theta -\cos (k+1)\theta )\\ & =& \frac{1}{2\sin \frac{\theta }{2}}(1-\cos (n+1)\theta )\\ & =& \frac{{\sin }^2\frac{n+1}{2}\theta }{\sin \frac{\theta }{2}},\end{array}$$

we have

$${\sigma }_n[f](x)={\sigma }_n(x)={\int }_{-\pi }^{\pi }{F}_n(x-y)f(y)dy=(F_n\ast f)(x)$$

where

$$F_n(x-y)=\frac{1}{2(n+1)\pi }\frac{{\sin }^2(\frac{n+1}{2}(x-y))}{{\sin }^2\frac{x-y}{2}}(n=0,1,\cdots ).$$

$F_n(x-y)$ is called the Fejér kernel. $F_n(\theta )$ is an even function and has a period of $2\pi$, so $F_n(x-y)$ is symmetric. The positive semi-definiteness is inherited from that of the Dirichlet kernel.

For the constant function $f(x)=1$, $S_n\equiv 1$ ($n=0,1,\cdots$) so that ${\sigma }_n(x)\equiv 1$, and hence

$$\begin{array}{}\text{(eq:Fnormal)}& {\int }_{-\pi }^{\pi }{F}_n(x-y)dy=1.\end{array}$$

Unlike $D_n$, $F_n$ is non-negative:

$$F_n(x-y)\ge 0(\mathrm{\forall }x,y).$$

The following figure shows the Fejér kernels for $n=0,1,5,20$.

Lemma (Uniform convergence of ${\sigma }_n[f]$)

If $f\in C_{2\pi }^0$, then ${\sigma }_n[f]$ converges uniformly to $f$. 

Proof. Using (eq:Fnormal), we have

$$f(x)={\int }_{-\pi }^{\pi }{F}_n(x-y)f(x)dy.$$

Therefore,

$$\begin{array}{}\text{(eq:fms)}& f(x)-{\sigma }_n(x)={\int }_{-\pi }^{\pi }{F}_n(x-y)\{f(x)-f(y)\}dy\end{array}$$

where the range of integration could be any one period.

For any $\epsilon >0$, there exists a $\delta >0$ such that 

$$|x-y|<\delta ⟹|f(x)-f(y)|<\epsilon$$

as $f$ is uniformly continuous. Fixing this $\delta$, we split the right-hand side of (eq:fms) into two parts:

$$f(x)-{\sigma }_n(x)=I_1+I_2$$

where

$$\begin{array}{rcl}{I}_1& =& {\int }_{|y-x|<\delta }{F}_n(x-y)\{f(x)-f(y)\}dy,\\ I_2& =& {\int }_{\delta \le |y-x|\le \pi }{F}_n(x-y)\{f(x)-f(y)\}dy.\end{array}$$

Since $F_n$ is non-negative and normalized (eq:Fnormal),

$$|I_1|\le \epsilon {\int }_{|y-x|<\delta }{F}_n(x-y)dy\le \epsilon {\int }_{|y-x|<\pi }{F}_n(x-y)dy=\epsilon .$$

Next, if $|x-y|\ge \delta$, noting that

$$F_n(x-y)\le \frac{1}{2(n+1)\pi }\cdot \frac{1}{{\sin }^2\frac{\delta }{2}}$$

and letting $M=\underset{|y|\le \pi }{max}|f(y)|$,

$$\begin{array}{rcl}|I_2|& \le & {\int }_{\delta \le |y-x|\le \pi }\frac{1}{2(n+1)\pi }\cdot \frac{1}{{\sin }^2\frac{\delta }{2}}\cdot 2Mdy\\ & \le & \frac{2M}{(n+1){\sin }^2\frac{\delta }{2}}.\end{array}$$

Thus, for a sufficiently large $N_1$, we have

$$n\ge N_1⟹|I_2|<\epsilon .$$

Putting these together, we have

$$|f(x)-{\sigma }_n(x)|<2\epsilon (n\ge N_1).$$

Here, $N_1$ does not depend on $x$, which proves that ${\sigma }_n$ uniformly converges to $f$ with respect to $x$. ■