Kernels arising from the Fourier series

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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 \(n \in \mathbb{N}\), \(\{x_1, x_2, \cdots, x_n\} \subset X\), \(\{c_1, c_2, \cdots, c_n\} \subset \mathbb{R}\), \[\sum_{i=1}^{n}\sum_{j=1}^{n}c_ic_jK(x_i,x_j) \geq 0.\]

We now consider kernel representations of the Fourier series.

The Dirichlet kernel

Consider the partial sum \(S_n(x)\) of \(S[f]\):
\[\begin{eqnarray*} S_n(x) &=& \sum_{|k|\leq n}c_ke^{ikx}\\ &=& \sum_{|k|\leq n}\left(\frac{1}{2\pi}\int_{-\pi}^{\pi}f(y)e^{-iky}dy\right)e^{ikx}\\ &=& \int_{-\pi}^{\pi}\frac{1}{2\pi}\sum_{k=-n}^{n}e^{ik(x-y)}f(y)dy. \end{eqnarray*}\]
Let us define \(D_n(\theta) ~ (\theta \in \mathbb{R})\) by
\[D_n(\theta) = \frac{1}{2\pi}\sum_{k=-n}^{n}e^{ik\theta}.\tag{eq:Ddef}\]
Noting
\[\begin{eqnarray*} \sum_{k=-n}^{n}e^{ik\theta} & = & \frac{e^{-in\theta} - e^{i(n+1)\theta}}{1 - e^{i\theta}}\\ &=& \frac{e^{i(n+1/2)\theta} - e^{-i(n+1/2)\theta}}{e^{i\theta/2} - e^{-i\theta/2}}\\ &=& \frac{\sin\left(n + \frac{1}{2}\right)\theta}{\sin\frac{\theta}{2}}, \end{eqnarray*}\]
we have
\[S_n[f](x) = S_n(x) = \int_{-\pi}^{\pi}D_n(x-y)f(y)dy\tag{eq:Dfconv}\]
where
\[D_n(x-y) = \frac{1}{2\pi}\frac{\sin\left(n + \frac{1}{2}\right)(x-y)}{\sin\frac{x-y}{2}} ~ (n = 0, 1, \cdots)\]
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 \(n \neq 0\) 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 \(\{a_n\}_{n=0}^{\infty}\), 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
\[\sigma_n[f](x) = \sigma_n(x) = \frac{1}{n+1}\sum_{k=0}^{n}S_k(x) ~~~ (n = 0, 1, \cdots).\tag{eq:fejer}\]
The integral representation of \(\sigma_n(x)\) can be obtained as follows.
\[\begin{eqnarray*} \sigma_n(x) &=& \frac{1}{n+1}\sum_{k=0}^{n}\int_{-\pi}^{\pi}D_k(x-y)f(y)dy\\ &=& \int_{-\pi}^{\pi}\left(\frac{1}{n+1}\sum_{k=0}^{n}D_k(x-y)\right)f(y)dy \end{eqnarray*}\]
where \(D_n\) is the Dirichlet kernel. Noting
\[\begin{eqnarray*} \sum_{k=0}^{n}\sin\left(k + \frac{1}{2}\right)\theta &=& \frac{1}{\sin\frac{\theta}{2}}\sum_{k=0}^{n}\sin\left(k + \frac{1}{2}\right)\theta\cdot {\sin\frac{\theta}{2}}\\ &=& \frac{1}{2\sin\frac{\theta}{2}}\sum_{k=0}^{n}(\cos k\theta - \cos(k+1)\theta)\\ &=& \frac{1}{2\sin\frac{\theta}{2}}(1 - \cos(n+1)\theta)\\ &=& \frac{\sin^2\frac{n+1}{2}\theta}{\sin\frac{\theta}{2}}, \end{eqnarray*}\]
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
\[F_n(x-y) = \frac{1}{2(n+1)\pi}\frac{\sin^2\left(\frac{n+1}{2}(x-y)\right)}{\sin^2\frac{x-y}{2}} ~~~ (n = 0, 1,\cdots).\]
\(F_n(x-y)\) 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
\[\int_{-\pi}^{\pi}F_n(x-y)dy = 1.\tag{eq:Fnormal}\]
Unlike \(D_n\), \(F_n\) is non-negative:
\[F_n(x-y) \geq 0 ~~~ (\forall x,y).\]
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,
\[f(x) - \sigma_n(x) = \int_{-\pi}^{\pi}F_n(x-y)\{f(x) - f(y)\}dy \tag{eq:fms}\]
where the range of integration could be any one period.

For any \(\varepsilon > 0\), there exists a \(\delta > 0\) such that 
\[|x - y| < \delta \implies |f(x) - f(y)| < \varepsilon\]
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
\[\begin{eqnarray*} I_1 &=& \int_{|y - x|<\delta}F_n(x-y)\{f(x) - f(y)\}dy,\\ I_2 &=& \int_{\delta\leq |y - x|\leq \pi}F_n(x-y)\{f(x) - f(y)\}dy. \end{eqnarray*}\]
Since \(F_n\) is non-negative and normalized (eq:Fnormal),
\[|I_1| \leq \varepsilon\int_{|y-x|< \delta}F_n(x-y)dy \leq \varepsilon\int_{|y-x|< \pi}F_n(x-y)dy = \varepsilon.\]

Next, if \(|x-y|\geq \delta\), noting that
\[F_n(x-y)\leq \frac{1}{2(n+1)\pi}\cdot\frac{1}{\sin^2\frac{\delta}{2}}\]
and letting \(M = \max_{|y|\leq \pi}|f(y)|\),
\[\begin{eqnarray*} |I_2| &\leq &\int_{\delta\leq |y-x|\leq \pi}\frac{1}{2(n+1)\pi}\cdot\frac{1}{\sin^2\frac{\delta}{2}}\cdot 2Mdy\\ &\leq& \frac{2M}{(n+1)\sin^2\frac{\delta}{2}}. \end{eqnarray*}\]
Thus, for a sufficiently large \(N_1\), we have
\[n \geq N_1 \implies |I_2| < \varepsilon.\]
Putting these together, we have
\[|f(x) - \sigma_n(x)| < 2\varepsilon ~~~ (n \geq N_1).\]
Here, \(N_1\) does not depend on \(x\), which proves that \(\sigma_n\) uniformly converges to \(f\) with respect to \(x\). ■


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