Introductory university-level calculus, linear algebra, abstract algebra, probability, statistics, and stochastic processes.
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:
(symmetry) For any \(x, y\in X\), \[K(x,y) = K(y,x).\]
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
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
\(F_n(x-y)\) 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
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