Definition.
A function \(f: [a,b] \to \mathbb{R}\) is called piecewise continuous if it is continuous on \([a,b]\) except possibly at finitely many points \(x_1,x_2,\ldots,x_k\). At each of these points, we require that the left- and right-hand limits of \(f\) exist. If one these points happens to be an endpoint of the interval, we only require that the appropriate one-sided limit exists.
We will write \(PC(a,b)\) for the set of piecewise continuous functions on \([a,b]\).
Definition.
A function \(f: [a,b] \to \mathbb{R}\) is called piecewise smooth if \(f\) and its first derivative \(f'\) are piecewise continuous on \([a,b]\).
We will write \(PS(a,b)\) for the set of piecewise smooth functions on \([a,b]\).
Return to a \(2\pi\)-periodic and integrable function \(f: \mathbb{R} \to \mathbb{R}\).
Recall the \(ab\)- and \(c\)-versions of its Fourier series are \[ \frac{a_0}{2} + \sum_{n=1}^\infty \left( a_n \cos(n\theta) + b_n \sin(n\theta) \right) = \sum_{n=-\infty}^\infty c_n e^{i n \theta}. \]
The coefficients are given by \[ a_n = \frac{1}{\pi} \int_{-\pi}^\pi f(\psi) \cos(n\psi) \, d\psi, \quad b_n = \frac{1}{\pi} \int_{-\pi}^\pi f(\psi) \sin(n\psi) \, d\psi, \quad c_n = \frac{1}{2\pi} \int_{-\pi}^\pi f(\psi) e^{-i n \psi} \, d\psi. \] where we use \(\psi\) instead of \(\theta\) to avoid a conflict of notation later.
What should the partial sums of the Fourier series look like? It’s easy, for the \(ab\)-version. The \(N\)-th partial sum is \[ S_N^f(\theta) = \frac{a_0}{2} + \sum_{n=1}^N \left( a_n \cos(n\theta) + b_n \sin(n\theta) \right). \]
For the \(c\)-version, the \(N\)-th partial sum is \[ S_N^f(\theta) = \sum_{n=-N}^N c_n e^{i n \theta}. \]
To say that the Fourier series of \(f\) converges to \(f\) at a point \(\theta\) means that \[ \lim_{N \to \infty} S_N^f(\theta) = f(\theta). \] This is pointwise convergence.
Inserting the formula for \(c_n\) into the \(c\)-version of the partial sum, we have \[ S_N^f(\theta) = \frac{1}{2\pi} \sum_{n=-N}^N \int_{-\pi}^\pi f(\psi) e^{i(\theta-\psi) n } \, d\psi = \frac{1}{2\pi} \sum_{n=-N}^N \int_{-\pi}^\pi f(\psi) e^{i(\psi-\theta) n } \, d\psi. \] Note that we have switched the order of \(\theta\) and \(\psi\) in the exponent in the second step. This is permissible because the sum ranges from \(n=-N\) to \(n=N\).
Now change variables by setting \(\phi = \psi - \theta\): \[ S_N^f(\theta) = \frac{1}{2\pi} \sum_{n=-N}^N \int_{-\pi-\theta}^{\pi-\theta} f(\theta + \phi) e^{i\phi n } \, d\phi. \]
But the integrand of the function is \(2\pi\)-periodic in \(\phi\), so we can shift the limits of integration without changing the value of the integral. In particular, we can shift the limits back to \([-\pi, \pi]\): \[ S_N^f(\theta) = \frac{1}{2\pi} \sum_{n=-N}^N \int_{-\pi}^{\pi} f(\theta + \phi) e^{i\phi n } \, d\phi. \]
We can then write \[ S_N^f(\theta) = \int_{-\pi}^{\pi} f(\theta + \phi) D_N(\phi) \, d\phi, \] where \[ D_N(\phi) = \frac{1}{2\pi} \sum_{n=-N}^N e^{i n \phi} \] is called the \(N\)-th Dirichlet kernel.
Note that \[ \begin{align*} D_N(\phi) &= \frac{1}{2\pi} ( e^{i (-N) \phi} + e^{i (-N+1) \phi} + \cdots + e^{i (N-1) \phi} + e^{i N \phi} ) \\ &= \frac{1}{2\pi} e^{-iN\phi} ( 1 + e^{i \phi} + e^{2 i \phi} + \cdots + e^{2N i \phi} ) \\ &= \frac{1}{2\pi} e^{-iN\phi} \frac{e^{i (2N+1) \phi} - 1}{e^{i \phi} - 1} \\ &= \frac{1}{2\pi} \frac{e^{i (N+1) \phi} - e^{-iN\phi} }{e^{i \phi} - 1 }. \end{align*} \] since the expression in parentheses is a geometric series.
But, after multiplying top and bottom by \(e^{-i\phi/2}\), we have \[ D_N(\phi) = \frac{1}{2\pi} \frac{e^{i (N+1/2) \phi} - e^{-i(N+1/2)\phi} }{e^{i \phi/2} - e^{-i \phi/2} } = \frac{1}{2\pi} \frac{\sin((N+1/2)\phi)}{\sin(\phi/2)}. \]
This formula lends insight into what the Dirichlet kernel looks like.
Lemma.
For each \(N\), we have \[ \int_{-\pi}^0 D_N(\phi) \, d\phi = \int_0^\pi D_N(\phi) \, d\phi = \frac{1}{2}. \]
For the proof, first recall \[ D_N(\phi) = \frac{1}{2\pi} \sum_{n=-N}^N e^{i n \phi}. \]
But \(2\cos(n\phi) = e^{i n \phi} + e^{-i n \phi}\), so we can write \[ D_N(\phi) = \frac{1}{2\pi} + \frac{1}{\pi}\sum_{n=1}^N \cos(n\phi). \]
Then \[ \int_0^\pi D_N(\phi) \, d\phi = \left[ \frac{\phi}{2\pi} + \frac{1}{\pi}\sum_{n=1}^N \frac{\sin(n\phi)}{n} \right]_0^\pi = \frac{1}{2}, \] and similarly for the other integral.