Collections of periodic functions

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To study the Fourier series, it is convenient to define collections of functions with a period of \(2\pi\).



Definition (Collection of periodic functions)

We denote by \(\mathcal{P}_{2\pi}\) the collection (set) of all functions on \(\mathbb{R}\) that have the period of \(2\pi\) and that have at most finitely many discontinuous points in each period. For each \(f\in \mathcal{P}_{2\pi}\), its value may not be defined at its discontinuous points.

Definition (Collections of integrable functions)

We denote by \(\mathcal{R}_{2\pi}^1\) the collection of all \(f \in \mathcal{P}_{2\pi}\) such that \(f\) absolutely integrable on the period, that is,

\[\int_0^{2\pi}|f(x)|dx < +\infty.\]

We denote by \(\mathcal{R}_{2\pi}^2\) the collection of all \(f \in \mathcal{P}_{2\pi}\) such that \(f\) is square-integrable, that is,

\[\int_0^{2\pi}|f(x)|^2dx < +\infty.\]

In these definitions, the integral may be improper.

Remark. The \(\mathcal{R}\) in \(\mathcal{R}_{2\pi}^1\) and \(\mathcal{R}_{2\pi}^2\) stands for Riemann-integrable. In the language of the Lebesgue integral, these are called \(L^1\)-integrable and \(L^2\)-integrable functions, respectively (and the "\(L\)" stands for "Lebesgue.") □

The Cauchy-Schwarz inequality

\[\int_0^{2\pi}|f(x)|dx \leq \sqrt{\int_0^{2\pi}|f(x)|^2dx}\sqrt{\int_0^{2\pi}1^2dx}\tag{eq:ineqR}\]

indicates that \(f\in \mathcal{R}_{2\pi}^2\) implies \(f\in \mathcal{R}_{2\pi}^1\) (i.e., \(\mathcal{R}_{2\pi}^2 \subset \mathcal{R}_{2\pi}^1\)).

Exercise. Prove the Cauchy-Schwarz inequality (eq:ineqR). Hint: For any \(t\in \mathbb{R}\), 

\[\int_0^{2\pi}(t - |f(x)|)^2\,dx \geq 0.\] □


Example. Consider

\[f(x) = \frac{1}{|\sin x|^\alpha} ~~~ (\alpha > 0).\]

If \(0 < \alpha < 1\), then \(f \in \mathcal{R}_{2\pi}^{1}\). If \(0 < \alpha < \frac{1}{2}\), then \(f \in \mathcal{R}_{2\pi}^{2}.\) Let's prove this. 

First, note that by symmetry (draw the graph!), we have

\[\int_{0}^{2\pi}\frac{1}{|\sin x|^\alpha}dx = 4\int_0^{\frac{\pi}{2}}\frac{1}{(\sin x)^\alpha}dx.\]

We can express this function in terms of the beta function. The beta function \(B(p,q)\)  is defined for \(p, q > 0\) as

\[B(p, q) = \int_0^1x^{p-1}(1-x)^{q-1}\,dx\]

In general, we can (and you should) show that 

\[\int_0^{\frac{\pi}{2}}\sin^a\theta\cos^b\theta\,d\theta = \frac{1}{2}B\left(\frac{a+1}{2},\frac{b + 1}{2}\right)\]

where \(a, b > -1\).

Thus, (you should show that)

\[\int_0^{2\pi}|f(x)|\,dx = 4\int_0^{\frac{\pi}{2}}\frac{1}{(\sin x)^\alpha}dx = 2B\left(\frac{1-\alpha}{2}, \frac{1}{2}\right) < +\infty\]

if \(\alpha < 1\), and

\[\int_0^{2\pi}|f(x)|^2\,dx = 4\int_0^{\frac{\pi}{2}}\frac{1}{(\sin x)^{2\alpha}}dx = 2B\left(\frac{1-2\alpha}{2}, \frac{1}{2}\right) < +\infty\]

if \(\alpha < \frac{1}{2}\). Otherwise, these integrals diverge. □

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