Collections of periodic functions

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,

$${\int }_0^{2\pi }|f(x)|dx<+\mathrm{\infty }.$$

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,

$${\int }_0^{2\pi }|f(x){|}^{2}dx<+\mathrm{\infty }.$$

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

$$\begin{array}{}\text{(eq:ineqR)}& {\int }_0^{2\pi }|f(x)|dx\le \sqrt{{\int }_0^{2\pi }|f(x){|}^{2}dx}\sqrt{{\int }_0^{2\pi }{1}^{2}dx}\end{array}$$

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 $t\in \mathbb{R}$, 

$${\int }_0^{2\pi }(t-|f(x)|{)}^{2}dx\ge 0.$$ □

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

$$B(p,q)={\int }_0^1{x}^{p-1}(1-x{)}^{q-1}dx$$

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

$${\int }_0^{\frac{\pi }{2}}{\sin }^a\theta {\cos }^b\theta d\theta =\frac{1}{2}B(\frac{a+1}{2},\frac{b+1}{2})$$

where $a,b>-1$.

Thus, (you should show that)

$${\int }_0^{2\pi }|f(x)|dx=4{\int }_0^{\frac{\pi }{2}}\frac{1}{(\sin x{)}^{\alpha }}dx=2B(\frac{1-\alpha }{2},\frac{1}{2})<+\mathrm{\infty }$$

if $\alpha <1$, and

$${\int }_0^{2\pi }|f(x){|}^{2}dx=4{\int }_0^{\frac{\pi }{2}}\frac{1}{(\sin x{)}^{2\alpha }}dx=2B(\frac{1-2\alpha }{2},\frac{1}{2})<+\mathrm{\infty }$$

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