Let $f$ be a function defined on some real interval $[a,b]$ . By a periodic extension of $f$ to the real line we mean a function $g$ such that

1. $g$ is defined on $\mathbb{R}$ except perhaps at points $a+n(b-a)$ , where $n\in\mathbb{Z}$ ;
2. $g(x)=f(x)$ for all $x\in (a,b)$ , and
3. $g(x+n(a-b))=g(x)$ for all $x\in (a,b)$ and all integers $n$ .

The best way to understand periodic extensions of a function is to look the graph of a periodic extension of a real-valued function. For example, let $f(x)=x$ be defined on $[-1,1]$ . The graph of $f$ looks like

Then the graph a periodic extension $g$ of $f$ may look like

or look like

Notice the two periodic extensions of $f$ are identical except at odd integer points on the $x$ -axis. The reason why we do not require $g$ to agree with $f$ on the end points of $[a,b]$ is because we do not know if $f(a)=f(b)$ . If they do not agree, requiring that $g=f$ on all of $[a,b]$ may result in points $a+n(b-a)$ getting mapped to two distinct values $f(a)$ and $f(b)$ , rendering $g$ not well-defined. In fact, if $f$ does not agree on its endpoints, no periodic extensions of $f$ are continuous.

Notice, also, that the domain of function $f$ does not have to be the entire closed interval $[a,b]$ . The domain of $f$ may very well be a subset $S\subseteq [a,b]$ . For example, $f(x)=x$ may be a function defined on the open interval $(-1,1)$ . The two graphs above are again graphs of periodic extensions of $f$ .

However, if $S$ is a proper subset of $[a,b]$ that is not the open interval $(a,b)$ , then the definition of a periodic extension needs to be modified: $g$ is a periodic extension of $f$ defined on $S\subseteq [a,b]$ if

1. $g$ is defined on a subset $T\subseteq \mathbb{R}$ except perhaps at points $a+n(b-a)$ , where $T=\lbrace x+n(a-b)\mid x\in S\rbrace$ and $n\in\mathbb{Z}$ ;
2. $g(x)=f(x)$ for all $x\in S-\lbrace a,b\rbrace$ , and
3. $g(x+n(a-b))=g(x)$ for all $x\in S-\lbrace a,b\rbrace$ and all integers $n$ .

For example, if $f(x)=x$ for all rational numbers $x\in [-1,1]$ , then a periodic extension of $f$ has its domain the set of all rational numbers except perhaps at are odd integers.

Remarks.

• Trigonometric functions defined on $\mathbb{R}$ are periodic extensions of the trigonometric functions defined for angles in the interval $[0,2\pi]$ .
• Suppose $f$ is defined either on a closed interval $[a,b]$ or an open interval $(a,b)$ , $f$ has a continuous periodic extension (defined on all of $\mathbb{R}$ ) iff $f$ is continuous and that
  1. either $f(a)=f(b)$ when $f$ is defined on a closed interval, or
  2. $f(a+)=f(b-)$ when $f$ is defined on an open interval, where $f(a+)$ is the one-sided limit approaching $a$ from the right, and $f(b-)$ is the one-sided limit approaching $b$ from the left.
  Simply define the periodic extension $g$ so that either $g(a)=f(a)$ , or $g(a)=f(a+)$ . For example, the following graph
  
  
  
  is the graph of the continuous periodic extension of a function $f_1$ given by $f_1(x)=|x|$ defined on $(-1,1)$ , or a function $f_2$ defined on $(0,2)$ , given by $f_2(x)=x$ for $0<x\le 1$ and $f_2(x)=2-x$ for $1\le x<2$ . With $f_1$ , we see that $f_1(-1+)=f_1(1-)=1$ , while with $f_2$ , we have $f_2(0+)=f_2(2-)=0$ .

  It is easy to see that if a continuous periodic extension of a function exists, then it is unique.

• Higher dimensional periodic extensions may also be defined for functions defined on a parallelepiped ($n$ -dimensional analog of a parallelogram). A periodic extension $g$ of a function $f$ defined on a parallelepiped is a function such that its projection $p_i(g)$ onto axis $i$ in $\mathbb{R}^n$ is a periodic extension of the projection $p_i(f)$ of $f$ onto axis $i$ .

Bibliography

1

G.P. Tolstov, Fourier Series, Prentice-Hall, 1962.