Periodic Function as Mapping from Unit Circle

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Theorem

Let $\SS$ denote the unit circle whose center is at the origin of the Cartesian plane $\R^2$.

Let $p: \R \to \SS$ be the mapping defined as:

$\forall x \in \R: \map p x = \tuple {\cos x, \sin x}$


Let $f: \R \to \R$ be a periodic real function whose period is $2 \pi$.


Then there exists a well-defined real-valued function $f': \SS \to \R$ such that:

$f = f' \circ p$

where $f' \circ p$ denotes the composition of $f'$ with $p$.


Proof

Let $f': \SS \to \R$ be defined as:

$\forall \tuple {x, y} \in \SS: \map {f'} {x, y} = \map f x$


Consider the inverse $p^{-1}: \SS \to \R$ of $p$:

$\forall \tuple {x', y'} \in \SS: p^{-1} \sqbrk {x', y'} = \set {x \in \R: \cos x = x', \sin x = y'}$

Let $\RR$ be the equivalence relation on $\R$ induced by $p$:

$\forall \tuple {x, y} \in \R \times \R: \tuple {x, y} \in \RR \iff \map p x = \map p y$

That is:

\(\ds \forall \tuple {x, y} \in \R \times \R: \, \) \(\ds \tuple {x, y} \in \RR\) \(\iff\) \(\ds \map p x = \map p y\)
\(\ds \leadsto \ \ \) \(\ds \tuple {x, y} \in \RR\) \(\iff\) \(\ds \tuple {\cos x, \sin x} = \tuple {\cos y, \sin y}\)
\(\ds \leadsto \ \ \) \(\ds \tuple {x, y} \in \RR\) \(\iff\) \(\ds x = y + 2 k \pi\)


Let $f'$ be defined as:

$f' = f \circ p^{-1}$

Then by the Quotient Theorem for Sets:

$f'$ is well-defined if and only if $f$ is periodic with period $2 \pi$.


It follows from Conditions for Commutative Diagram on Quotient Mappings between Mappings that $f$ and $f'$ are related by the commutative diagram:

$\begin{xy} \xymatrix@L+2mu@+1em{ \R \ar[r]^*{p} \ar@{-->}[rd]_*{f := f' \circ p} & \SS \ar[d]^*{f'} \\ & \R }\end{xy}$

$\blacksquare$


Motivation

By this technique, the study of continuous periodic real functions on $\R$ can be reduced to the study of continuous real functions on the unit circle $\SS$.


Also see

Sources

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