Periodic Function is Continuous iff Mapping from Unit Circle is Continuous

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

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 $f': \SS \to \R$ be the well-defined real-valued function such that:
 * $f = f' \circ p$

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

Then $f$ is continuous $f'$ is continuous.

Proof
The existence and well-definedness of $f'$ are demonstrated in Periodic Function as Mapping from Unit Circle.

Necessary Condition
Let $f'$ be continuous.

We have that Real Sine Function is Continuous and Cosine Function is Continuous.

Hence from Continuity of Composite with Inclusion: Inclusion on Mapping and Continuous Mapping to Product Space it follows that $p$ is continuous.

Hence $f = f \circ p$ is continuous.

Sufficient Condition

 * Continuity-of-Periodic-Function.png

Let $f$ be continuous.

Let $U \subseteq \SS$.

Let $p^{-1} \sqbrk U$ be open in $\R$.

Let $b \in U$.

Then:
 * $\exists a \in p^{-1} \sqbrk U: b = \tuple {\cos a, \sin a}$

Because $p^{-1} \sqbrk U$ be open:
 * $\exists \delta \in \R_{>0}: \map {B_\delta} a \subseteq p^{-1} \sqbrk U$

Let us assume that $\delta < \pi$.

Thus:
 * $b \in p \sqbrk {\map {B_\delta} a} \subseteq U$

But we have that:
 * $p \sqbrk {\map {B_\delta} a} = S \cap \map {B_\delta} b$

as can be seen from the diagram on the right

It follows by calculation that:
 * $\epsilon = 2 \sin \dfrac \delta 2$

Thus $p \sqbrk {\map {B_\delta} a}$ is an open ball in $\SS$ which has a center at $b$.

Hence it follows that $U$ is open in $\SS$.

We have asserted that $f$ is continuous.

Let $W$ be open in $\RR$.

Then $p^{-1} \sqbrk {f'^{-1} \sqbrk W} = f^{-1} \sqbrk W$ is open in $\RR$.

Hence $f'^{-1} \sqbrk W$ is open in $\SS$.

Thus it follows that $f'$ is continuous.