TITLE: In what sense is a function on a circle the same as a $2 \pi$ periodic function on $\mathbb{R}$?
QUESTION [6 upvotes]: I was reading the appendix of Elias M Stein's Fourier Analysis and before proving the approximation lemma the author mentions the following 
Recall that a function on a circle is the same as a $2 \pi$ periodic function on $\mathbb{R}$
Could someone explain me what this exactly means and how are the functions are the same?

REPLY [7 votes]: Here's a more abstract explanation. This is really an example of the universal property of a quotient construction. The circle group
$$S^1 := \{ z \in \mathbb{C} \mid |z|=1 \}$$ 
can be identified as the quotient 
$S^1=\mathbb{R}/\mathbb{Z}$, 
where the quotient map is $\theta \mapsto e^{2\pi i\theta}$. In other words, for real numbers $r, r' \in \mathbb{R}$, we have an equivalence relation $r \sim r'$ if and only if $r-r' = n$ for some integer $n \in \mathbb{Z}$. 
By the universal property of a quotient, any function on $\mathbb{R}$ which is compatible with the equivalence relation (in the sense of producing the same output when the inputs are equivalent under the relation) corresponds bijectively to a function on the circle. In other words, any function with period 1 on $\mathbb{R}$ corresponds to a function on $S^1$.
But what about a function on $\mathbb{R}$ with period $T$? Well, we can just pick a different quotient map $\mathbb{R} \to S^1$, namely $\theta \mapsto \exp(\frac{2\pi i}{T} \theta)$. And again, we get a bijective correspondence betweens functions on $S^1$ and functions with period $T$ on $\mathbb{R}$.