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?


Consider a circle of some radius r with center at origin (0,0). circle with radius r

Any circle can totally be described by two parameters in x and y in 2-D space.

so if you consider the opposite it is given by $$y=r*sin(\theta)$$

and if you consider the adjacent it is given by $$x=r*cos(\theta)$$

which implies


which is nothig but the equation of a circle.

Both of these trigonometric functions sin and cos have a period of $2 \pi$.

(the center of the circle is taken at origin simply for easy explanation)

So the functions here are y and x which has the same period as that of 2$\pi$

  • $\begingroup$ I haven't read the book but i think the statement describes the above. $\endgroup$ – Jasser Sep 14 '14 at 12:36
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    $\begingroup$ Thank you. It makes perfect sense. I just wrote my own solution just to get some Latex practice $\endgroup$ – user3503589 Sep 14 '14 at 12:58

What is meant here is that any $2\pi$-periodic function $f : \Bbb{R} \to \Bbb{C}$ yields a well-defined function

$$ g : S^1 \to \Bbb{C}, e^{i x} \mapsto f(x). $$

Note that this is well-defined, because $e^{i x } = e^{i y}$ implies $x-y \in 2\pi \Bbb{Z}$ and hence $f(x) = f(y)$, because $f$ is $2\pi$-periodic.

Conversely, any function $g : S^1 \to \Bbb{C}$ yields a $2\pi$-periodic function

$$ f : \Bbb{R} \to \Bbb{C}, x \mapsto g(e^{i x}). $$

Finally, you should verify that the two "transformations" $f \mapsto g$ and $g \mapsto f$ defined above are inverse to each other.

Hence, we have constructed a natural identification between both classes of functions.

  • $\begingroup$ What do you mean by $S^1$ $\endgroup$ – user3503589 Sep 14 '14 at 12:57
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    $\begingroup$ The unit circle $S^1 = \{z \in \Bbb{C} \mid |z|=1\}$. $\endgroup$ – PhoemueX Sep 14 '14 at 13:47

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 $\mathbb{R}$.

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    $\begingroup$ Wow almost two years later. Thank you. :) $\endgroup$ – user3503589 Nov 6 '16 at 1:14

Circle of radius 1 is given by parametric equations $ (x,y) = (\cos \theta, \sin \theta), \theta \in [0,2\pi[)$ so a function $f$ defined on the circle is a function $f(x,y) = f(\cos \theta, \sin \theta) = g(\theta)$, with $ g:[0,2\pi) \rightarrow R $ for certain uniquely determined $g$.

Periodicity follows from the fact that both $\sin $ and $\cos$ are periodic , so that $g(\theta + 2\pi) = g(\theta)$, for every $\theta$.


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