5
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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

$$x^2+y^2=r^2$$

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$\pi$

$\endgroup$
  • $\begingroup$ I haven't read the book but i think the statement describes the above. $\endgroup$ – Jasser Sep 14 '14 at 12:36
  • 1
    $\begingroup$ 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
8
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ What do you mean by $S^1$ $\endgroup$ – user3503589 Sep 14 '14 at 12:57
  • 4
    $\begingroup$ The unit circle $S^1 = \{z \in \Bbb{C} \mid |z|=1\}$. $\endgroup$ – PhoemueX Sep 14 '14 at 13:47
4
$\begingroup$

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 $\mathbb{R}$.

$\endgroup$
  • 1
    $\begingroup$ Wow almost two years later. Thank you. :) $\endgroup$ – user3503589 Nov 6 '16 at 1:14
3
$\begingroup$

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 $\theta$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.