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Let $\mathbb S^1:=\mathbb R/\mathbb Z$. What does $f$ continuous on $\mathbb S^1$ mean ? That it's continuous over $[0,1)$ or $[0,1]$ ? I would say $[0,1)$ but I have doubt since we sometimes take the norm $\|f\|_{L^\infty }$ and if $\lim_{x\to 1}f(x)$ doesn't exist, the norm $l^\infty $ wouldn't be defined. By the way, sometimes we prolonge $f$ to $\mathbb R$ by $1-$periodicity, and thus, $f$ must be continuous on each point of $\mathbb Z$, which is often not the case. What do you think ?

user349449
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2 Answers2

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You need to think of $\mathbb{S}^1$ as a topological space, and use the definition of continuous maps from topology.

Using the formula $\mathbb{S}^1 = \mathbb{R} / \mathbb{Z}$ you can use the quotient topology on $\mathbb{S}^1$. In this case you would have to prove that the periodic function on $\mathbb{R}$ that you get from $f$ is a continuous function on $\mathbb{R}$.

Alternatively, you could use the formula $\mathbb{S}^1 = \{(x,y) \in \mathbb{R}^2 \,\bigm|\, x^2 + y^2 = 1\}$ and use the subspace topology on $\mathbb{R}^1$ relative to $\mathbb{R}^2$.

There is a third option, where you can think of $\mathbb{S}^1$ as the quotient space obtained from the interval $[0,1]$ by identifying the points $0$ and $1$ to a single point. In this case a continuous function on $\mathbb{S}^1$ is the same as a continuous function on $[0,1]$ whose values at $0$ and at $1$ are equal to each other.

Lee Mosher
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  • so is it $[0,1)$ or $[0,1]$ ? – user349449 Jul 05 '16 at 20:14
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    It is neither $[0,1)$ nor $[0,1]$. It is $\mathbb{S}^1$. – Lee Mosher Jul 05 '16 at 20:14
  • But, when we write $\int_{S^1}f$, we write $\int_0^1 f$, is it false ? And on this space, $L^\infty $ norm is always well defined ? – user349449 Jul 05 '16 at 20:15
  • I added another paragraph to address your question. – Lee Mosher Jul 05 '16 at 20:18
  • Just to be sure, the function f(x)=x on $[0,1)$ extend on all $\mathbb R$ by $1-$ periodicity is not a function on the circle, is it? And what about $f(x)=\tan(\frac{\pi}{2}x)$ on $[0,1)$ extend by $1-$periodicity ? – user349449 Jul 05 '16 at 20:26
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    The first function, extended by 1-periodicity, is indeed a function on the circle...but not a continuous function. Same for the second one. But if you took $h(x) = \tan(\pi x)$, then you'd have a function on $S^1$ that's actually continuous on $S^1$. – John Hughes Jul 05 '16 at 20:58
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Since $f$ is defined on $\mathbb R / \mathbb Z$, you can think of a new function $g$ defined on all of $\mathbb R$ via $f$: simply define $g(x)$ to be $f(t)$, where $t$ is the fractional part of $x$.

Now we say that $f$ is continuous if this "lift" $g$ is a continuous function from the reals to the reals (and do the same for "differentiable", etc.)

John Hughes
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  • So is it $[0,1)$ or $[0,1]$ ? – user349449 Jul 05 '16 at 20:14
  • @MathBeginner Try $(-\epsilon, 1)$ instead – Hagen von Eitzen Jul 05 '16 at 20:15
  • To be honest, an element of $\mathbb R / \mathbb Z$ is a set of the form ${r + n \mid n \in \mathbb Z}$, called a "coset." Each such set contains a single element $x$ with $0 \le x < 1$; it also contains a single element $u$ with $0 < u \le 1$. So either $x$ or $u$ will serve as an identifier for the coset. For the coset whose $x$ value is $0$, the $u$-value is 1. I personally tend to favor the "x" definition: for a given element $r \in \mathbb R$, the associated $x$ value is $x = floor(r)$. And I'm saying to define $g(x) = f( C)$, where $C_x$ is the (unique) coset containing $x$. – John Hughes Jul 05 '16 at 20:38