What does $f$ continue on $\mathbb S^1$ mean?

Question

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 ?

Answer 1

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.)

Answer 2

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.