Is a function $f:\mathbb R/\mathbb Z\to \mathbb R$ bounded ? Let $f:\mathbb R/\mathbb Z\to \mathbb R$ a function (1-periodic). Is such a function bounded ? (it's the fact that f is defined on the circle that disturb me). Indeed, for such a function (usually at least $L^1$), I often see in my course $\|f\|_{L^\infty }$, but I don't see why it would be well defined if $f$ is not bounded. 
For example, does $f(x)=\tan(\pi/2 x)$ is defined on $S^1$ ? I really have problem with this $\mathbb R/\mathbb Z$.
 A: There are certainly plenty of unbounded functions $f:\mathbb{R}/\mathbb{Z}\to\mathbb{R}$.  For instance, let $x_0,x_1,x_2,\ldots\in\mathbb{R}/\mathbb{Z}$ be any infinite sequence of distinct points.  Then you could define $f(x_n)=n$ and $f(x)=0$ if $x\neq x_n$ for all $n$.  This $f$ is unbounded, since its image is all of $\mathbb{N}$.
For a slightly less artificial example, you could take a stereographic projection: identify $\mathbb{R}/\mathbb{Z}$ with the unit circle in $\mathbb{R}^2$, and define $f(p)$ for $p\neq (0,1)$ as follows (you can then define $f(0,1)$ however you want).  Draw the line from $(0,1)$ to $p$.  This line will intersect the $x$-axis at a point $(q,0)$; define $f(p)=q$.  This map is in fact surjective, since for any $q\in\mathbb{R}$, the line from $(q,0)$ to $(0,1)$ will intersect the circle at another point $p$, and then $f(p)=q$.
A: As Eric Wofsey has said, such functions need not be bounded. But I think it is important to note as well that if the function is continuous, then it will be bounded. This follows from the fact that $\mathbb{R/Z}$ is compact, and so its image under a continuous map is also compact. Since compact subsets of $\mathbb{R}^N$ are closed and bounded...
Anyhow, his function is not continuous, and nor is the stereographic projection as described (at least, it is not continuous over all of $\mathbb{R/Z}$), so this is not a concern.
