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 $\mathbb R/\mathbb Z$.

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    $\begingroup$ To be clear, you aren't assuming your functions are continuous? $\endgroup$ – Eric Wofsey Jul 7 '16 at 7:07
  • $\begingroup$ @EricWofsey: no I don't. $\endgroup$ – MathBeginner Jul 7 '16 at 7:09
  • $\begingroup$ $tan(x/(2\pi))$ is unbounded but still defined on (0,1). You could define a function to equal tan(x/2pi) when $x\neq 0$ and something else when $x=0$. $\endgroup$ – mathematician Jul 7 '16 at 7:13
  • $\begingroup$ Eric Wofsey's very nice answer shows this need not be true. However, if $f$ is continuous, then since $S^{1}$ is compact, $f$ must have compact image. Since every compact subset of $\mathbb{R}$ is bounded, $f$ is bounded. Of course, you probably already knew this. :) $\endgroup$ – Alex Wertheim Jul 7 '16 at 7:20

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.

  • $\begingroup$ so $f(x)=\tan(\frac{\pi}{2}x)$ is not continuous on $\mathbb R/\mathbb Z$ ? $\endgroup$ – MathBeginner Jul 7 '16 at 7:33
  • $\begingroup$ Well, it's not even technically well-defined on $\mathbb{R/Z}$ (since we would need to have $0 = f(0) = f(1) "=" \infty$, but those do not agree. But once you fix that concern (e.g. define instead $f(x) = \tan(\pi x)$), then what is $f(1/2)$? $\endgroup$ – Simon Rose Jul 7 '16 at 7:35
  • $\begingroup$ Thank you for this nice answer ! I understand now. Just to be, $f(x)=\tan(\pi x)$ is well defined on $\mathbb R/\mathbb Z$, isn't it ? (even if it not continuous). $\endgroup$ – MathBeginner Jul 7 '16 at 7:37
  • $\begingroup$ It's almost well-defined. What you would have to do is define $$ f(x) = \begin{cases} \tan(\pi x) & x \neq 1/2 \\ c & x = 1/2 \end{cases} $$ for some choice of a real number $c$. Then it is defined on all of $\mathbb{R/Z}$, but not continuous. $\endgroup$ – Simon Rose Jul 7 '16 at 14:39

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 $f(p)=q$.

  • $\begingroup$ Great examples! Stereographic projection is a natural example which, to my chagrin, didn't immediately come to mind before more bizarre functions. +1 $\endgroup$ – Alex Wertheim Jul 7 '16 at 7:18

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