This comes from the proof of the following lemma in Jost's Compact Riemann Surfaces (Lemma 2.3.3).

Lemma 2.3.3 Every compact Riemann surface $\Sigma$ admits a conformal Riemann metric.

proof. ... For a disk $D\subset\mathbb C$ we choose a smooth function $\eta:\mathbb C\to\mathbb R$ with $$\eta>0\text{ on }D,\quad\eta=0\text{ on }\mathbb C\backslash D$$ ...

My questions:

(1) Does "smooth" here mean "infinitely differentiable as a $\mathbb R^2\to\mathbb R$ function (just to make sure)?

(2) How to guarantee the existence of such functions?

For (2) I know such functions must be smooth but non-analytic. The only example I know is $$f(x)=\left\{\begin{array}{lll}e^{-1/x}&,&x>0\\0&,&x\leq0\end{array}\right.$$ But how to generalize this to an $\mathbb R^2\to\mathbb R$ function?

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    $\begingroup$ Replace $x$ by $\vert x\vert$. $\endgroup$ – Chris Custer Jul 18 at 10:06
  • $\begingroup$ You mean like $f(1-|x|)$ with $x\in\mathbb R^2$? $\endgroup$ – trisct Jul 18 at 10:11
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    $\begingroup$ (1): yes. For (2), consider, say, the function $g(x)=f(1-|x|^2)$ for your $f$. $\endgroup$ – Mindlack Jul 18 at 10:12
  • $\begingroup$ It seems I left out the square. $\endgroup$ – Chris Custer Jul 18 at 10:40
  • $\begingroup$ You're looking for a [bump function][1]. [1]: en.m.wikipedia.org/wiki/Bump_function $\endgroup$ – Chris Custer Jul 18 at 10:49

$f(x)=e^{-\frac 1 {1-x}}$ for $x<1$ and $0$ for $x \geq 1$ defines a smooth function which is positive on $(-\infty,1)$ and $0$ outside it. So $f(\|x\|^{2})$ is a smooth function on $\mathbb R^{2}$ which is positive for $\|x\|<1$ and $0$ elsewhere. For any other disk in $\mathbb R^{2}$ use an appropriate affine transformation.


Define$$\eta(z)=\begin{cases}e^{\frac1{\lvert z\rvert^2-1}}&\text{ if }\lvert z\rvert<1\\0&\text{ otherwise.}\end{cases}$$

  • $\begingroup$ is it smooth at $0$? $\endgroup$ – trisct Jul 18 at 10:20
  • $\begingroup$ Now it is smooth. $\endgroup$ – José Carlos Santos Jul 18 at 10:23

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