# Find a smooth function $\eta:\mathbb{C}\to\mathbb{R}$ whose support is a disk.

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?

• Replace $x$ by $\vert x\vert$. – Chris Custer Jul 18 at 10:06
• You mean like $f(1-|x|)$ with $x\in\mathbb R^2$? – trisct Jul 18 at 10:11
• (1): yes. For (2), consider, say, the function $g(x)=f(1-|x|^2)$ for your $f$. – Mindlack Jul 18 at 10:12
• It seems I left out the square. – Chris Custer Jul 18 at 10:40
• You're looking for a [bump function][1]. [1]: en.m.wikipedia.org/wiki/Bump_function – 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$$\eta(z)=\begin{cases}e^{\frac1{\lvert z\rvert^2-1}}&\text{ if }\lvert z\rvert<1\\0&\text{ otherwise.}\end{cases}$$
• is it smooth at $0$? – trisct Jul 18 at 10:20