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Let $f : \{\mathbb{R}^2 - (1, 0) - (-1, 0)\} \to \mathbb{R}^2$ be the function $f(z) = \frac{1}{z-1} - \frac{1}{\bar{z}+1}$. Show that $f$ extends to a smooth map $\tilde{f} : S^2 \to S^2$.

We define $\hat{f}(\infty) = 0, \hat{f}(\pm 1,0) = \infty$. And we show \begin{align*} \lim_{z \to \infty} f(z) & = \frac{1}{z-1} - \frac{1}{\bar{z}+1} = 0.\\ \lim_{z \to \pm 1 + 0i} f(z) & = \frac{1}{z-1} - \frac{1}{\bar{z}+1} = \infty. \end{align*} Hence, $f$ extends smoothly into $\hat{f}$.

Thanks~very much!

1LiterTears
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2 Answers2

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Supposing what you did is correct**, you've shown it extends continuously, not that it extends smoothly. To prove that it extends smoothly you need to show that derivatives of all orders exist at the new points.

This matters. Consider the function $g:\mathbb{R} \setminus \{0\} \to \mathbb{R}$ given by $g(x)= |x|$. This is smooth, and extends uniquely to a continuous function $\widehat g : \mathbb{R} \to \mathbb{R}$ by setting $\widehat g(0) = 0$, and yet $\widehat g$ is not even differentiable (let alone smooth).

**I say this because your notation has confused me a bit. Does $(1,0)$ mean $1+0i$? Or are you identifying $S^2$ with $\mathbb{R}^2 \cup \infty$ by stereographic projection? Or what? Is $f$ even smooth?

Clive Newstead
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This is not analytic. The mapping $z\mapsto {\overline z}$ is not analytic. Smoothly? No.

ncmathsadist
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