I am trying to understand how a radial extension of a continuous function on $S^1$ would be. Consider $f \in S^1$ such that $f$ is continuous. It can be extended radially to a function on $\mathbb{R}^2$ such that the function is constant along the radius. We need to remove $(0,0)$ from $\mathbb{R}^2$ for it to be well defined. However, I just don't get how this radial extension would look like.
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Convert to polar coordinates $(r, \theta)$ and define $g((r, \theta)) = f((1, \theta))$, then $g$ extends $f$ and is constant along each radius (here you can assume $r > 0$ because you have removed the origin from the domain of $f$). – Rob Arthan Sep 30 '21 at 20:06
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Here's a plot of the continuous function
$$(\cos(\theta), \sin(\theta)) \mapsto \frac{2 + \sin(3\theta)}{3}$$
on the unit circle (hopefully you can see that this map is well-defined!):
Now, let's look at the radial extension:
The radial extension also extends beyond the unit disc, though I don't show it here. Basically, the graph is extended via horizontal rays, beginning on the $z$-axis, passing through the original graph. Of course, the function cannot possibly be defined at $0$ in this way (unless the original function on the circle were a constant function).
In general terms, if we have a function $f$ from the unit sphere to $\Bbb{R}$, then we define $g$ from the whole space except $0$ to $\Bbb{R}$, defined by $$g(x) = f\left(\frac{x}{\|x\|}\right).$$
Theo Bendit
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Thank you! In this case the function g(x) would be constant along the radius, right? – User987931 Sep 30 '21 at 21:46
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Yes, $g$ is constant along the radial ray (which is like the radius of the circle, but not restricted just to being inside the circle). – Theo Bendit Sep 30 '21 at 21:47