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Let $\mathbb{R}^n$ be $n$-dimensional Euclidean space and $S^n$ be $n$-sphere.

Then, it is well-known that $S^n$ is the one-point compactification of $\mathbb{R}^n$.

Now consider $C^\infty(S^n)$, the Frechet space of smooth functions on $S^n$.

Then, my question is

Which function space on $\mathbb{R}^n$ is identified with $C^\infty(S^n)$?

According to this link, the Schwartz space $\mathcal{S}(\mathbb{R}^n)$ is identified with a subspace of $C^\infty(S^n)$ comprising functions whose derivatives all vanish at a fixed point of $S^n$.

So, my current question is sort of inverse to the above link. I suspect that the answer to be

Smooth functions on $\mathbb{R}^n$ with bounded derivatives.

However, I am not sure if this is right. Could anyone help me?

Keith
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    A necessary condition is for all order derivatives to have a limit at $\infty$ under a suitable change of coordinates but this is almost tautological – ronno Mar 07 '24 at 05:48
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    I’m not sure much more than ronno’s comment can be said. Bounded derivatives are definitely not enough. For example, $\sin(x)$ is smooth on $\mathbb{R}$ with all orders of derivatives bounded, but there’s obviously no way to extend it to the circle. – David Gao Mar 07 '24 at 06:18
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    All the derivatives having a limit at $\infty$ seems also sufficient for smooth extendability to $S^n$ via composing with inversion $x\mapsto |x|^{-2}x$ and using the multivariable version of https://math.stackexchange.com/questions/3301610/.... – P. P. Tuong Mar 07 '24 at 07:57

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