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Let $\mathcal{S}(\mathbb{R}^n)$ be the Schwartz space on $\mathbb{R}^n$ and $C^\infty(S^n)$ be the space of smooth functions on $n$-sphere.

Now fix a point $x \in S^n$ and define \begin{equation} C^\infty_x(S^n) := \{ f \in C^\infty(S^n) \mid f(x)=0\} \end{equation}

Then I heard that $\mathcal{S}(\mathbb{R}^n)$ may be identified with $C^\infty_x(S^n)$. That is, a Schwartz function is one extending smoothly to $S^n$ while vanishing at infinity.

However, I wonder why we need such a strong decay condition of Schwartz functions to extend it smoothly to $S^n$?

Could anyone please clarify for me?

Keith
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1 Answers1

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You have heard of a wrong identification.

In his book Théorie des distributions, Laurent Schwartz mentions that by identifying $S^n$ with the one point compactification of $\mathbb R^n$, the space $\mathcal S(\mathbb R^n)$ may be identified with the space $\dot{\mathscr D}(S^n)$ of all scalar smooth functions on $S^n$ whose derivatives of any order all vanish at infinity (in the latest edition (Zbl 0962.46025), see chap. VII, § 3, th. II).

Identifying $x$ with the point at infinity, the space $\dot{\mathscr D}(S^n)$ is a lot smaller than $C^\infty_x(S^n)$ as there are many extensions of rational functions with no singularity in $\mathbb R^n$ that are in $C^\infty_x(S^n)$ but not in $\dot{\mathscr D}(S^n)$.

The identification between $\dot{\mathscr D}(S^n)$ and $\mathcal S(\mathbb R^n)$ is quite intuitive.

If $f\in\mathcal S(\mathbb R^n)$, then by composing with the inversion $y\mapsto\|y\|^{-2}y$, you obtain a smooth function $g:S^n\setminus\{0\}\to\mathbb C$ whose derivatives all decay at $0$ faster than any $\|\cdot\|^k$. So $g$ may be extended at $0$ into a smooth function whose derivatives all vanish at $0$ and by composing again with the inversion $y\mapsto\|y\|^{-2}y$, you can see that the continuous extension of $f$ to $S^n$ is smooth and all its derivatives vanish at infinity.

Conversely, if $f\in\dot{\mathscr D}(S^n)$, then by composing with the inversion $y\mapsto\|y\|^{-2}y$, you obtain a smooth function $g:S^n\to\mathbb C$ whose derivatives all vanish at $0$. The Taylor expansion of $g$ then tells you that the derivatives of $g$ all decay at $0$ faster than any $\|\cdot\|^k$ and by composing with the inversion $y\mapsto\|y\|^{-2}y$ again, you can see that the derivatives of $f$ all decay at infinity faster than any $\|\cdot\|^{-k}$.

P. P. Tuong
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  • Thank you. I guess the latest version as in your link has English translation? – Keith Mar 04 '24 at 16:13
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    @Keith Unfortunately, it does not and I couldn't find the result in any of the "classical" English analysis texts I have in my collection. – P. P. Tuong Mar 04 '24 at 17:02
  • By the way, what do youn mean by $S^n - {0}$ in your answer? Does $0$ denote a point on $S^n$ (for example the north pole)? – Keith Jul 08 '24 at 23:51
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    @Keith It's been a while, but I believe that in my head, $0$ was indeed a point on $S^n$ and was the point at the origin of $\mathbb R^n$ when its one point compactification is identified with $S^n$. – P. P. Tuong Sep 02 '24 at 09:29