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I want to show the following theorem in a lecture:

Let $F \in C^{\infty}(\mathbb{C}^{k}, \mathbb{C})$ such that $F(0)=0.$

Let $G: \mathbb{R}^n \rightarrow \mathbb{C}^{k}$, $x \mapsto (f_1(x),..,f_k(x))$ where $f_1,...,f_k \in \mathcal{S}(\mathbb{R}^n;\mathbb{C}).$ (the space of rapidly decaying functions.)

Then I want to show that $F \circ G: \mathbb{R}^n \rightarrow \mathbb{C}$ is a map in $ \mathcal{S}(\mathbb{R}^n;\mathbb{C}).$

Moreover, I want to argue then that in the standard semi-norm topology we have that the map $\Phi : \mathcal{S}(\mathbb{R}^n;\mathbb{C})^k \rightarrow \mathcal{S}(\mathbb{R}^n;\mathbb{C})$ , $(f_1,...,f_k) \mapsto F\circ (f_1,...,f_k)$is continuous.

The result is rather obvious and I also have a proof for this involving the chain-rule for arbitrary expressions of the form $\partial^{\alpha}(F\circ G)$, as they appear in the seminorms $p_{\alpha,\beta}((F\circ G))= \sup_{x}|x^{\beta} \partial^{\alpha}(F\circ G)(x)|$. But this chain-rule formula is a nightmare to write down which is why I started wondering whether there is a more direct argument why this holds avoiding the cumbersome calculation.

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    $\begingroup$ The chain rule isn't that complicated. See en.wikipedia.org/wiki/Faà_di_Bruno's_formula $\endgroup$ Apr 16 '16 at 22:28
  • $\begingroup$ @JohannesHahn in 1d, i agree $\endgroup$
    – Jonathan
    Apr 16 '16 at 23:18
  • $\begingroup$ I mean in every dimension. Faa di Bruno's formula is exactly the same for arbitrary dimensions, including for Banach-space-valued functions. The mistake is to use partial derivatives. Use the Frechet differential and everything looks exactly the same as in 1D. $\endgroup$ Apr 17 '16 at 13:26
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This outline may help. After multiplying $F$ by a smooth cutoff, you may assume $F$ and all of its partial derivatives are bounded in a neighborhood of the image of $G$ without changing the composition $F(G(x))$. This operation does not change the composition $F\circ \tilde{G}$ for $\tilde{G}$ that are $C^0$ close to $G$. Note the following Lemma:

Lemma: If $F$ is smooth and $G$ is a Schwartz function with $\| F \|_{C^k} + \| G \|_{C^k} \leq A$. Then there exists $C$ depending only on $A$ and $k$ such that $\| F(G(x)) \|_{C^k} \leq C$.

Then you can write $F(G(x)) = \sum_{i=1}^n \left( \int_0^1 \partial_i F( \sigma G(x) ) d\sigma \right) f^i(x)$. For each $i$, the expression in parentheses is a smooth function with all partial derivatives bounded (by the Lemma), so the product with $f^i(x)$ is Schwartz. Similarly, for Schwartz $\tilde{G}$ close to $G$, write $F(\tilde{G}(x)) - F(G(x)) = \sum_{i=1}^n \left( \int_0^1 \partial_i F(G(x) + \sigma ( \tilde{G} - G)) d\sigma \right) (\tilde{f}^i(x) - f^i(x) ) $. I haven't entirely thought this through, but I think that if $\tilde{G}$ ranges in a Schwartz neighborhood of $G$ then the $C^k$ norms of the expression in the parentheses may be bounded (using the Lemma). The resulting bounds on the Schwartz seminorms of the difference should give you continuity in the Schwartz topology. (Are you sure you need this continuity for your applications? In my experience it can usually be avoided...)

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  • $\begingroup$ thanks, I think this is completely sufficient and a much nicer approach. $\endgroup$
    – Jonathan
    Apr 16 '16 at 21:23

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