Let $g,f$ be real-valued functions defined on the real line. Let $s$ be a real number. Assuming that $g,f$ are both in the Gevrey class $G^{s}$, it is true that $g\circ f$ belongs to $G^{s}$ if $s\ge 1$. Here we define $G^{s}$ as the class $h$ of smooth functions on $\mathbb R$ such that for all $R>0$, there exists $\rho_R$ so that $$ \sup_{\vert x\vert\le R, k\in \mathbb N}\vert h^{(k)}(x)\vert (k!)^{-s} \rho_R^k<+\infty. $$ Note that $G^1$ stands for analytic functions. A sketch of the proof goes as follows. We have for $I, J$ open subsets of $\mathbb R$, $f: I\rightarrow J$, $g: J\rightarrow \mathbb R$, smooth functions, $k\in \mathbb N^{*}$, the Faà de Bruno formula $$ \frac{(g\circ f)^{(k)}}{k!}=\sum_{1\le r\le k}\frac{g^{(r)}\circ f}{r!} \sum_{\substack{(k_{1},\dots, k_{r})\in {(\mathbb N^{*})}^{r}\\k_1+\dots+k_r=k}}\prod_{1\le j\le r}\frac{f^{(k_{j})}}{k_{j}!}. \tag{$\ast$}$$ From this, we get for $K$ compact set, $L=f(K)$ (also a compact set), \begin{multline} \sup_{K} {\vert{(g\circ f)^{(k)}}\vert}\\ \le (k!)^{s}\rho_{K, f}^{-k}\sigma_{L, g} \sum_{1\le r\le k} { \bigl({\rho_{L, g}^{-1} \sigma_{K,f}\bigr)^{r} (r!)^{s-1}} } \sum_{\substack{(k_{1},\dots, k_{r})\in {(\mathbb N^{*})}^{r}\\k_{1}+\dots+k_{r}=k}} \Bigl(\frac{k_{1}!\dots k_{r}!}{k!}\Bigr)^{s-1}. \end{multline} We can prove that the number of terms in the sum over $(k_{1}, \dots, k_{r})$ above is $$ \binom{k-r+r-1}{r-1}=\binom{k-1}{r-1}, $$ and that for $(k_{1},\dots, k_{r})\in {(\mathbb N^{*})}^{r}$ such that $k_{1}+\dots+k_{r}=k$, we have the inequality
$$ r!\le \frac{k!}{k_{1}!\dots k_{r}!}, $$ and this entails the composition algebra result for $s\ge 1$.

Now a new version of my question: I believe that this result is not true for $s<1$, but I do not see a simple counterexample: are there some "explicit" $g,f$ in some $G^s$ for $s<1$ so that $g\circ f$ does not belong to $G^s$?

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    $\begingroup$ There is a result (Cartan?) on Carleman-Denjoy classes saying that they are stable under composition if the sequence $(M_n/n!)^{1/n}$ is increasing. Letting M_n = (n!)^{s}, this should be true. See this paper. Also Gevrey himself seems to have proved that. $\endgroup$
    – CPJ
    Oct 9, 2015 at 16:14
  • $\begingroup$ If I remember correctly, the answer in general is no. However if you compose a Gevrey function of order s'>s with a Gevrey function of order s you get a Gevrey function of order s. $\endgroup$ Oct 11, 2015 at 20:31
  • $\begingroup$ @Piero D'Ancona You mean $s'<s$ since the Gevrey space $G^{(s)}$ with the notation above is increasing with $s$, e.g. analytic is $G^{(1)}$ is included in $G^{(2)}$ which contains compactly supported functions. On the other hand, I believe that the answer to the question is positive and is a matter of writing a precise Faà de Bruno formula. $\endgroup$
    – Bazin
    Oct 13, 2015 at 7:23
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    $\begingroup$ @CPJ Thanks for your comment. In fact with $M_n=n^{ns}$, we get $(M_n/n^n)^{1/n}=n^{s-1}$ which is increasing for $s\ge 1$. This suggests that the composition algebra property holds for $s\ge 1$ but not for $s<1$. $\endgroup$
    – Bazin
    Oct 14, 2015 at 14:32
  • $\begingroup$ @CPJ thanks for the reference to the original work of Gervey. $\endgroup$
    – BigM
    Dec 23, 2018 at 3:58

1 Answer 1


This is probably not relevant for the OP anymore, but as I ran into this problem myself as well, I reckoned giving an answer for future reference might be of use for someone else some other time.

Let's take $f : \mathbb{R} \to \mathbb{R}$, $f(x) = e^x$. We will show $f \in G^0$, but $f \circ f \not\in G^0$.

Following OP's definition of Gevrey-class $G^{\sigma}$, we easily find that $f \in G^0$. Indeed, take $R > 0$ arbitrarily, and take $\rho_R = e^{-R}$. As all derivatives of $f$ are equal to $f$, we find: $$\sup_{|x| \leq R, k \in \mathbb{N}} |f^{(k)}(x)| \cdot \rho_R^k = \sup_{|x| \leq R, k \in \mathbb{N}} e^x \cdot e^{-kR} \leq \sup_{k \in \mathbb{N}} e^{-(k-1)R} < \infty$$ Now we prove that $f \circ f \not\in G^0$. To that end, suppose that it is; we will derive a contradiction. For any $R > 0$ we have: $$L_R := \sup_{|x| \leq R, k \in \mathbb{N}} |(f \circ f)^{(k)}(x)| \cdot \rho_R^k < \infty$$ Obviously, this implies, for all $k \in \mathbb{N}$: $$|(f \circ f)^{(k)}(0)| \leq L_R\rho_R^{-k}$$ And hence: $$\log|(f \circ f)^{(k)}(0)| \leq \log(L_R) - k\log(\rho_R) = \log(L_R) + k\log(1/\rho_R)$$ But we can explicitly write out $(f \circ f)^{(k)}(0)$ (e.g. using the combinatorial form of Faà di Bruno's formula, on wikipedia), to find that for all $k \in \mathbb{N}$: $$(f \circ f)^{(k)}(0) = |\Pi_k| = B_k$$ where $\Pi_k$ is the set of partitions of $\{1, \dots, k\}$, and $(B_k)_{k \in \mathbb{N}}$ are the Bell numbers. Several approximation formula's to $B_k$ are available, but we will be using a weaker version of a result from de Bruijn (1981), which is displayed on the wikipedia page as well: $$\log |(f \circ f)^{(k)}(0)| = \log(B_k) = k\log(k) + \mathcal{O}(\log\log(k))$$ But it is apparent that $k\log(k)$ grows more quickly than $k\log(1/\rho_R)$, and hence we reach a contradiction.

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    $\begingroup$ You can do it in a bit cheaper way if you notice that for $s<1$ any function in $G_s$ is entire of a certain order of growth and that that order can be read from the Taylor series coefficients at $0$. In particular, $G_0$ allows only exponential growth while it is fairly clear that $e^{e^z}$ goes up a bit faster. $\endgroup$
    – fedja
    May 27, 2018 at 1:51

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