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 $G^s$?