Assume that $f\in C^{\infty}$ and that $M\_n$ is a sequence such that $$\sum\_{n=0}^{\infty}\frac{M\_n}{(n+1)M\_{n+1}}=\infty$$


and for certain compact neighborhood of the origin $U$ of $\mathbb{R}$, there is a constant $A$ such that for every $x\in U$, and $n\in\mathbb{N}$, \begin{equation}|g^{(n)}(x)|\leq n!A^nM\_n,\end{equation}


where $g(x):=f(x)+f(2x)$. We say that $g$ is [*quasianalytic*](http://en.wikipedia.org/wiki/Quasi-analytic_function) or that it belongs to the Denjoy-Carleman quasianalytic class $C\_M(U)$.


Does it follow that $f\in C\_M(V)$ for certain compact neighborhood of the origin $V$, i.e. $$|f^{(n)}(x)|\leq n!B^nM\_n,$$ for certain constant $B$ and any $x\in V$, and $n\in\mathbb{N}$?


**Note:** For example, if $\forall n,\ M\_n=1$ then the condition on $g$ implies it is analytic. Then it follows $f$ is also analytic and therefore $f\in C\_M(V)$ for some $V$and some $B$.



If $f(x)+f(2x)$ is quasianalytic, is $f(x)$ necessarily quasianalytic?
Assuming $V:=U=(-\alpha,+\alpha)$, we want the sequence of real numbers
$$B\_n:=\bigg( \frac{\| f^{(n)} \|\_{\infty,U}} {n!M\_n} \bigg) ^{1/n}$$
to be bounded (in which case $B:=\sup\_{n\in\mathbb{N}}B\_n$ is the best constant for the quasianalytic bounds to hold for $f$ on $U$ ). 


Indeed, since $f(x)=g(x/2)-f(x/2)$, differentiating $n$ times we get
$B^n\_n\le2^{-n}\big( A^n+B^n\_n\big)$, so $B\_n\le A/2\big(1+o(1)\big)$.


