# If $f(x)+f(2x)$ is quasianalytic, is $f(x)$ necessarily quasianalytic?

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 $n\in\mathbb{N}$, $$|g^{(n)}(x)|\leq n!A^nM_n,$$

where $g(x):=f(x)+f(2x)$. We say that $g$ is quasianalytic 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 $B$.

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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 $B_n\le A/2\big(1+o(1)\big)$.