Let $f$ be a function of a real variable expandable in power series on $\mathbb R$: there exists a sequence $(a\_n)\_{n\in\mathbb N}$ of reals such that for all $x\in\mathbb R$, one has
$$f(x)=\sum\_{n\ge0}a\_nx^n.$$


Let $(P\_n)\_{n\in\mathbb N}$ be a sequence of polynomials that converges uniformly towards $f$ on every compact of $\mathbb R$. One assumes that for all $n\in\mathbb N$, one has $P'\_n(0)=0$. Is $a\_1=0$ ?


If we would work in $\mathbb C$ instead of $\mathbb R$, it would be obvious, but in this setting, I do not know if it is true or not.


Thanks in advance for any hint.



derivatives and uniformly convergence
No. Start out with any entire $f$ with $f(0)=0$, $f'(0)\not= 0$. Then modify the (natural) approximation of $f$ by its Taylor polynomials $p\_n$ as follows. Let $\varphi\_n\in C(\mathbb R)$, $0\le \varphi\_n\le 1$, $\varphi\_n(0)=0$ and $\varphi\_n(x)=1$ for $|x|>1/n$. Then $\varphi\_n p\_n$ is still uniformly close to $f$ because we're only modifying close to zero where $f$ was small to start with. Now approximate $\varphi\_n$ uniformly by a polynomial $q\_n$ with $q\_n(0)=q'\_n(0)=0$. (It is convenient to make both $\varphi\_n$ and $q\_n$ even functions.) Then $P\_n\equiv q\_np\_n\to f$ locally uniformly and $P'\_n(0)=0$.


