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.


closed as off-topic by Deane Yang, Will Jagy, S. Carnahan Jul 1 '14 at 5:01

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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 $P'_n(0)=0$.

  • $\begingroup$ Thanks for this very nice proof. But can you explain the difference with the complex case where it is working? $\endgroup$ – joaopa Jul 1 '14 at 2:52
  • 2
    $\begingroup$ These $P_n$'s will not approximate $f$ on $\mathbb C$. On $\mathbb R$, you can bend and twist polynomials at will, they can be close to anything that's continuous. On $\mathbb C$, you have much greater rigidity because the functions you can be close to are holomorphic; so local changes may have unintended effects somewhere else (don't know if that helps...). $\endgroup$ – Christian Remling Jul 1 '14 at 3:01

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