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I am trying to show that

there is a sequence $(P_{n})_{n}$ of polynomials such that $P'_{n}(0)=1$ for all $n$, $P'_{n}(z)\rightarrow0$ if $z \in \mathbb{C}^{\times}$ and $P_{n}(z)\rightarrow0$ if $z \in \mathbb{C}$

but I could not able to do that. I will appreciate for any help.

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  • $\begingroup$ Where does this extraordinary statement come from? $\endgroup$ – Georges Elencwajg Apr 10 '14 at 19:57
  • $\begingroup$ I do not know exactly where the question come from, a friend of mine ask me. $\endgroup$ – user135582 Apr 10 '14 at 20:04
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Let $K_n=\{0\}\cup\{z\in\Bbb C:\frac1n\le |z|\le n\}\setminus(\Bbb R^+\times(0,\frac1n))$. This is an annulus of inner radius $\frac1n$, outer radius $n$, with $0$ added and with a strip of width $\frac1n$ just above the positive real axis deleted. This peculiar sequence of sets was chosen because:

  • $\bigcup_{n\in\Bbb N}K_n=\Bbb C$
  • $K_n$ is compact
  • $\Bbb C\setminus K_n$ is connected
  • $K_n^\times$ is connected

Now define $f_n$ to be $0$ on a neighborhood of $K_n^\times$ and $1$ on a neighborhood of $0$. This is a holomorphic function, so by Runge's theorem, there is a polynomial $g_n$ that is within $\frac1{2n(n+1)}$ of $f_n$ on $K_n$, so letting $Q_n(x)=g_n(x)-g_n(0)+1$, $Q_n$ is a polynomial that is less than $\frac1{n(n+1)}$ on $K_n^\times$ and such that $Q_n(0)=1$. Thus $Q_n(x)\to0$ pointwise on $\Bbb C^\times$.

Now define $P_n(z)=\int_{-1}^zQ_n(t)\,dt$. Then for any $z\in K_n^\times$, $$|P_n(z)|\le|z+1|\sup_{t\in K_n^\times}|Q_n(t)|\le(n+1)\cdot\frac1{n(n+1)}=\frac1n,$$ so $P_n(z)\to0$ pointwise on $\Bbb C^\times$, and also $P_n'=Q_n$, hence the other properties.

But there is one piece missing, namely $P_n(0)\to0$. To do this, define $R_n(z)=\frac12(P_n(z)-P_n(-z))$. That way, the limits are preserved, the derivative is preserved, and $R_n(0)=0$ for each $n$.

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  • $\begingroup$ Exactly, How to find such a "holomorphic function $f_n$" ? $\endgroup$ – Omran Kouba Apr 11 '14 at 5:42
  • $\begingroup$ @Omran Since $K_n$ is a compact disconnected union of the pieces $\{0\}$ and $K_n^\times$, it can be separated by disjoint open sets $A,B$ around $\{0\}, K_n^\times$ respectively. The function $f_n$ doesn't have to be defined on $\Bbb C$, just an open set containing $K_n$, for Runge's theorem to apply. Thus the function is equal to $1$ on $A$ and $0$ on $B$, and constant functions are holomorphic, so $f_n$ is holomorphic on $A\cup B$. $\endgroup$ – Mario Carneiro Apr 12 '14 at 3:58
  • $\begingroup$ and you think that $\Bbb{C}\setminus K_n$ is connected ? $\endgroup$ – Omran Kouba Apr 12 '14 at 16:51
  • $\begingroup$ @Omran Yes. Obviously the interior and exterior of the annulus are themselves connected, and they are connected to each other by the strip above the positive real line. Note that the strip (which I wrote as a cartesian product) can also be written as $\{z\in\Bbb C:\Re z>0\wedge 0<\Im z<1/n\}$. $\endgroup$ – Mario Carneiro Apr 12 '14 at 22:41
  • $\begingroup$ Excellent Solution, see "Pointwise Limits of Analytic Functions", Kennith Davidson, Amer. Math. Monthly, Vol 90, No. 6 (1983). $\endgroup$ – Omran Kouba Apr 13 '14 at 10:21

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