# A sequence of polynomials that converges to $0$ pointwise except at $z=0$

This is an exercise from John Conway's book on complex analysis:

Investigate if there exists a sequence of polynomials $(P_n)$ that fulfills the conditions $P_{n}(0)=1$ for all natural numbers $n$ and $\lim_{n\rightarrow\infty}P_n(z)=0$ for all $z\neq0$

Polynomials obey the maximum principle, but I don't see how to apply it if all we know is point-wise convergence. (Uniform convergence would imply $|P_n|<\epsilon$ on the unit circle for large $n$, contradicting $P_n(0)=1$.)

• What happens if you apply the maximum moduus theorem to the poynomials? – copper.hat Jul 2 '12 at 17:25
• @copper.hat: I don't see how to apply it if all we know is point-wise convergence – user8268 Jul 2 '12 at 20:22
• It was a bad idea... – copper.hat Jul 2 '12 at 21:52

I think this can be done using Mergelyan's theorem: Let $K_n$ be a compact set of the form $\{0\}\cup \{z: 1/n\le |z|\le n, |\arg z-\theta_n|\ge \epsilon_n \}$. On this set you can approximate the function $\chi_{\{0\}}$ by a polynomial $P_n$ within $1/n$. Make sure to choose $\theta_n$ and $\epsilon_n$ so that no point of the plane belongs to $K_n^c$ for infinitely many $n$.
• I can see how to use this to get a sequence such that $p_n(z) \to 0$ for $z \neq 0$ and $p_n(0) \to 1$; but is there a way to ensure that $p_n(0)$ is exactly $1$ for all $n$? – ec92 Jul 7 '12 at 5:13
• @ec52 Divide by $p_n(0)$. – user31373 Jul 7 '12 at 11:11