Is there a sequence of polynomials $p_n $such that $p_n(0)=1$ for all $n$ but $\lim_{n \rightarrow \infty} p_n(z)=0$?

Question

I want to solve the following problem:

Is there a sequence of polynomials $p_n$, such that $p_n(0)=1$, $n \in \mathbb{N}$, but $\lim_{n \rightarrow \infty} p_n(z)=0$ for all $z \neq 0$.

As a hint I got: “Consider $K_n:=(\{ z \in \mathbb{C}:|z|\leq n \} \setminus \{ z \in \mathbb{C}: d(z,[0,n]) <\frac{1}{n}\}) \cup \{0\} \cup [\frac{1}{n},n]$, then show that $K_n $is compact and $\mathbb{C}\setminus K$ is connected. Then use Runge theorem to find a polynomial.”

My approach: Obviously $\{ \ z \in \mathbb{C}:|z|\leq n \}$ is compact, and $\{ z \in \mathbb{C}: d(z,[0,n]) <\frac{1}{n}\}$ is open. Now subtracting an open set from a compact set yields a compact set. Further $\{0\}$ and $\{[\frac{1}{n},n]\}$ are both compact. Thus $K_n$ is compact.

Now I am struggling to show that $\mathbb{C}\setminus K$ is connected. The only thing that comes to mind is to use the definition of connected sets. (A set $E$ is called connected, if it can not be writen as the union of two disjoint nonempty sets $A$ and $B$ which statisfy $A \cap E \neq \emptyset$ and $B \cap E \neq \emptyset$).

Further I am not sure on how to use Runge’s theorem to “find a polynomial”.

The version of Runge’s theorem that I know (and think may be useful) is: Let $\Omega$ be an open subset of $\mathbb{C}$. Then $\mathbb{\hat{C}}\setminus \Omega$ is connected $\iff$ all $f \in H(\Omega)$ can be approximated uniformly on compact sets in $\Omega$ by polynomials.

Answer 1

Edited again: Thanks to @stochasticboy321 for pointing out that I originally misread the question (and for providing a beautiful fix).

To show that $\mathbb C \setminus K_n$ is connected, it is helpful to use the following fact:

If $X$ is a topological space and $S_1$, $S_2$ are two connected subsets of $X$ such that $S_1 \cap S_2 \neq \emptyset$, then $S_1 \cup S_2$ is connected.

With this in mind, I suggest you try to write $\mathbb C \setminus K_n$ as the union of some simpler sets that are obviously connected, and are not disjoint?

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To apply Runge's theorem in a useful way, define $$ \Omega_n = \{ z \in \mathbb C : |z| < \tfrac 1 {3n} \} \cup \{ z \in \mathbb C : |z| > \tfrac 2 {3n} \} . $$

Now consider the holomorphic function $f_n : \Omega_n \to \mathbb C$, defined by: $$ f_n(z) = \begin{cases} 0 & \text{if } |z| < \tfrac 1 {3n} \\ 1/z & \text{if } |z| > \tfrac 2 {3n} \end{cases}. $$

Then Runge's theorem says that there exists a polynomial $\widetilde p_n$ such that $| \widetilde p_n(z) - f_n(z) | < 1/n$ for all $z \in K_n$.

Then consider $p_n(z) := 1 - z \widetilde p_n(z)$. Notice that $p_n(0) = 1$, and $|p_n(z)| < z/n$ for $z \in K_n \setminus \{ 0 \}$. You can go from there.

Note: that the version of Runge's theorem that I'm using is Theorem 13.7 in Rudin:

Let $K$ be a compact subset of $\mathbb C$ such that $\mathbb C \setminus K$ is connected, and let $\Omega$ be an open subset of $\mathbb C$ containing $K$. If $f : \Omega \to \mathbb C$ is a holomorphic function, then for all $\epsilon > 0$, there exists a polynomial $p$ such that $|p(z) - f(z)| < \epsilon$ for all $z \in K$.

The version you're using is a corollary of this.