Complex polynomials converging in the compact-uniform topology Suppose that $p_n$ are polynomials with degree $n$, with $p_n(0)=1$ and which converges uniformly on compact sets of $\mathbb{D}$ towards an analytic function $f$.
Now, suppose that the $p_n$ have no roots in a disk of radius strictly greater than 1, say - for simplicity - $D(0,2)$.
Is it true that one can extend the uc convergence to all compacts of the disk $D(0,2)$ ?  Or in any intermediary disk $D(0, r)$ for some $1<r<2$ ? What would be simple assumptions on the $p_n$ to ensure such a behaviour (on the coefficients, the roots, on $f$...) ?
Answer. As answered above by Sangchul, the answer to the first question is no. He gave a counterexample.
Edit: a possible other formulation. By taking the inverses $z \to p_n(z)^{-1}$, whose radii of convergence are greater than $2$,  and using the Hurwitz theorem, the question becomes: we have a sequence of rational functions with no poles in $D(0, 2)$ which takes the value $1$ at zero, and which converges uniformly on compact sets of $D(0, 1)$ towards an analytic function $g=f^{-1}$. The radius of convergence $R$ of $g$ is greater than or equal to $1$. Is it possible that $R=1$ ? If not, $R>1$; then, does the uniform convergence of $1/p_n$ towards $g$ holds (uniformly on compact sets) for some disk $D(0, s)$ with $1<s<R$ ?
 A: In this answer, we will construct a counter-example. First, note that it suffices to find a sequence $(q_k)_{k\geq 1}$ of polynomials satisfying:

*

*$0 < \deg q_1 < \deg q_2 < \deg q_3 < \dots $;

*$q_k(0) = 1$;

*Each $q_k$ has no roots in $D(0, 2)$;

*$q_k$ converges uniformly on compact subsets of $\mathbb{D}$;

*$q_k(1)$ does not converge.

Then we can construct $(p_n)_{n\geq 1}$ by letting $p_n(z) = q_k(z) + \epsilon_n z^n$ for $\deg q_k \leq n < \deg q_{k+1}$ and for some $\epsilon_n \geq 0$ chosen sufficiently small.
Construction. For each $k \geq 1$, choose $q_k(z)$ as the Taylor polynomial of $e^{z+z^2+\dots+z^k}$ with degree large enough so that
$$ \sup_{z \in D(0,2)} \left| q_k(z) - e^{z+z^2+\dots+z^k} \right| < e^{-2^{k+1}} $$
as well as $\deg q_k > \deg q_{k-1}$ if $k \geq 2$. Then for $z \in D(0, 2)$,
$$ \left| q_k(z) \right| > \left| e^{z+z^2+\dots+z^k} \right| - e^{-2^{k+1}} \geq e^{-2-2^2-\dots-2^k} - e^{-2^{k+1}} > 0, $$
and so, $q_k$ has on zero on $D(0, 2)$. Moreover, since $e^{z+z^2+\dots+z^k}$ converges to $e^{z/(1-z)}$ uniformly on compact subsets of $\mathbb{D}$, the same is true for $q_k$. Finally, since $\left| q_k(1) - e^{k} \right| < e^{-2^{k+1}}$, it follows that $q_k(1)$ diverges as $k\to\infty$.
