Let $K\_1 \subsetneq K\_2$ be two non-empty compact sets and let $D = (d\_n)\_{n \in \mathbb{N}}$ be a dense sequence on $K\_2\smallsetminus K\_1.$ Consider $f\_n : \mathbb{C}\smallsetminus K\_1 \rightarrow \mathbb{C}$ to be a sequence of analytic functions. Assume there exists an analytic function $f : \mathbb{C} \smallsetminus K\_2 \rightarrow \mathbb{C}$ such that $f\_n$ converges uniformly on compact sets of $\mathbb{C} \smallsetminus K\_2$ to $f$ and that $f\_n$ converges pointwisely on the dense set $D$, but $f\_n$ does not converge for the rest of points of its domain.


I would like to construct such an example or to prove that this is impossible, or at least to get a better understanding of this situation: it seems quite a strange behaviour for a sequence of analytic functions but I am not able to get a contradiction. If you drop some assumptions yo can get both results: if $K\_1$ is empty then $f\_n$ are entire functions and you can apply the maximum modulus principle to show that $f\_n$ must converges uniformly on the whole complex plane. If you just ask $f\_n$ to converges pointwisely on some dense sequence, you can get an infinite product with zeros on the sequence such that diverges on the rest of $\mathbb{C}.$


Can anyone help me?


Thank you very much.



Looking for a sequence of analytic functions with strange behaviour
Here is an attempt to construct an example.


I am going to let $K\_1$ and $K\_2$ be compact subsets in the sphere $\hat{\mathbb{C}}$, rather than the plane (of course, we can change coordinates to move infinity to a point outside of $K\_2$).


I will let $K\_1$ be the complement of the unit disc, and $K\_2$ the union of $K\_1$ with the interval $[0,1]$. Then $\hat{\mathbb{C}}\setminus K\_1$ is the slit disc $U = \mathbb{D}\setminus [0,1]$. Let $d\_n$ be an enumeration of the rationals (or any other countable dense subset) in $[0,1]$.


Let $g\_k:[0,1]\to [0,1]$ be a sequence of functions such that $g\_k(d\_n)=0$ for all $n$ and sufficienly large $k$, and such that for all $x\notin D$, $g\_k(x)$ does not converge. (The existence thereof is surely well-known, but for completeness I give a construction below.)


Now let $(U\_k)$ be an increasing sequence of simply-connected open subsets with $\overline{U\_k}\subset U$ whose union is $U$. Set $g\_k(z) = 0$ on $U\_k$.


Apply Arakelyan's approximation theorem to obtain a holomorphic function $f\_k$ on $\mathbb{D}$ such that $|g\_k(z)-f\_k(z)|\leq 1/k$ on $U\_k\cup [0,1)$. These functions converge locally uniformly to zero on $U$, they converge pointwise to zero on $D$, but they do not converge on any point outside $D$.


To construct the functions $g\_k$, we inductively choose closed intervals $I\_n^k$ for $k\geq n$ around $d\_n$, such that:


1. $d\_n\in I\_n^{k+1}\subset I\_n^k$ for all $n$ and all $k\geq n$;
2. $I\_n^k$ does not contain $d\_{k+1}$;
3. $I\_k^k$ is disjoint from $I\_n^{k-1}$, for $k\geq 1$ and $n<k$.


(3. is possible due to 2.) Observe that it follows that, for fixed $k$, the $I\_n^k$, $n=0,\dots,k$, are pairwise disjoint. In particular, for fixed $n$ the length of $I\_n^k$ tends to zero as $k\to\infty$.


Now define $G\_k$ continuous (say, piecewise linear) so that $g\_k(d\_n)=0$ for $n\leq k$ and $G\_k(x)=1$ when $x$ does not belong to $I\_n^k$ for some $n\leq k$. Now let $x$ be irrational. If $x\in I\_n^k$ for some $n$ and $k$, then there is $k\_1\geq k$ such that $x\in I\_n^{k\_1}$ but $x\notin I\_n^{k\_1+1}$. So $x\notin I\_{n'}^{k\_1+1}$ for $n'\leq k\_1$. But also, by 3., $x\notin I\_{k\_1+1}^{k\_1+1}$. It follows that $G\_k(x)=1$ for infinitely many $k$. Now we can just set $g\_{2k} = G\_k$ and $g\_{2k+1}\equiv 0$ to obtain a sequence of functions $g\_k$ that does not converge pointwise at any irrational, but is eventually zero at every rational number $d\_n$.


