# Looking for a sequence of analytic functions with strange behaviour

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.

• I guess that you want $K_2\setminus K_1$ to be uncountable, or at least not agree with $D$. Otherwise, you just take a sequence that converges uniformly on compact subsets of $\C\setminus K_1$. :) Dec 15 '20 at 20:53

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.

(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 $$d_n$$.