3
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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 $\C\setminus K_1$. :) $\endgroup$ Dec 15 '20 at 20:53
2
$\begingroup$

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 $d_n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.