Question: Let $\{f_n\}$ be a sequence of analytic functions on the unit disk $\Delta$ and suppose that $f_n$ converges to a continuous function $f$ on $\Delta$ pointwisely. (1) Can we say that $f$ is analytic on $\Delta$? (2) If $f$ is analytic, is the convergence $\underline{locally}$ uniform on $\Delta$? (Note: I add the words "locally" due to obvious reason.)

If we do not assume that the limit function $f$ is continuous (of course $f$ is measurable) in advance, (3) can we say that $f$ is continuous? [I number this new question by (3)]

Note that evrey measurbale functon on $\Delta$ can be the limit of a sequence of analytic functions in the Lebesgue sense (i.e. almost everywhere).

  • $\begingroup$ Of course, not. The set of analytic functions is dense in $C(\Delta)$, so for any continuous non-analytic function ($f(z)=\overline z$, say) there exists a sequence of analytic functions which converges to $f$ uniformly and hence, pointwise. $\endgroup$ – Nikita Sidorov Dec 30 '12 at 14:47
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    $\begingroup$ Actually analytic functions aren't dense (Morera's theorem) $\endgroup$ – Anthony Quas Dec 30 '12 at 14:59
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    $\begingroup$ @Nikita: A uniform limit of analytic functions is analytic. $\endgroup$ – GH from MO Dec 30 '12 at 15:01
  • $\begingroup$ I don't see why someone proposed to close it. A very reasonable, non-trivial question. It was completely solved by Lavrentiev (see my ans below). $\endgroup$ – Alexandre Eremenko Dec 30 '12 at 16:12
  • $\begingroup$ I voted to close - misread the question as the uniform limit of analytic functions $\endgroup$ – Anthony Quas Dec 30 '12 at 19:00

These questions were investigated by Osgood, Montel and Lavrentiev, Sur les fonctions d'une variable complexe representable par des series de polynomes, Paris 1936. (There is a Russian translation in his selected Works available free on Internet). If you prefer German, see Hartogs and Rosenthal, Uber Folgen analytischer Funktionen, Math Ann., 1928, 100, 212-263, and 1932, 104, 606-610.

In general, if a sequence of polynomials converges pointwise in a region $D$, then the limit function is analytic except for a closed nowhere dense set $E$ (Osgood). Montel proved that $E$ is a perfect set whose union with the complement of the disc is connected. Lavrentiev completely characterized the sets $E$ that can occur, and proved that every function of the first Baire class which is analytic outside $E$ is a pointwise limit of polynomials.

Thus every continuous function, analytic outside $E$ can be obtained as a limit of polynomials. Convergence outside $E$ is locally uniform.

This answers all your questions.

By the way, similar problems for harmonic functions (characterization of their pointwise limits) is still not solved completely.

EDIT. For example, any simple curve in the unit disc, going from $0$ to $1$, satisfies the Lavrentiev condition. Taking this curve with positive area, we can construct a continuous function $f$ in the unit disc, which is analytic outside the curve but not analytic in the unit disc. This function will be a pointwise limit of polynomials.

  • $\begingroup$ What if one says that such continuous function, analytic outside $E$ is necessarily analytic on $E$? So, we should prove that $E$ can be the singularity set of some continuous function, analytic outside $E$. $\endgroup$ – woodbass Dec 30 '12 at 16:22
  • $\begingroup$ As I said, Lavrentiev completely characterized the sets $E$ which can occur. The limit function is NOT continuous on $E$. $\endgroup$ – Alexandre Eremenko Dec 30 '12 at 16:34
  • $\begingroup$ Yes, this answers quesion (3) negatively, but not a final solution to question (1) since I assume thaf $f$ is contiuous. $\endgroup$ – woodbass Dec 30 '12 at 16:45
  • $\begingroup$ To answer question (1) negatively, we must prove $E$ can be the singularity set of some continuous function, analytic outside $E$. Otherwise, "Thus every continuous function, analytic outside $E$ can be obtained as a limit of polynomials." does not imply that the answer to (1) is no. $\endgroup$ – woodbass Dec 30 '12 at 16:54
  • $\begingroup$ See my EDIT in the answer. $\endgroup$ – Alexandre Eremenko Dec 30 '12 at 17:13

Of course (1) does not imply (2) : the functions $f_n:=z^n$ converge pointwisely to $0$ on the unit disk, but the convergence is not uniform.

In fact, assuming (1), the convergence need not be locally uniform : using Runge's Theorem, it is possible to find a sequence of polynomials $p_n$ such that $p_n \rightarrow 0$ pointwisely on the unit disk, but the convergence is not uniform in any neighborhood of $0$ : see e.g. this question


A similar argument (using Runge's Theorem) answers your question (3) negatively : It is possible to construct a sequence of polynomials $p_n$'s with $p_n(z) \rightarrow 0$ for every $z \neq 0$ but $p_n(0) \rightarrow 1$.

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    $\begingroup$ @AlexandreEremenko: I don't understand your objection. The answer given there exhibits a nested increasing sequence of compact sets $K_n$ whose union is the closed unit disk. The polynomials $p_n$ satisfy $|p_n| < 1/n$ on $K_n$. Any point of the disk is in $K_n$ for sufficiently large $n$, so $p_n$ converges to $0$ there. $\endgroup$ – Robert Israel Dec 30 '12 at 19:18
  • $\begingroup$ You are right. I removed my objection. $\endgroup$ – Alexandre Eremenko Dec 30 '12 at 20:41

I suspect the answer is no in general, but if you assume local boundedness then (1) is true and, for (2), convergence is locally uniform. This is a special case of the Vitali-Porter theorem, see here.

  • $\begingroup$ Thank you! Can you tell me from which book you adapt the Vitali-Porter theorem? (the bookname and the editor?) I woulk like to get it. $\endgroup$ – woodbass Dec 30 '12 at 15:39
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    $\begingroup$ Of course if the sequence is locally bounded, then the convergence is locally uniform, and the limit is analytic. For the proof, see any textbook on complex variables, which has a chapter "Normal families", for example Ahlfors. $\endgroup$ – Alexandre Eremenko Dec 30 '12 at 17:27

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