On the set of divergence to infinity for sequences of positive continuous functions

Hi,

I have asked this question on math.stackexchange but it has not received much attention, so I ask it here.

This question is partly motivated by this one, which contains an example of a sequence $(f_n)$ of positive continuous functions on $\mathbb{R}$ such that $$f_n(x) \rightarrow \infty$$ if and only if $x \in \mathbb{Q}$.

My question is the following :

For a given sequence of positive continuous functions $(f_n)$ on $\mathbb{R}$, denote by $S((f_n))$ the set of divergence to $\infty$ :

$$S((f_n)):={ x \in \mathbb{R}: f_n(x) \rightarrow \infty }$$

Is there a necessary and sufficient condition for a given set $S$ to be $S((f_n))$ for some sequence $(f_n)$?

As noted in the comments at math.stackexchange, a necessary condition is that $S$ must be a countable interesection of $F_{\sigma}$'s... But is this sufficient?

Thank you, Malik

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Nice question! What about the Cantor set - is the answer known for it? – Felix Goldberg May 22 2012 at 15:04
@Felix Goldberg : Thank you! If $S=F$ is closed, I think one can take $f_n(x)=\frac{1}{d(x,F)+1/n}$, where $d(x,F)$ is the distance between $x$ and $F$... – Malik Younsi May 22 2012 at 16:15
We can equivalently replace "convergence to $\infty$" with "convergence to $0$", right? – Pietro Majer May 22 2012 at 16:54
@Pietro Majer : Yes, indeed we can. – Malik Younsi May 22 2012 at 17:00

The condition that the set be an $F_{\sigma\delta}$ is necessary and sufficient. This is a result proved by Hahn in 1919. The reference is H. Hahn, Ueber die Menge der Konvergenzpunkte einer Funktionenfolge, Archiv. der Math. und Physik 28 (1919), 34-45.

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 For German readers, here's the ZMath link: zentralblatt-math.org/zmath/en/search/… – Willie Wong May 23 2012 at 7:35 Nice, thank you for the reference! – Malik Younsi May 23 2012 at 13:26

Sierpinski's 1921 paper was written without knowledge of Hahn's 1919 paper, this being a time (end of WWI) when the flow of information and journals was intermittent and/or temporarily suspended.

p. 348 in Volume 2 of Hans Hahn's Collected Works (1996) includes these remarks about Hahn's 1919 paper:

"... Hahn sets himself in Über die Menge der Konvergenzpunkte einer Funktionenfolge the task of finding out if this property gives a complete characterization of such sets. He not only proves it, but also enters the study of Baire functions by finding a characterization of the sets of convergence of functions of any given Baire class.

Jolanta Wesolowska has published versions of Hahn/Sierpinski's result for sequences of functions belonging to various other classes of functions, for example On sets of convergence points of sequences of some real functions (MR 2001d:26003; Zbl 1035.26006) and On sets determined by sequences of quasi-continuous functions (MR 2002i:26002; Zbl 1002.26004) and On sets of discrete convergence points of sequences of real functions (MR 2005f:26010; Zbl 1070.26005). Incidentally, I was in attendance when she first presented her work (represented by the first paper above) to people outside her immediate research group [1] (this work was her 2000 Ph.D. Dissertation at Uniwersytet Gdański), and it created a minor buzz among those attending, who found it simply amazing that the kinds of questions she was working on had not been thoroughly worked over before. (I should point out that the literature of real functions and point set theory is pretty much everywhere dense with minutia on most anything you can imagine, and more.)

[1] She gave this talk on 26 May 2000 at Summer Symposium in Real Analysis XXIV, held at The University of North Texas (Denton, Texas).

(Next Day) Yesterday I forgot to post a couple of references in English to a proof of the Hahn/Sierpinski result. A proof can be found on pp. 307-308 of the 1978 3rd English edition (and presumably, on the same pages for any of the other English editions) of the 1935 3rd edition of Hausdorff’s Set Theory [1957 (MR 19,111a; Zbl 81.04601); 1962 (MR 25 #4999); 1978 (Zbl 488.04001); 1991 (Zbl 896.04001); 2005] and on pp. 185-186 of Kechris’ Classical Descriptive Set Theory [MR 96e:03057; Zbl 819.04002].

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 That is very interesting, thank you for the additional remarks. – Malik Younsi May 23 2012 at 13:32 @Dave L Renfro : Thank you for the reference. – Malik Younsi May 24 2012 at 14:29

In addition to Michael Renardy's reference, another proof (obtained perhaps independently?) was given by Sierpinski in 1921.

Sierpiński, W. Sur l'ensemble des points de convergence d'une suite de fonctions continues. (French) [J] Fundamenta math. 2, 41-49 (1921). ZMath Link

(It has the advantage of (a) being in French so I can actually understand it and (b) published in a Journal which now offers [I think] open access to the old articles; one can obtain a copy by searching here.)

The main theorem in the paper states the following (all sets are subsets of $\mathbb{R}$):

Theorem For a set $E$ to be the set of convergence points of a sequence of continuous functions $f_1, f_2, \ldots$ (meaning that $f_n(x)$ converges if and only if $x\in E$) it is necessary and sufficient that $E$ be $F_{\sigma\delta}$

The proof proceeds via a series of Lemmas.

Lemma 1 Any $F_\sigma$ can be decomposed into the sum of an $F_\sigma$ with no interior points which we call $P$ with an at most countable collection of mutually disjoint intervals whose endpoints all appear in $P$.

Lemma 2 An $F_\sigma$ which contains no interior points can be written as a sum of at most countably many disjoint closed sets.

Together the above yields

Lemma 3 An $F_\sigma$ can be written as a union of $P\cup Q$ where $P$ can be written as an at most countable union of mutually disjoint closed sets and $Q$ can be written as an at most countable union of mutually disjoint intervals whose endpoints lie in $P$.

Given the above we have

Lemma 4 For $E$ an $F_\sigma$ there exists a sequence of bounded continuous functions which converge to 0 on $E$ and does not converge otherwise.

Sketch of construction:

Write $E = P \cup Q$ as above. Write $P = F_1 \cup F_2\cdots$ where the $F_i$ are mutually disjoint. Let $S_n = \cup_1^n F_i$. Let $\delta_n = \mathrm{dist}(S_n,F_{n+1}) > 0$ since we have disjoint closed sets. Let $T_n = \{ x : \mathrm{dist}(S_n,x) \geq \delta_n / (3 + n \delta_n) \}$.

Define a sequence of functions $\varphi_n(x)$ such that for $n$ odd, $\varphi_n \equiv 0$. For $n = 2k$ even, let $\varphi_n(x) = 1$ on $T_k$ and $0$ on $S_k$, and linearly interpolate in between (we can do this because $T_k\cup S_k$ is closed and its complement is a union of open intervals).

Now define $f_n(x)$ to be equal to $\varphi_n(x)$ on the complement of the interior of $Q$. The remaining portion is again a union of open intervals so we can interpolate linearly on it.

It is clear that $f_n$ converges to 0 on $E$ by definition. It takes a little bit of computation to show that for $x\not\in E$, $\limsup f_n(x) = \limsup \varphi_n(x) = 1$.

Sketch of construction for the Theorem

Now let $E = E_1 \cap E_2 \cap \cdots$ where $E_k$ are $F_\sigma$. Let $\tilde{f}_{k,n}(x)$ denote the sequence found by applying Lemma 4 to $E_k$. Let $\bar{f}_{k,n}(x) = \frac{1}{k} \tilde{f}_{k,n}$. The sequence we want in the theorem can be obtained by the diagonal method:

$$f_1 = \bar{f}_{1,1} \quad f_2 = \bar{f}_{2,1} \quad f_3 = \bar{f}_{1,2} \quad f_4 = \bar{f}_{3,1} \ldots$$

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Note that for the sequence constructed for the main theorem the $f_n$ converges to 0 on $E$ and does not converge otherwise. Since $f_n$ are all non-negative, you can easily modify it to be strictly positive. – Willie Wong May 23 2012 at 8:49
Note also that the construction is very similar to what was given by Gerry in the post that inspired your question. The trick is essentially all contained in Lemmas 1 and 2, which gives a good decomposition for any $F_\sigma$. – Willie Wong May 23 2012 at 8:51
@Willie Wong : I don't know german, so thank you very much for pointing out Sierpinski's article and for giving the details here! Just a remark : In Lemma 1 to 4, do you mean $F_{\sigma \delta}$ instead of $F_{\sigma}$? – Malik Younsi May 23 2012 at 21:01
No, Lemma 4 is for $F_\sigma$s. Then you apply Lemma 4 to $F_{\sigma\delta}$ by writing an $F_{\sigma\delta}$ as intersection of $F_{\sigma}$s, apply Lemma 4 to each of the component $F_\sigma$, and then "diagonalizing". – Willie Wong May 24 2012 at 8:46
And Lemmas 1 through 3 are all also facts about $F_\sigma$ sets. Note that the real line with a point removed is $F_\sigma$. Thus the set of irrationals is a countable intersection of $F_\sigma$ and hence $F_{\sigma\delta}$. You'd find that Lemma 2 is false for the set of irrational numbers. – Willie Wong May 24 2012 at 8:53
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