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I have been pondering over something for quite a sometime, and as a part of understanding it, I had to formulate a problem like the one given below. It is quite lengthy and in case it turns out to be totally absurd or trivially wrong, I sincerely apologize for that.

Let $s_n : (0,1) \to \mathbb{R}\quad \forall \; n \in \mathbb{N}$ be a set of smooth functions. The sequence of real numbers $\{s_n(x)\}$ is always positive and increasing for all $x \in (0,1)$. Let $D$ be a countable dense subset of $(0,1)$ and $h : D \to \mathbb{N}$ be an enumeration. The sequence $\{s_n(x)\}$ diverges and $$\{s_n(x)\} \in O(\log n) \forall x \in (0,1)\setminus D$$The sequence $$\{s_n(x)\} \in O(n^{\frac{1}{h(x)}}) \wedge \{s_n(x)\} \in \Omega(n^{\frac{1}{h(x)+1}}) \quad \forall \; x \in D$$Additionally $s_n(x)$ is smooth in $(0,1)$ and $s_n(x)$ has finite number of maxima and minima, for all $n \in \mathbb{N}$.

My question is : Is the set of all such sequences nonempty ?

This has been posted on MO as well here

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    When you write $s_n(x) \in O(\log n)$, there's an implied constant. Are we to assume that this constant may depend on $x$? Also, I think it's much more common to write $s_n(x) = O(\log n)$ rather than the more precise $\in$.2011-06-12
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    @Theo Buehler : Dear Theo Buehler, it is $O(n^{\frac{1}{h(x)}})$, please see the question again, seems to me like you might have misread this time, or there is some problem with your viewer (web browser). Is there anyone facing the same please let me know.2011-06-12
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    @Theo Buehler : it is like for example 10nth root of $n$.2011-06-12
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    Oh, yes, you're right. I simply misread, no problem with my browser. Sorry about that.2011-06-12
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    @Alon Amit : nothing can be said about the constants, which means they could depend on $x$.2011-06-13
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    note: cross-posted at http://mathoverflow.net/questions/676682011-06-13
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    That post has been deleted, and there is now another one: http://mathoverflow.net/questions/67731. Rajesh, could you please explain all this?2011-06-14
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    @Joriki : I had taken comments from here http://tea.mathoverflow.net/discussion/1067/about-a-question-posted-on-mathse/#Item_0 and later went ahead to delete the post and wait for more time on math.se, I really do not understand the thing with cross posting and would like to know the etiquette behind this.2011-06-14
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    @Rajesh: I don't know much about etiquette, but there are some things you can deduce from basic principles of cooperation: If you crosspost, the attempts to solve the question will be split up in two different threads. You should make these threads aware of each other so they don't duplicate efforts. It seems that you did this on MO but not here. Also, when Anton points out the cross-post and you delete it and then re-ask the question on MO, clearly you should point this out here, for the same reasons as above and also so it doesn't look like Anton got it wrong.2011-06-14
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    @joriki : thank you for the comment, I really forgot to post about it here for which i really apologize.2011-06-14
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    @Rajesh: Thanks for the edit.2011-06-14
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    @joriki : I guess you are talking about the edit of the equations (and not the reframing of question), The equations were edited in readable form by yayu, It was at the same time i changed the question a bit so not that visible, you may check the edit history. So i guess the thanks should be for yayu.2011-06-14
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    @Rajesh: Sorry if that was unclear; I was talking about the edit pointing out the cross-post; I didn't realize the rest was edited.2011-06-14

1 Answers 1

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There are no sequences satisfying the required properties.

If there did exist such $s_n$, then $f_n(x)=s_n(x)/(1+\log n)$ would be a sequence of continuous functions tending to infinity at each point $x\in D$ and bounded at each point $x\not\in D$, contradicting the following statement.

Let $f_n\colon(0,1)\to\mathbb{R}$ be a sequence of continuous functions and $D$ be the set of $x\in(0,1)$ for which $\{f_1(x),f_2(x),\ldots\}$ is unbounded. If $D$ is dense in $(0,1)$ then it is uncountable.

This follows from the Baire category theorem. If $D$ is dense then, for each $n\in\mathbb{N}$, $\bigcup_{m=n}^\infty\{x\colon \vert f_m(x)\vert > n\}$ contains $D$ and hence is a dense open set. Then, $$ D=\bigcap_{n=1}^\infty\bigcup_{m=n}^\infty\left\{x\in(0,1)\colon\vert f_m(x)\vert > n\right\} $$ is a countable intersection of dense open sets (i.e., it is comeagre), so $D$ is uncountable.