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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 $\in$. –  Alon Amit Jun 12 '11 at 7:28
    
@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. –  Rajesh D Jun 12 '11 at 12:17
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note: cross-posted at mathoverflow.net/questions/67668 –  Anton Geraschenko Jun 13 '11 at 17:12
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That post has been deleted, and there is now another one: mathoverflow.net/questions/67731. Rajesh, could you please explain all this? –  joriki Jun 14 '11 at 6:38
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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. –  joriki Jun 14 '11 at 7:08

1 Answer 1

up vote 7 down vote accepted
+50

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.

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