5
$\begingroup$

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

  • 0
    @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

7

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.