If $\forall x \in (0,1] \ \ f_n(x) \to 0, n \to \infty$ then $\exists x_n \downarrow 0$ such that $f_n(x_n) \to 0$.

Question

In the second edition of the book "Probability" (A. N. Shiryaev, R.P. Boas, 1996) there is a problem 2, p. 553 ($\S$ 8, chapter VII) which is equivalent to the next one. Suppose that for every $n \ge 1$ $f_n(x): (0,1] \to \mathbb{R}$.

If $\forall x \in (0,1] \ \ f_n(x) \to 0, n \to \infty$ then $\exists x_n \downarrow 0$ (strictly monotone) such that $f_n(x_n) \to 0$.

This problem is an important part of the proof of CLT for martingales.

Without loss of generality we can assume that $f_n(x) \ge 0$ (because we may consider $\tilde{f}_n(x) = |f_n(x)|$). Let's consider this case.

It is easy to prove that if conditions of the problem hold true then $\exists x_n \to 0$ such that $f_n(x_n) \to 0$. Sketch of this proof: we show that $\inf_{x \in (0,1]} (x + f_n(x)) \to 0$ (by contradiction) and find $x_n$ such that $x_n \le \inf_{x \in (0,1]} (x + f_n(x)) + \frac1{n}$. Unfortunately, $x_n$ from this proof may be not strictly monotone.

I also can solve the problem in case when there is the next additional condition: $\forall x \in (0,1]$ $f_n(x)$ has the set $E_n$ of discontinuities with Lebesgue measure $\mu(E_n) = 0$ [e.g. $f_n(x)$ is monotone function of $x$]. Idea of the proof: $E = \bigcup_{n=1}^{\infty} E_n$. We will work with $(0,1] \backslash E \ $ $ \ $ instead of $(0,1]$. Denote by $\tilde{x}_n \to 0$ the sequence from the previous proof, which is identical if we have $(0,1] \backslash E$ instead of $(0,1]$. It's easy to find $n_1 < n_2 < \ldots$ such that $\tilde{x}_{n_k} \downarrow 0$. Put $x_{n_k} =\tilde{x}_{n_k}$. We can find $x_{n_1 + 1} > x_{n_1 + 2} > \ldots > x_{n_2-1}$ in a little left neighborhood of $x_{n_1}$, then we can find $x_{n_2 + 1} > x_{n_2 + 2} > \ldots > x_{n_3-1}$ in a little left neighborhood of $x_{n_2}$ and so on and make a sequence $x_{n}$ such that $f_n(x_n) \to 0$.

My question: how to solve the problem without any additional assumptions?

Answer 1

Set $g(1) = 1$ and for $n\geq 2$ let $g(n)$ denote the minimal $n$ strictly greater than $g(n-1)$ for which there exists an infinite subset $A_n \subset (\frac{1}{n+1}, \frac{1}{n})$ such that for $m \geq g(n)$, for every $x \in A_n$ $$f_m(x) \leq \frac{1}{n}$$ Then for all $n > 1$ and $g(n) \leq k < g(n+1)$, Let $x_k$ be a strictly decreasing sequence in $A_n$.

Answer 2

Let $n_m$ be indices such that $$ f_n(1/m) < 1/m \qquad \forall n\ge n_m $$ and notice that you can take $n_m$ strictly crescent in $m$. Now take $$ x_n = 1/m \qquad n_{m+1}>n\ge n_m $$ and notice that $$ f_n(x_n) < 1/m \to 0 $$

Answer 3

I elaborate and clarify cha21's solution. Let $f_n:(0,1]\rightarrow [0,\infty)$ be a function such that for each $x\in(0,1], f_n(x)\rightarrow 0$.

We define a sequence of positive integers $(n_{k})$ such that $n_{1}<n_{2}<\ldots$ by recursion: Define $n_{1}=1$. Suppose that $n_{1},n_{2},\ldots,n_{k}$ have been specified. Note that $$ (\frac{1}{n_{k}+1},\frac{1}{n_{k}})=\bigcup_{L=1}^{\infty}A_{L}^{(k)}, $$ where $A_{L}^{(k)}=\left\{ x\in(\frac{1}{n_{k}+1},\frac{1}{n_{k}})\mid\forall l\geq L\,\,\,f_{l}(x)\leq\frac{1}{n_{k}}\right\} $. Since $(\frac{1}{n_{k}+1},\frac{1}{n_{k}})$ is uncountable, there exists $L$ such that $A_{L}^{(k)}$ is uncountable. Clearly, $A_{1}^{(k)}\subseteq A_{2}^{(k)}\subseteq\ldots$, so we may choose $L$ such that $L>n_{k}$ and $A_{L}^{(k)}$ is uncountable. Let $n_{k+1}$ be any such $L$. Also observe that $A_{n_{k+1}}^{(k)}\subseteq(\frac{1}{n_{k}+1},\frac{1}{n_{k}})$, so $A_{n_{2}}^{(1)},A_{n_{3}}^{(2)},\ldots$ are pairwisely disjoint. Moreover, for any $x\in A_{n_{i+1}}^{(i)}$, $y\in A_{n_{j+1}}^{(j)}$, we have $x>y$ if $i<j$.

Partition the set $$ \mathbb{N}=\bigcup_{k=1}^{\infty}\{m\mid n_{k}\leq m<n_{k+1}\}. $$ For $m\in\{m\mid n_{k+1}\leq m<n_{k+2}\}$, we choose $x_{m}\in A_{n_{k+1}}^{(k)}$ such that $x_{n_{k+1}}>x_{n_{k+1}+1}>\dots$. For $m\in\{m\mid1\leq m<n_{2}\}$, Choose $x_{1},x_{2},\ldots,x_{n_{2}-1}$ in an arbitrary way such that $x_{1}>x_{2}>\ldots>x_{n_{2}-1}>x_{n_{2}}$. This is possible because $x_{n_{2}}<1$. Clearly $(x_{m})$ is a strictly decreasing sequence in $(0,1)$, $x_{m}\rightarrow0$. Moreover, if $n_{k+1}\leq m<n_{k+2}$, then $x_{m}\in A_{n_{k+1}}^{(k)}$ and hence $f_{m}(x_{m})\leq\frac{1}{n_{k}}$ because the condition $m\geq n_{k+1}$ is satisfied. It follows that $f_{m}(x_{m})\rightarrow0$.