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Let $f_n: [0, 1] \to \mathbb R$ be a sequence of functions.

Given a measurable subset $E$ of $[0, 1]$, we say that the sequence $f_n$ is equicontinuous on $E$ if for every $x \in E$, and $\varepsilon > 0$ there exists a $\delta > 0$ and all $n \in \mathbb N$ we have $|f_n(x) - f_n(y)| < \varepsilon$ for all $y \in [0, 1]$ such that $|y - x| < \delta$.

We say a sequence of functions $f_n: [0, 1] \to \mathbb R$ has uniformly bounded variation if there exists some $M > 0$ such that each $f_n$ has total variation less than or equal to $M$.

Question: Given any sequence of functions $f_n: [0, 1] \to \mathbb R$ of uniformly bounded variation, is it true that there exists a measurable subset $E$ of $[0, 1]$ with Lebesgue measure $1$ and a subsequence $f_{n_k}$ of lower density $1$ such that $f_{n_k}$ is equicontinuous on $E$?

Note: We define the lower density of a subsequence $f_{n_k}$ to be the number

$$\liminf_n\frac {\#\{k \, | \, k \in \mathbb N, \, n_k < n\}}{n}$$

where $\#$ denotes the cardinality of a finite set.

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  • $\begingroup$ I think if you take the pairs $(n,k)\in\mathbb{N}^2;0\leq k<n$ in lexicographic order and for each pair $(k,n)$ you define the function $f_{n,k}$ defined by $0$ in $[0,\frac{k}{n}]$, $1$ in $[\frac{k+1}{n},1]$ and interpolating linearly in $[\frac{k}{n},\frac{k+1}{n}]$ that gives a counterexample (you don't even need the subsequences to have lower density $1$) $\endgroup$
    – Saúl RM
    Jun 4 at 15:00
  • $\begingroup$ Hmm, the points at which equicontinuity fail would depend a lot on the chosen subsequence in that case. Do you have a particular subsequence in mind? $\endgroup$
    – Nate River
    Jun 4 at 15:31
  • $\begingroup$ Oh sorry, I was thinking of uniformly equicontinuous (it was the only type of equicontinuity I had seen). Now it makes sense why the subsequence needs to have lower density $1$ $\endgroup$
    – Saúl RM
    Jun 4 at 15:41
  • $\begingroup$ Ah it does so happen that pointwise equicontinuity implies uniform equicontinuity (if I’m not mistaken), so it makes sense your counterexample would work. $\endgroup$
    – Nate River
    Jun 5 at 0:29
  • $\begingroup$ That is true if we are talking about the whole interval $[0,1]$. With some subset $E$ of measure $1$ both conditions may not be equivalent, for example we can construct a family of functions pointwise equicontinuous in $(0,1)$ but not uniformly equicontinuous in $(0,1)$: the functions $f_n$ given by $0$ in $[0,\frac{n-1}{n}]$ and $nx-(n-1)$ in $[\frac{n-1}{n},1]$ $\endgroup$
    – Saúl RM
    Jun 5 at 0:36

1 Answer 1

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Consider the pairs $(n,k)\in\mathbb{N}^2$ with $0\leq k<2^n$, they form a sequence in lexicographic order. Consider now the functions $f_{n,k}$ which are defined as $0$ in $[0,\frac{k}{2^n}]$, $1$ in $[\frac{k+1}{2^n},1]$, and $2^nx-k$ in $[\frac{k}{2^n},\frac{k+1}{2^n}]$, and suppose there is a set $E\subseteq [0,1]$ and a subsequence of $f_{n,k}$ as in the question.

If we define $A_n=\bigcup\{(\frac{k}{2^n},\frac{k+1}{2^n});(n,k)\text{ is in the subsequence}\}$, then the fact that the subsequence has density $1$ implies that when $n$ tends to $\infty$, the measure of $A_n$ tends to $1$. This implies that there is a point $x\in E$ which is in infinitely many of the $A_n$: for example, take a subsequence $A_{n_i}$ such that $\sum_i(1-m(A_{n_i}))<1$; then $m(\cap_k A_{n_i})>0$), so $E\cap(\cap_k A_{n_i})\neq\emptyset$.

Suppose then that $x\in E\cap(\cap_i A_{n_i})$ and we are given $\varepsilon<\frac{1}{2}$; there is no $\delta$ that satisfies the definition of equicontinuity at $x$: indeed, take $i$ such that $2^{-n_i}<\delta$, then there is some interval $(\frac{k}{2^{n_i}},\frac{k+1}{2^{n_i}})$ contained in $(x-\delta,x+\delta)$ and such that $(n_i,k)$ is in the subsequence. So the image of $(x-\delta,x+\delta)$ under the function $f_{n_i,k}$ is all $[0,1]$, thus it contains values at distance $>\varepsilon$ of $f_{n_i,k}(x)$.

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