# Is every sequence of functions with uniformly bounded variation almost equicontinuous?

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.

• 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 $1$) Jun 4 at 15:00
• 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? Jun 4 at 15:31
• 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 $1$ Jun 4 at 15:41
• Ah it does so happen that pointwise equicontinuity implies uniform equicontinuity (if I’m not mistaken), so it makes sense your counterexample would work. Jun 5 at 0:29
• 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 $[\frac{n-1}{n},1]$ Jun 5 at 0:36

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 $$f_{n_i,k}(x)$$.