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Does there exist a continuous function $f:[0,1]\to\mathbb{R}$, along with a sequence $(\pi_n)_{n\geq 0}$ of subdivisions of $[0,1]$ $$\pi_n\equiv\Big(0=t_0^n<t_1^n<\cdots <t_{p_n}^n=1\Big)$$ with $\|\pi_n\|=\sup_{1\leq k\leq p_n}(t_{k}^n-t_{k-1}^n)\longrightarrow 0$ such that $$\sup_n V(\pi_n,f)<+\infty$$ yet $f$ is not of bounded variation on $[0,1]$ ?

As usual, $$V(\pi_n,f)=\sum_{k=1}^{p_n}|f(t^n_{k})-f(t^n_{k-1})|.$$


My guess is that such functions could exist, and I have an idea of how to construct one. The idea will be that $f$ will be expressed as a uniform limit of affine functions such that along well chosen subdivisions $\pi_n$, $f$ coïncides with well behaved functions of uniformly bounded total variation.

The following does not work, the sequence of functions does not converge uniformly, but maybe someone can find a modification that does, or provide a different solution (or a proof of the contrary).

Let me define $$\forall n\geq 1,\quad f_n(x)=\frac{(-1)^{n-1}}{\lceil n/2\rceil}\cdot x(x-1)$$ These functions have uniformly bounded total variation : $$\forall n\geq 1,\quad TV_{[0,1]}(f_n)\leq 1/2$$ me put $p_n=$ the $n$-th prime number : $p_1=2,p_2=3,\dots$ Let me define, for $n\geq 1$, and $k\in\lbrace 0,1,\dots,p_n\rbrace$ $$t^n_k:=\frac{k}{p_n}$$ and $$\pi_n\equiv\Big(t^n_k\Big)_{k\in\lbrace 0,1,\,\dots\,,p_n\rbrace}$$ Then the idea would be to define affine functions $a_n$ by imposing

  1. $a_n$ is affine on every interval of the subdivision $\cup_{i=1}^n\pi_i=\Big(\frac{k}{p_1p_2\cdots p_n}\Big)_{k\in\lbrace 0,1,\,\dots\,,\,p_1p_2\cdots p_n\rbrace}$,
  2. for $n\geq 2$, $a_n$ coincides with $a_{n-1}$ on the vertices of the subdivision $\cup_{i=1}^{n-1}\pi_i$
  3. on the $t^n_k$, we have $a_n(t^n_k)=f_n(t^n_k)$.
  4. $f$ is the uniform limit of the $a_n$

Screen Shot

As stated this does not work : the sequence $a_n$ does not converge uniformly. But I'm hopeful that there is a work around, although I can't quite find it yet. Maybe by asking that the $a_n$ only change outside of $[\epsilon_n,1-\epsilon_n]$ (for instance, in its current incarnation, one will have $\lim_{1/2^-}a=0\neq a(1/2)=1$, and this could be avoided if one didn't change the $a_n$ on $[\epsilon_n,1-\epsilon_n]$.

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  • $\begingroup$ My guess is that something like $\epsilon_n=\frac1 {\ln (n)} $ could work... $\endgroup$ – Olivier Bégassat Jan 6 '18 at 20:11
  • $\begingroup$ (The picture is not quite right, but it gives the idea). $\endgroup$ – Olivier Bégassat Jan 6 '18 at 22:12
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Such a function and sequence of partitions don't exist. Suppose to the contrary that $f$ and $(\pi_n)$ had the desired properties, and

$$S = \sup_n V(\pi_n,f) < +\infty\,.$$

Since $f$ has unbounded variation, we can find a partition $\rho$, given by partition points

$$0 = r_0 < r_1 < \dotsc < r_{m-1} < r_m = 1$$

such that $V(\rho,f) > S + 1$. Let $\varepsilon \leqslant \lVert \rho\rVert/4$ so small that $\lvert x-y\rvert \leqslant \varepsilon$ implies $\lvert f(x) - f(y)\rvert < \frac{1}{4m}$. Then pick $n$ so large that $\lVert \pi_n\rVert < \varepsilon$. For $0 \leqslant \mu \leqslant m$, let $\ell_{\mu}$ be the largest point of $\pi_n$ not greater than $r_{\mu}$, and $u_{\mu}$ the smallest point of $\pi_n$ not smaller than $r_{\mu}$. Then

\begin{align} \sum_{u_{\mu} \leqslant t^n_k < \ell_{\mu+1}} \lvert f(t^n_{k+1}) - f(t^n_k)\rvert &\geqslant \lvert f(\ell_{\mu+1}) - f(u_{\mu})\rvert \\ &\geqslant \lvert f(r_{\mu+1}) - f(r_{\mu})\rvert - \lvert f(r_{\mu+1}) - f(\ell_{\mu+1})\rvert - \lvert f(u_{\mu}) - f(r_{\mu})\rvert \\ &\geqslant \lvert f(r_{\mu+1}) - f(r_{\mu})\rvert - \frac{1}{2m} \end{align}

for $0 \leqslant \mu < m$. Adding those, we find

$$V(\pi_n,f) \geqslant \sum_{\mu = 0}^{m-1}\sum_{u_{\mu} \leqslant t^n_k < \ell_{\mu+1}} \lvert f(t^n_{k+1}) - f(t^n_k)\rvert \geqslant \sum_{\mu = 0}^{m-1} \biggl(\lvert f(r_{\mu+1}) - f(r_{\mu})\rvert - \frac{1}{2m}\biggr) = V(\rho,f) - \frac{1}{2} > S\,,$$

contradicting the definition of $S$.

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    $\begingroup$ Corollary: If $f$ is continuous on a compact interval $[a,b]$, then $$V([a,b],f) = \lim_{n\to\infty} V(\pi_n,f)$$ for every sequence $(\pi_n)$ of partitions with $\lVert \pi_n\rVert \to 0$. $\endgroup$ – Daniel Fischer Jan 6 '18 at 22:56
  • $\begingroup$ Excellent, that looks very convincing ! Is this a classical result ? $\endgroup$ – Olivier Bégassat Jan 6 '18 at 23:10
  • $\begingroup$ Do you think this result could be extended to $f$ with limits on the left and right at every point, and $f(x)$ sandwiched between $f(x^-)$ and $f(x^+)$ ? You are using uniform continuity, but I wonder whether the result would still hold. Also, I think you are possibly counting certain intervals twice (those where overlap happens) which won't matter since there at most $m$ or so such overlaps). Am I right ? $\endgroup$ – Olivier Bégassat Jan 6 '18 at 23:30
  • $\begingroup$ I think you could do the proof without the $u_\mu$'s, thus eliminating potential overlap. $\endgroup$ – Olivier Bégassat Jan 6 '18 at 23:42
  • $\begingroup$ I expect it is classical, but I don't know. One could eliminate either the $\ell_{\mu}$ or the $u_{\mu}$, but there's no overlap. We have $\ell_{\mu} \leqslant r_{\mu} \leqslant u_{\mu} < \ell_{\mu+1} \leqslant r_{\mu+1} \leqslant u_{\mu+1}$, so at points of $\rho$ that aren't in $\pi_n$ we even have a gap. We can modify the proof to work in the more general situation you describe. We only need to look at what happens near the $r_{\mu}$, so we need only satisfy finitely many conditions. For $0 < \mu < m$, we must choose $\varepsilon_{\mu}$ so small that the difference between … $\endgroup$ – Daniel Fischer Jan 7 '18 at 13:44
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This is just tying up the loose ends from the comments in Daniel Fischer's answer.


Let $f:[0,1]\to\mathbb{R}$ be a function. Let us say that $x_0\in[0,1]$ is a good point if $f$ has both left and right limits at $x_0$ : $$f(x_0^-)=\lim_{\substack{x\to x_0\\x<x_0}}f(x) \quad\text{and}\quad f(x_0^+)=\lim_{\substack{x\to x_0\\x>x_0}}f(x)$$ both exist and are finite (we set $f(0^-):=f(0)$ and $f(1^+):=f(1)$. Define the total jump of $f$ at $x_0$ as $$J(x_0,f)=|f(x_0)-f(x_0^-)|+|f(x_0)-f(x_0^+)|.$$ Given a good subdivision $\sigma=(0=s_0<s_1<\cdots<s_{m-1}<s_m=1)$ of $[0,1]$, that is all the $s_k$ are good points, we define $$J(\sigma,f)=\sum_{k=0}^mJ(s_k,f).$$

Lemma. Let $\sigma$ be a good subdivision of $[0,1]$. Then $$ \forall~\epsilon>0,~\exists~\delta>0,\forall~\pi\in\mathfrak{S}[0,1],~\Big(\|\pi\|<\delta\implies V(\sigma,f)\leq V(\pi,f)+J(\sigma,f)+\epsilon\Big) $$

Here $\mathfrak{S}[0,1]$ is the set of all subdivisions of $[0,1]$, partially ordered by refinement.

Proof. Let $\epsilon>0$ be given. Since $\sigma=(0=s_0<s_1<\cdots<s_{m-1}<s_m=1)$ is a good subdivision of $[0,1]$, there exists $\delta_0>0$ such that for all $k\in\lbrace 1,\dots,m\}$ and for all $x\in[0,1]$ $$s_k-\delta_0<x<s_k\implies |f(x)-f(s_k^-)|<\frac\epsilon{2m}$$ and for all $k\in\lbrace 0,\dots,m-1\}$ and for all $x\in[0,1]$ $$s_k<x<s_k+\delta_0\implies |f(x)-f(s_k^+)|<\frac\epsilon{2m}$$ Set $\delta_1=\inf_{k=0,\dots,m}|s_{k+1}-s_k|$, and $\delta=\min\{\delta_0,\delta_1\}$.

Now let $\pi=(0=t_0<t_1<\cdots< t_{n-1}<t_n=1)\in \mathfrak{S}[0,1]$ be a subdivision such that $\|\pi\|<\delta$. Consider the subdivision $\sigma\cup\pi$ : since it is a refinement of $\sigma$, one has $$ V(\sigma,f)\leq V(\sigma\cup\pi,f) $$

  • Since $\|\pi\|<\delta_1$, for every $0\leq k<m$ there exists $0<l<n$ with $s_k<t_l<s_{k+1}$.
  • For $0\leq k<m$, define $0<a_k\leq b_k<m$ such that $a_k\leq l\leq b_k \iff s_k<t_l<s_{k+1}$.

Then $$ V(\sigma\cup\pi,f) = \sum_{k=0}^{m-1} \left( |f(s_k)-f(t_{a_k})| + \sum_{a_k\leq l<b_k} |f(t_{l+1})-f(t_{l})| + |f(s_{k+1})-f(t_{b_k})| \right) $$ Now for all $0\leq k<m$, $|f(s_k)-f(t_{a_k})|\leq|f(s_k)-f(s_k^+)|+|f(s_k^+)-f(a_k)|$ and since $\|\sigma\cup\pi\|\leq \|\pi\|<\delta_0$, we get $|f(s_k^+)-f(a_k)|<\frac\epsilon{2m}$ (and similarly for $|f(s_{k+1})-f(t_{b_k})|$), thus $$V(\sigma\cup\pi,f)\leq V(\pi,f)+J(\sigma,f)+\epsilon$$


It is known that any real valued function $[0,1]\to\mathbb{R}$ can only have countably many jump discontinuities. Thus, define, for any real valued function $f:I\to\mathbb{R}$ defined on some interval $I$ $$ J(f)= \sum_{x} J(x,f) $$ where the sum is taken over the set of jump discontinuities of $f$. Furthermore, it is known that a function has both a left and a right limit at all points iff it is regulated. Since monotone functions are regulated, and any function of bounded variation can be expressed as a sum of monotone functions, all functions of bounded variation are regulated. Also, one verifies that for a function of bounded variation $f$, $$J(f)\leq TV(f)<+\infty$$ Hence all functions of bounded variation are regulated and have finite $J$.

Proposition. Let $f:[0,1]\to\mathbb{R}$ be a regulated function such that $J(f)<+\infty$. Suppose there exists a sequence of subdivisions $(\pi_n)_{n\geq 0}$ with $\|\pi_n\|\to 0$ and $C=\sup_n V(\pi_n,f)<+\infty$, then $f$ is of bounded variation.

Proof. Let $\sigma$ be a subdivision of $[0,1]$. Since $f$ is regulated, it is a good subdivision. By the lemma, there exists $\delta>0$ such that for any subdivision $\pi$ with $\|\pi\|<\delta$, $$ V(\sigma,f)\leq V(\pi,f)+J(\sigma,f)+1 $$ Since $\|\pi_n\|\to 0$, there exists $N$ with $\|\pi_N\|<\delta$. Thus $$ \begin{array}{rcl} V(\sigma,f) & \leq & V(\pi_N,f)+J(\sigma,f)+1 \\ & \leq & C+J(f)+1. \end{array} $$ and $f$ has finite variation.

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  • $\begingroup$ The lemma from my previous answer follows from the lemma in this answer. $\endgroup$ – Olivier Bégassat Jan 8 '18 at 21:53
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This is just me rewriting Daniel Fischer's answer and comment. The map $f$ is continuous.

Lemma. Let $\sigma$ be a subdivision of $[0,1]$ and let $\epsilon>0$ be given. There exists $\delta>0$ such that for any subdivision $\pi$ of $[0,1]$, $$\|\pi\|\leq\delta\implies V(\sigma,f)\leq V(\pi,f)+\epsilon$$

Proof. Let $\sigma=(0=s_0<s_1<\cdots<s_{m-1}<s_m=1)$ be a subdivision of $[0,1]$. If $m=1$, there is nothing to prove, so assume $m\geq 2$. By uniform continuity of $f$, there exists $\delta>0$ such that $$\forall~x,y\in[0,1],\Big(|x-y|\leq\delta\implies|f(x)-f(y)|\leq\frac{\epsilon}{2(m-1)}\Big)$$ Let $\pi=(0=t_0<t_1<\cdots<t_{n-1}<t_n=1)$ be a subdivision such that $\|\pi\|\leq\delta$. Consider the subdivision $\sigma\cup\pi$. Since it refines $\sigma$, one has $$ V(\sigma,f)\leq V(\sigma\cup\pi,f) $$ Let's set $\sigma\cup\pi=(0=u_0<u_1<\cdots<u_{N-1}<u_N=1)$, and for all $0\leq k\leq n$, $t_k=u_{N_k}$. We will sum $V(\sigma\cup\pi,f)$ in stages : $$ V(\sigma\cup\pi,f) = \sum_{k=0}^{n-1} \sum_{l=N_k}^{N_{k+1}-1} |f(u_{l+1})-f(u_l)| $$ and analyse the packets $\sum_{l=N_k}^{N_{k+1}-1} |f(u_{l+1})-f(u_l)|$ separately :

  • if $t_k$ and $t_{k+1}$ are consecutive in $\sigma\cup\pi$, then $$ \sum_{l=N_k}^{N_{k+1}-1} |f(u_{l+1})-f(u_l)| = |f(t_{k+1})-f(t_k)|\,, $$
  • otherwise there are $0<a_k\leq b_k<m$ such that the nodes of $\sigma\cup\pi$ between $t_k$ and $t_{k+1}$ are precisely $t_k<s_{a_k}<\cdots<s_{b_k}<t_{k+1}$. Since $|t_{k+1}-t_k|\leq\|\pi\|\leq\delta$, we have $$ \begin{array}{rcl} \displaystyle \sum_{l=N_k}^{N_{k+1}-1} |f(u_{l+1})-f(u_l)| & \leq & \displaystyle \frac{b_k-a_k+1+1}{2(m-1)}\epsilon \\ & = & \displaystyle \frac{\#\{a_k,\dots,b_k\}+1}{2(m-1)}\epsilon \end{array} $$

Since the $(\{a_k,\dots,b_k\})_k$ are pairwise disjoint subsets of $\{1,\dots,m-1\}$, there are at most $m-1$ non empty ones, and the sum of their cardinalities is at most $m-1$. Hence $$V(\sigma\cup\pi,f)\leq V(\pi,f)+\epsilon$$

This is an illustration of the idea : Screen Shot


Proposition Let $f$ be continuous on $[0,1]$. Then $f$ is of bounded variation iff there exists a sequence $(\pi_n)$ of subdivisions with $\|\pi_n\|\to0$ and $C=\sup_n V(\pi_n,f)<+\infty$. Furthermore, given such a sequence, $$ TV(f) = \lim_n V(\pi_n,f). $$

Proof. Let $\sigma$ be a subdivision of $[0,1]$. By the preceding lemma, and since $\|\pi_n\|\to0$, there exists $n_0\geq 0$ such that for all $n\geq n_0$, $$ \begin{array}{rcl} V(\sigma,f) & \leq & V(\pi_n,f)+1 \\ & \leq & C+1 \end{array} $$ so that $f$ has bounded variation.

Let $\epsilon>0$. Again by the lemma, there exists $N$ such that for all $n\geq N$, $ V(\sigma,f) \leq V(\pi_n,f)+\epsilon $ and thus $$ \begin{array}{rcl} V(\sigma,f) & \leq & \displaystyle \inf_{n\geq N}V(\pi_n,f)+\epsilon \\ & \leq & \displaystyle \liminf_{n}V(\pi_n,f)+\epsilon \end{array} $$ Since $\sigma$ and $\epsilon >0$ are arbitrary, we get $$TV(f) \leq \liminf_{n}V(\pi_n,f)$$ On the other hand, $$\limsup_{n}V(\pi_n,f)\leq TV(f)$$ so that $$ TV(f) = \liminf_{n}V(\pi_n,f) = \limsup_{n}V(\pi_n,f) = TV(f) $$ and $V(\pi_n,f)$ converges to $TV(f)$.


To my surprise, the analoguous property for quadratic variation fails miserably. This is discussed after Corollary 2.5 in On a class of generalized Takagi functions with linear pathwise quadratic variation. The situation is much more complicated, and I might ask a question about it later.

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